Class 12

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

Direction (1 – 4): Prove the following.

Question 1.
3 sin^-1 x = sin^-1 (3x – 4x³), x ∈ [- \(\frac{1}{2}\), \(\frac{1}{2}\)]
Solution.
Let x = sin θ. Then, sin^-1 x = θ.
We have,
R.H.S. = sin^-1 (3x – 4x³) = sin^-1(3 sin θ – 4 sin³ θ)
= sin^-1 (sin 3θ) = 3θ = 3 sin^-1 x
= L.H.S.
Hence proved.

Question 2.
3 cos^-1 x = cos^-1 (4x³ – 3x), x ∈ [\(\frac{1}{2}\), 1]
Solution.
Let x = cos θ. Then, cos^-1 x = θ.
We have, R.H.S. = cos^-1 (4x³ – 3x)
= cos^-1 (4cos 3θ – 3 cos θ)
= cos^-1 (cos 3θ) = 3θ = 3 cos^-1 x
= L.H.S.
Hence proved.

Question 3.
tan^-1 \(\frac{2}{11}\) + tan^-1 \(\frac{7}{24}\) = tan^-1 \(\frac{1}{2}\).
Solution.
Given, tan^-1 \(\frac{2}{11}\) + tan^-1 \(\frac{7}{24}\) = tan^-1 \(\frac{1}{2}\)

Question 4.
2 tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{7}\) = tan^-1 \(\frac{31}{17}\)
Solution.
Given, 2 tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{7}\) = tan^-1 \(\frac{31}{17}\)
L.H.S. = 2 tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{7}\)
= \(\tan ^{-1}\left[\frac{2 \cdot \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right]+\tan ^{-1}\left(\frac{1}{7}\right)\) [∵ 2 tan^-1 x = tan^-1 (\(\frac{2 x}{1-x^{2}}\))]

Direction (5 – 10):- Write the following functions in the simplest form:

Question 5.
tan^-1 \(\frac{\sqrt{1+x^{2}}-1}{x}\), x ≠ 0.
Solution.
We have, tan^-1 \(\frac{\sqrt{1+x^{2}}-1}{x}\)
put x = tan θ
⇒ θ = tan^-1 x

Question 6.
tan^-1 \(\frac{1}{\sqrt{x^{2}-1}}\), |x| > 1
Solution.
Let x = sec θ, then θ = sec^-1 x
∴ tan^-1 \(\frac{1}{\sqrt{x^{2}-1}}\) = tan^-1 \(\left(\frac{1}{\sqrt{\sec ^{2} \theta-1}}\right)\)
= tan^-1 \(\left(\frac{1}{\sqrt{\tan ^{2} \theta}}\right)\) [∵ sec² θ – 1 = tan² θ]
= tan^-1 \(\left(\frac{1}{\tan \theta}\right)\)
= tan^-1 (cot θ)
= tan^-1 [tan (\(\frac{\pi}{2}\) – θ)] [∵ tan (\(\frac{\pi}{2}\) – θ) = cot θ]
= \(\frac{\pi}{2}\) – θ
= \(\frac{\pi}{2}\) – sec^-1 x.

Question 7.
tan^-1 \(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\), x < π.
Solution.
We have, tan^-1 \(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\)
= tan^-1 \(\left(\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}\right)\)
= tan^-1 (tan \(\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)\))
= tan^-1 (tan \(\frac{x}{2}\))
= \(\frac{x}{2}\)

Question 8.
tan^-1 (\(\frac{\cos x-\sin x}{\cos x+\sin x}\)), 0 < x < π.
Solution.

Question 9.
tan^-1 \(\frac{x}{\sqrt{a^{2}-x^{2}}}\), |x| < a.
Solution.
We have, tan^-1 \(\frac{x}{\sqrt{a^{2}-x^{2}}}\)
Let x = a sin θ
⇒ \(\frac{x}{a}\) = sin θ
⇒ θ = sin^-1 (\(\frac{x}{a}\))
∴ tan^-1 \(\frac{x}{\sqrt{a^{2}-x^{2}}}\) = tan^-1 \(\left(\frac{a \sin \theta}{\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right)\)
= tan^-1 \(\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^{2} \theta}}\right)\)
= tan^-1 \(\left(\frac{a \sin \theta}{a \cos \theta}\right)\)
= tan^-1 (tan θ)
= θ = sin^-1 \(\frac{x}{a}\).

Question 10.
tan^-1 \(\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)\), a > 0; \(\frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}\).
Solution.
We have, tan^-1 \(\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)\), a > 0; \(\frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}\)
Let x = a tan θ
⇒ \(\frac{x}{a}\) = tan θ
⇒ θ = tan^-1 \(\frac{x}{a}\)

∴ tan^-1 \(\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)\), a > 0; \(\frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}\) = tan^-1 \(\left(\frac{3 a^{2} \cdot(a \tan \theta)-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot\left(a^{2} \tan ^{2} \theta\right)}\right)\)

= tan ^-1 \(\left(\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}\right)\)

= tan^-1 \(\left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)\)

= tan^-1 (tan 3θ) [∵ tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)]

= 3θ = 3 tan^-1 \(\frac{x}{a}\).

Direction (11 – 15) : Find the value of each of the following.

Question 11.
tan^-1 [2 cos(2 sin^-1 \(\frac{1}{2}\))].
Solution.
Let sin^-1 \(\frac{1}{2}\) = x
Then, sin x = \(\frac{1}{2}\) = sin (\(\frac{\pi}{6}\)))
Now, tan^-1 [2 cos(2 sin^-1 \(\frac{1}{2}\))] = tan^-1 [2 cos(2 × \(\frac{\pi}{6}\))]
= tan^-1 [2 cos \(\frac{\pi}{3}\)]
= tan^-1 [2 × \(\frac{1}{2}\)] [∵ cos (\(\frac{\pi}{3}\)) = \(\frac{1}{2}\))
= tan^-1 1 = \(\frac{\pi}{4}\)

Question 12.
cot(tan^-1 a + cot^-1 a).
Solution.
We have, cot(tan^-1 a + cot^-1 a)
= cot (\(\frac{\pi}{2}\))
= 0 [∵ tan^-1 x + cot^-1 x = \(\frac{\pi}{2}\)].

Question 13.
tan \(\frac{1}{2}\) [sin^-1 \(\frac{2 x}{1+x^{2}}\) + cos^-1 \(\frac{1-y^{2}}{1+y^{2}}\)], |x| < 1, y > 0 and xy < 1.
Solution.
Let x = tan θ.
Then, θ = tan^-1 x.
∴ sin^-1 \(\frac{2 x}{1+x^{2}}\) = sin^-1 \(\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)\)
= sin^-1 (sin 2θ) = 2θ = 2 tan^-1 x
Again, let y = tan φ.
Then, φ = tan^-1 y
∴ cos^-1 \(\frac{1-y^{2}}{1+y^{2}}\) = cos^-1 \(\left(\frac{1-\tan ^{2} \varphi}{1+\tan ^{2} \varphi}\right)\)
= cos^-1 (cos 2φ) = 2φ = 2 tan^-1 y
Now, tan \(\frac{1}{2}\) [sin^-1 \(\frac{2 x}{1+x^{2}}\) + cos^-1 \(\frac{1-y^{2}}{1+y^{2}}\)]
= tan \(\frac{1}{2}\) [2 tan^-1 x + tan^-1 y]
= tan [tan^-1 x + tan^-1 y]
= tan[tan^-1 \(\left(\frac{x+y}{1-x y}\right)\)]
[∵ tan^-1 x + tan^-1 y = tan^-1 \(\left(\frac{x+y}{1-x y}\right)\)]
= \(\frac{x+y}{1-x y}\)

Question 14.
If sin(sin^-1 \(\frac{1}{5}\) + cos^-1 x) = 1, then find the value of x.
Solution.
Given, sin(sin^-1 \(\frac{1}{5}\) + cos^-1 x) = 1
⇒ sin^-1 \(\frac{1}{5}\) + cos^-1 x = sin^-1 (1)
[∵ sin θ = x ⇒ θ = sin^-1 x]
⇒ sin^-1 \(\frac{1}{5}\) + cos^-1 x = sin^-1 (sin \(\frac{\pi}{2}\))
[∵ sin (\(\frac{\pi}{2}\)) = 1]
⇒ sin^-1 \(\frac{1}{5}\) + cos^-1 x = \(\frac{\pi}{2}\)
⇒ sin^-1 \(\frac{1}{5}\) = \(\frac{\pi}{2}\) – cos^-1 x
sin^-1 \(\frac{1}{5}\) = sin^-1 x
[∵ sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\)]
⇒ \(\frac{1}{5}\) = x
Hence, the value of x is \(\frac{1}{5}\).

Question 15.
If tan^-1 \(\frac{x-1}{x-2}\) + tan^-1 \(\frac{x+1}{x+2}=\frac{\pi}{4}\), then find the value of x.
Solution.
We have, tan^-1 \(\frac{x-1}{x-2}\) + tan^-1 \(\frac{x+1}{x+2}=\frac{\pi}{4}\)

⇒

Direction (16 – 18): Find the value of each expression.

Question 16.
sin^-1 (sin \(\frac{2 \pi}{3}\))
Solution.
We have, sin^-1 (sin \(\frac{2 \pi}{3}\))
We know that sin^-1(sin x) = x , if x ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)) which is the principal value branch of sin^-1 x.
Here, \(\frac{2 \pi}{3}\) ∉ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)
Now, sin^-1 (sin \(\frac{2 \pi}{3}\)) can be written as
sin^-1 (sin \(\frac{2 \pi}{3}\)) = \(\sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{3}\right)\right]=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)\) where \(\frac{\pi}{3}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
∴ sin^-1 (sin \(\frac{2 \pi}{3}\)) = sin^-1 (sin \(\frac{\pi}{3}\)) = \(\frac{\pi}{3}\).

Question 17.
tan^-1 (tan \(\frac{3 \pi}{4}\))
Solution.
We have, tan^-1 (tan \(\frac{3 \pi}{4}\))
We know that tan^-1 (tan x) = x, if x ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)), which is the principal value branch of tan^-1 x.
Here, \(\frac{3 \pi}{4}\) ∉ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
Now, tan^-1 (tan \(\frac{3 \pi}{4}\)) can be written as
tan^-1 (tan \(\frac{3 \pi}{4}\)) = tan^-1 [tan(π – \(\frac{\pi}{4}\))]
= tan^-1 [- tan \(\frac{\pi}{4}\)]
= tan^-1 [tan (- \(\frac{\pi}{4}\))]
where – \(\frac{\pi}{4}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
[∵ – tan θ = tan(- θ)]
∴ tan^-1 (tan \(\frac{3 \pi}{4}\)) = tan^-1 [tan (- \(\frac{\pi}{4}\))]
= – \(\frac{\pi}{4}\)

Question 18.
tan (sin^-1 \(\frac{3}{5}\) + cot^-1 \(\frac{3}{2}\))
Solution.
Let sin^-1 \(\frac{3}{5}\) = x.
Then, sin x = \(\frac{3}{5}\)
⇒ cos x = \(\sqrt{1-\sin ^{2} x}\) = \(\frac{4}{5}\)
⇒ sec x = \(\frac{5}{4}\)
∴ tan x = \(\sqrt{\sec ^{2} x-1}=\sqrt{\frac{25}{16}-1}=\frac{3}{4}\)
∴ x = tan^-1 \(\frac{3}{4}\)
∴ sin^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{3}{4}\) ………..(i)
Now, cot^-1 \(\frac{3}{2}\) = tan^-1 \(\frac{2}{3}\)
[∵ tan^-1 \(\frac{1}{x}\) = cot^-1 x] ……………(ii)
Hence, tan (sin^-1 \(\frac{3}{5}\) + cot^-1 \(\frac{3}{2}\))
= tan (tan^-1 \(\frac{3}{4}\) + tan^-1 \(\frac{2}{3}\))
= \(\tan \left(\tan ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\)
[∵ tan^-1 x + tan^-1 y = tan^-1 \(\left(\frac{x+y}{1-x y}\right)\)]
= tan (tan^-1 \(\frac{9+8}{12-6}\))
= tan (tan^-1 \(\frac{17}{6}\))
= \(\frac{17}{6}\).

Question 19.
cos^-1 (cos \(\frac{7 \pi}{6}\)) is equal to
(A) \(\frac{7 \pi}{6}\)

(B) \(\frac{5 \pi}{6}\)

(C) \(\frac{\pi}{3}\)

(D) \(\frac{\pi}{6}\)

Solution.
We know that cos^-1 (cos x) = x if x ∈ [0, x], which is the principal value branch of cos^-1 x.
Here, \(\frac{7 \pi}{6}\) ∉ x ∈ [0, π]
Now, cos^-1 (cos \(\frac{7 \pi}{6}\)) can be written as
cos^-1 (cos \(\frac{7 \pi}{6}\)) = cos^-1 [cos(2π – \(\frac{5 \pi}{6}\))]

= cos^-1 [cos \(\frac{5 \pi}{6}\)], where \(\frac{5 \pi}{6}\) ∈ [0, π]
[∵ cos(2π – x) = cos x]
∴ cos^-1 (cos \(\frac{7 \pi}{6}\)) = cos^-1 (cos \(\frac{5 \pi}{6}\))
= \(\frac{5 \pi}{6}\)
The correct option is (B).

Question 20.
sin[\(\frac{\pi}{3}\) – sin^-1 (- \(\frac{1}{2}\))] is equal to
(A) \(\frac{1}{2}\)

(B) \(\frac{1}{3}\)

(C) \(\frac{1}{4}\)

(D) 1
Solution.
Let sin^-1 (- \(\frac{1}{2}\)) = x.
Then, sin x = – \(\frac{1}{2}\) = – sin \(\frac{\pi}{6}\) = sin(-\(\frac{\pi}{6}\))
We know that the range of the principal value of sin^-1 x is (- \(\frac{\pi}{2}\), –\(\frac{\pi}{2}\))
∴ sin^-1 (- \(\frac{1}{2}\)) = – \(\frac{\pi}{6}\)
Now, sin[\(\frac{\pi}{6}\) – sin^-1 (- \(\frac{1}{2}\))] = sin \(\left[\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right]\)
= sin \(\left(\frac{\pi}{3}+\frac{\pi}{6}\right)\)
= sin (\(\frac{3 \pi}{6}\))
= sin (\(\frac{\pi}{2}\)) = 1
The correct option is (D).

Question 21.
tan^-1 (- √3) – cot^-1 (- √3) is equal to
(A) π
(B) – \(\frac{\pi}{2}\)
(C) 0
(D) 2√3
Solution.
Let tan^-1 √3 = x
⇒ tan x = √3 = tan \(\frac{\pi}{3}\)
∴ tan^-1 √3 = \(\frac{\pi}{3}\)
Again, let cos^-1(- √3) = x
⇒ cot x = – √3 = – cot \(\frac{\pi}{6}\)
= cot (π – \(\frac{\pi}{6}\))
= cot \(\frac{5 \pi}{6}\)
∴ cot^-1 (- √3) = \(\frac{5 \pi}{6}\)
Now, tan^-1 (- √3) – cot^-1 (- √3) = \(\frac{\pi}{3}\) – \(\frac{5 \pi}{6}\)
= \(\frac{2 \pi-5 \pi}{6}=\frac{-3 \pi}{6}=-\frac{\pi}{2}\)
Hence, correct option is (B).

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 6 General Principles and Processes of Isolation of Elements Textbook Exercise Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements

Very Short Answer Type Questions

Question 1.
Zinc acts as a reducing agent in the extraction of silver. Comment.
Answer:
Zinc acts as a reducing agent in the extraction of silver. It reduces Ag⁺ to Ag and itself get oxidised to Zn²⁺.
2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Ag↓

Question 2.
Winch reducing agent is employed to get copper from the leached low grade copper ore?
Answer:
Scrap iron, Cu²⁺(aq) + Fe(s) → Cu(s) + Fe²⁺(aq)
or H₂ gas, Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)

Question 3.
Name the method used for refining of zirconium.
Answer:
Van Arkel method

Question 4.
Name the method that is used for refining of nickel.
Answer:
Mond process (Vapour phase refining)

Question 5.
Name the method used for refining of copper metal.
Answer:
Electrolytic refining

Question 6.
Although carbon and hydrogen are better reducing agents but they are not used to reduce metallic oxides at high temperatures. Why?
Answer:
At high temperature carbon and hydrogen react with metals to form carbides and hydrides respectively.

Question 7.
What is the function of collectors in the froth floatation process for the concentration of ores?
Answer:
Collectors (e.g., pine oil, xanthates etc.) enhance non-wettability of the ore particles.

Question 8.
Why is it that only sulphide ores are concentrated by froth floatation process?
Answer:
This is because the sulphide ore particles are preferentially wetted by oil and gangue particles are preferentially wetted by water.

Question 9.
At temperatures above 1073 K, coke can be used to reduce FeO to Fe. How can you justify this reduction with Ellingham diagram?
Answer:
Using Ellingham diagram, we observe that at temperature greater than 1073 K; △G(C, CO) < △G (Fe, FeO).
Hence, coke can reduce FeO to Fe.

Question 10.
The mixture of compounds A and B is passed through a column of Al₂O₃ by using alcohol as eluant. Compound A is eluted in preference to compound B. Which of the compounds A or B, is more readily adsorbed on the column?
Answer:
Since, compound ‘A’ comes out before compound ‘B’ the compound ‘B’ is more readily adsorbed on the column.

Short Answer Type Questions

Question 1.
Write the role of:
(i) I₂ in the van Arkel method of refining.
(ii) Dilute NaCN in the extraction of silver.
Answer:
(i) Impure titanium is heated with iodine to form volatile TiI₄, which decomposes on tungsten filament at high temperature to give pure titanium.

(ii) Dilute NaCN forms a soluble complex with Ag or Ag₂S while the impurities remain unaffected which are filtered off.
4Ag + 8NaCN + O₂ + 2H₂O → 4Na[Ag(CN)₂] + 4NaOH
or

Question 2.
Describe the role of
(i) Iodine in the refining of zirconium.
(ii) NaCN in the extraction of gold from gold ore.
Write chemical equations for the involved reactions.
Answer:
(i) Impure zirconium is heated with iodine to form volatile compound ZrI₄ which on further heating over tungsten filament decomposes to give pure zirconium.

(ii) Gold ore is leached with dilute solution of NaCN in the presence of air from which the metal is obtained later by replacement.
4Au + 8NaCN + O₂ + 2H₂O → 4Na[Au(CN)₂] + 4NaOH

Question 3.
Explain the role of each of the following in the extraction of metals from their ores:
(i) CO in the extraction of nickel.
(ii) Zinc in the extraction of silver.
Answer:
(i) CO in the extraction of nickel: Impure nickel is heated in a stream of carbon monoxide when volatile nickel tetracarbonyl is formed and the impurities are left behind in the solid state. The vapour of nickel tetracarbonyl is taken to a decomposer chamber maintained at 450-470 K where it decomposes to give pure nickel metal and carbon monoxide.

(ii) Zinc in the extraction of silver : Silver present in the ore is leached with dilute solution of NaCN in the presence of air or oxygen to form a soluble complex.

Silver is then recovered from the complex by displacement method using more electropositive zinc metal.
2[Ag(CN)₂]^– (aq) + Zn(s) → 2Ag(s) + [Zn(CN)₂]^2- (aq)

Question 4.
Wrought iron is the purest form of iron. Write a reaction used for the preparation of wrought iron from cast iron. How can the impurities of sulphur, silicon and phosphorus be removed from cast iron?
Answer:

This reaction takes place in reverberatory furnace lined with haematite.

(b) Limestone is added as flux. Impurities of S, Si and P oxidise and pass into slag. The metal is removed and freed from slag by passing through rollers.

Question 5.
Write the chemical reactions involved in the extraction of gold by cyanide process. Also give the role of zinc in the reaction.
Answer:
(i) 4Au(s) + 8CN^– (aq) + 2H₂O(aq) + O₂(g) → 4[Au(CN)₂]^– (aq) + 4OH^–(aq)
(ii) 2[Au(CN)₂]^– (aq) + Zn(s) → 2Au(s) + [Zn(CN)₄]^2- (aq)
Zinc acts as a reducing agent in this reaction.

Question 6.
Describe the role of
(i) NaCN in the extraction of gold from its ore.
(ii) Cryolite in the extraction of aluminium from pure alumina.
(iii) CO in the purification of nickel.
Answer:
(i) Gold is leached with a dilute solution of NaCN in the presence of air.
(ii) Cryolite lowers the high melting point of alumina and makes it a good conductor of electricity.
(iii) CO forms a volatile complex with metal nickel which is further decomposed to give pure Ni metal.

Long Answer Type Questions

Question 1.
(a) Explain how an element can be extracted using an oxidation reaction?
(b) What do you mean by refining? Mention some of the methods used for refining of metals.
Answer:
(a) Some of the extractions, particularly of non-metals are based upon oxidation.
A very common example of extraction based on oxidation is the extraction of chlorine from brine (Chlorine is abundant in sea water as common salt).
2Cl^–(aq) + 2H₂O(l) → 2OH^–(aq) + H₂(g) + Cl₂(g)
The △G⁰ for this reaction is + 422 kJ. When it is converted to E⁰ (using △G⁰ = -nE⁰F), we get E⁰ = -2.2 V. Naturally, it will require an external e.m.f. that is greater than 2.2 V. But the electrolysis requires an excess potential to overcome some other hindering reactions. Thus, Cl₂ is obtained by electrolysis giving out H₂ and aqueous NaOH as by products. Electrolysis of molten NaCl is also carried out. But in that case, Na metal is produced and not NaOH.

The extraction of gold and silver involves leaching the metal with CN^–. This is also an oxidation reaction (Ag → Ag⁺ or Au → Au⁺). The metal is later recovered by displacement method.
4Au(s) + 8CN^–(aq) + 2H₂O(aq) + O₂(g) → 4[Au(CN₂)]^–(aq) + 4OH(aq)
2[Au(CN)₂]^–(aq) + Zn(s) → 2Au(s) + [Zn(CN)₄]^2- (aq)
In this reaction zinc acts as a reducing agent.

(b) A metal extracted by any method is usually contaiminated with some impurity. For obtaining metals of high purity, several techniques are used depending upon the difference in properties of the metal and the impurity. The process is called refining. Some of them are listed below :

1. Distillation,
2. Liquation,
3. Electrolysis,
4. Zone-refining,
5. Vapour phase refining,
6. Chromatographic methods.

Question 2.
How is the concept of coupling reactions useful in explaining the occurrence of non-spontaneous thermochemical reactions? Explain giving an example?
Answer:
Coupled reactions : Many reactions which are non-spontaneous (△G^⊖ is positive) can be made to occur spontaneously if these are coupled with reactions having larger negative free energy. By coupling means carrying out simultaneously both non- spontaneous and spontaneous reactions. For example, decomposition of Fe₂O₃into iron is a non-spontaneous reaction (△G^⊖ = +1487 kJ mol^-1). However, this decomposition can take place spontaneously if carbon monoxide is simultaneously burnt in oxygen (△G^⊖ = – 514.4 kJ mol^-1).
2Fe₂O₃(s) → 4Fe(s) + 3O₂(g); …(i);
△G^⊖ = + 1487.0 kJmol^-1
2CO(g) + O₂(g) → 2CO₂(g); … (ii);
△G^⊖ = -514.4 kJmol^-1
Multiplying equation (ii) by 3 and then adding to equation (i), we get
6CO(g) + 3O₂(g) → 6CO₂(g) △G^⊖ = -1543.2 kJ mol^-1
2Fe₂O₃ (s) → 4Fe(s) + 3O₂(s) △G^⊖ = +1487.0 kJ mol^-1
2Fe₂O₃(s) + 6CO(g) → 4Fe(s) + 6CO₂(g) △G^⊖ = – 56.2 kJ mol^-1
Since, △G^⊖ in the reduction of Fe₂O₃ with CO is negative, therefore, the reaction is feasible and spontaneous.

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 6 General Principles and Processes of Isolation of Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

PSEB 12th Class Chemistry Guide General Principles and Processes of Isolation of Elements InText Questions and Answers

Question 1.
Copper can be extracted by hydrometallurgy but not zinc% Explain.
Answer:
The E^⊖ value of zinc (Zn²⁺/Zn = – 0.76 V) is lower than that of copper (Cu2+/Cu = 0.34 V). This means that .zinc is a stronger reducing agent and can displace copper from solution of Cu²⁺ ions.
Zn(s) + Cu²⁺ (aq) → Zn²⁺(aq) + Cu(s)
In order to extract zinc by hydrometallurgy, we need stronger reducing agent like

etc. However, all these metals reduce water to hydrogen gas. Therefore, these metals cannot be used to displace Zn from solution of Zn²⁺ ions. Thus, copper can be extracted by hydrometallurgy but not zinc.

Question 2.
What is the role of depressant in froth floatation process?
Answer:
In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. For example, to separate two sulphide ores (ZnS and PbS), NaCN is used as a depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because NaCN reacts with ZnS to form Na₂[Zn(CN)₄].
4NaCN + ZnS → Na₂[Zn(CN)₄] + Na₂S

Question 3.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Answer:
The Gibbs free energy of formation (△_fG) of Cu₂S is less than that of H₂S and CS₂. Therefore, H₂ and C cannot reduce Cu₂S to Cu.

On the other hand, the Gibbs free energy of formation of Cu₂O is greater than that of CO. Hence, C can reduce Cu₂O to Cu.
C(s) + Cu₂O(s) → 2Cu(s) + CO(g)
Hence, the extraction of copper from its pyrite ore is difficult than from its oxide ore through reduction.

Question 4.
Explain:
(i) Zone refining
(ii) Column chromatography.
Answer:
(i) Zone refining : This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.

(ii) Column chromatography : Column chromatography is a technique used to separate different components of a mixture. It is a very useful technique used for the purification of elements available in minute quantities. It is also used to remove the impurities that are not very different in chemical properties from the element to be purified. Chromatography is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. In chromatography, there are two phases: mobile phase and stationary phase. The stationary phase is immobile and immiscible. Al₂O₃ column is usually used as the stationary phase in column chromatography. The mobile phase may be a gas, liquid, or supercritical fluid in which the sample extract is dissolved. Then, the mobile phase is forced to move through the stationary phase. The component that is more strongly adsorbed on the column takes a longer time to travel through it than the component that is weakly adsorbed. The adsorbed components are then removed (eluted) using a suitable solvent (eluant).

Question 5.
Out of C and CO, which is a better reducing agent at 673 K?
Answer:
At 673 K, the value of △G_(CO,CO2) is less than that of △G_(C,CO).
Therefore, CO can be oxidised more easily to CO₂ than C to CO. Hence, CO is a better reducing agent than C at 673 K.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present ?
Answer:
In electrolytic refining of copper, the common elements present in anode mud are selenium, tellurium, silver, gold, platinum, and antimony.

These elements are very less reactive and are not affected during the purification process. Hence, they settle down below the anode as anode mud.

Question 7.
Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Answer:
Reactions in blast furnace are as follows :
(a) Reactions at lower temperature range (500 to 800 K)
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂

(b) Reactions at higher temperature range (900-1500 K)
C + CO₂ → 2CO
FeO + CO → Fe + CO₂

Question 8.
Write chemical reactions taking place in the extraction of zinc from zinc blende. .
Answer:
The different steps involved in the extraction of zinc from zinc blende (ZnS) are given below:
(i) Concentration of ore : First, the gangue from zinc blende is removed by the froth floatation method.
(ii) Conversion to oxide (Roasting) : Sulphide ore is converted into oxide by the process of roasting. In this process, ZnS is heated in a regular supply of air in a furnace at a temperature, which is below the melting point of Zn.
2ZnS + 3O₂2 → 2ZnO + 2SO₂

(iii) Extraction of zinc from zinc oxide (Reduction) : Zinc is extracted from zinc oxide by the process of reduction. The reduction of zinc oxide is carried out by mixing it with powdered coke and then, heating it at 1673 K.

(iv) Electrolytic refining: Zinc can be refined by the process of electrolytic refining. In this process, impure zinc is made the anode while a pure copper strip is made the cathode. The electrolyte used is an acidified solution of zinc sulphate (ZnSO₄). Electrolysis results in the transfer of zinc in pure form from the anode to the cathode.
Anode : Zn → Zn²⁺ + 2e^–
Cathode : Zn²⁺ + 2e^– → Zn

Question 9.
State the role of silica in the metallurgy of copper.
Answer:
During the roasting of pyrite ore, a mixture of FeO and Cu₂O is obtained.

The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of rosting as ‘slag’. If the sulphide ore of copper contains iron, then silica (SiO₂) is added as flux before roasting. Then, FeO combines with silica to form iron silicate, FeSiO₃ (slag).

Question 10.
What is meant by the term “chromatography”?
Answer:
Chromatography is a collective term used for a family of laboratory techniques for the separation of mixtures. The term is derived from Greek words ‘chroma’ meaning ‘colour’ and ‘graphy5 meaning ‘writing’. Chromatographic techniques are based on the principle that different components are absorbed differently on an absorbent. There are several chromatographic techniques such as paper chromatography, column chromatography, gas chromatography, etc.

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Answer:
The stationary phase is selected in such a way that the components of the sample have different solubility’s in the phase. Hence, different components have different rates of movement through the stationary phase and as a result, can be separated from each other.

Question 12.
Describe a method for refining nickel.
Answer:
Nickel is refined by Mond’s process. In this process, nickel is heated in the presence of carbon monoxide to form nickel tetracarbonyl, which is a volatile complex.

Then, the obtained nickel tetracarbonyl is decomposed by subjecting it to a higher temperature (450-470 K) to obtain pure nickel metal.

Question 13.
How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Answer:
(i) To separate alumina from silica in a bauxite ore associated with silica, first the powdered ore is digested with a concentrated NaOH solution at 473-523 K and 35-36 bar pressure. This results in the leaching out of alumina (Al₂O₃) as sodium aluminate and silica (SiO₂) as sodium silicate leaving the impurities behind.

(ii) Then, CO₂gas is passed through the resulting solution to neutralise the aluminate in the solution, which results in the precipitation of hydrated alumina. To induce precipitation, the solution is seeded with freshly prepared samples of hydrated alumina.

(iii) During this process, sodium silicate remains in the solution. The obtained hydrated alumina is filtered, dried, and heated to get back pure alumina.

Question 14.
Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Answer:

Roasting	Calcination
1.  Sulphur dioxide is produced along with metal oxide.	Carbon dioxide is produced along with metal oxide.
2.  Ore is heated in the presence of excess of air or oxygen.	Ore is heated in the absence or limited supply of air or O2.
3. Volatile impurities are removed as oxides, such as SO2, As2O3, etc.	Water and organic impurities are removed.

Question 15.
How is ‘cast iron’ different from ‘pig iron”?
Answer:
The iron obtained from blast furnaces is known as pig iron. It contains around 4% carbon and many impurities such as S, P, Si, Mn in smaller amounts.

Cast iron is obtained by melting pig iron and coke using a hot air blast. It contains a lower amount of carbon (3%) than pig iron. Unlike pig iron, cast iron is extremely hard and brittle.

Question 16.
Differentiate between “minerals” and “ores”.
Answer:

Mineral	Ore
Naturally occurring substances of metals present in the earth’s crust are called minerals.	Minerals which can be used to obtain the metal profitably are cahed ores.
All minerals are not ores.	All ores are essentially minerals too.
e.g,, bauxite (Al2O3-xH2O) and clay (A12O3 -2SiO2 -2H2O)	e.g., bauxite (A12O3 ∙xH2O)

Question 17.
Why copper matte is put in silica lined converter?
Answer:
Copper matte contains Cu₂S and FeS. Copper matte is put in a silica-lined converter to remove the remaining FeO and FeS present in the matte as slag (FeSiO₃). Also, some silica is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the remaining FeS and FeO are converted to iron silicate (FeSiO₃) and Cu₂S is converted into metallic copper.
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → FeSiO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂O + Cu₂S → 6Cu + SO₂

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Answer:
Cryolite (Na₃AlF₆) has two roles in the metallurgy of aluminium :

1. To decrease the melting point of the mixture from 2323 K to 1140 K.
2. To increase the electrical conductivity of Al₂O₃.

Question 19.
How is leaching carried out in case of low grade copper ores?
Answer:
In case of low grade copper ores, leaching is carried out using acid or bacteria in the presence of air. In this process, copper goes into the solution as Cu²⁺ ions.
Cu(s) + 2H⁺(aq) + \(\frac{1}{2}\)O₂(g) → Cu²⁺(aq) + 2H₂O(l)
The resulting solution is treated with scrap iron or H₂ to get metallic copper.
Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Answer:
The standard free energy of formation (△_fG^⊖) of CO₂ from CO is
higher than that of the formation of ZnO from Zn. Therefore, CO cannot be used to reduce ZnO to Zn.

Question 21.
The value of △_fG^⊖ for formation of Cr₂O₃ is – 540 kJmol^-1 and that of A₂O₃ is – 827 kJ mol^-1. Is the reduction of Cr₂O₃ possible with Al?
Answer:
The two thermochemial equations may be written as
(i) 2Al + \(\frac{1}{2}\)O₂ → Al₂O₃ △_fG^⊖ = -827kJmol^-1
(ii) 2Cr + \(\frac{1}{2}\)O₂ → Cr₂O₃ △_fG^⊖ = -540kJmol^-1
Subtracting equation (ii) from (i), we have
2Al + Cr₂O₃ → Al₂O₃ + 2Cr
△_fG^⊖ = -827-(-540)
= -287kJmol^-1
As △_fG^⊖ for the reduction reaction of Cr₂O₃ by Al is negative, this reaction is possible.

Question 22.
Out of C and CO, which is a better reducing agent for ZnO?
Answer:
The free energy of formation (△_fG^⊖) of CO from C becomes lower at temperatures above 1120 K whereas that of CO₂ from C becomes lower above 1323 K than △_fG^⊖ of ZnO. However, △_fG^⊖ of CO₂ from CO is always higher than that of ZnO. Therefore, C and reduce ZnO to Zn but not CO. Therefore, out of C can CO, C is a better reducing agent than CO for ZnO.

Question 23.
The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Answer:
For any spontaneous reaction, the Gibbs free energy change (△G) must be negative. △G = △H – T△S where △H is the enthalpy change during the reaction, T is the absolute temperature and △S is the change in entropy.

Consider the Ellingham diagram (given below) for some metal oxides. From the diagram, it is evident that metals for which the free energy of formation of their oxides is more negative can reduce those metal oxides for which the free energy of formation of their respective oxides is less negative. In other words, any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram because the free energy will become more negative by an amount equal to the difference in the two graphs at that particular temperature. Thus, Al reduces FeO, Cr₂O₃ and NiO in Thermite reaction, but Al will not reduce MgO at a temperature below 1773 K.

It can be followed that:
(i) 2Al + Cr₂O₃→ Al₂O₃ + 2Cr
(Aluminothermic process)
(ii) 2Al + Fe₂O₃ → Al₂O₃ + 2Fe are spontaneous.
But Al can’t be used to reduce MgO below 1500°C. From the above it is clear that thermodynamic considerations help us in choosing a suitable reducing agent in metallurgy.

Question 24.
Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?
Answer:
(i) Down’s process for the manufacture of Na metal: When molten NaCl is subjected to electrolysis, chlorine is obtained as a by product at anode because in molten state only Na⁺ and Cl^– ions are present.
NaCl (melt) → Na⁺ (melt) + Cl^– (melt)
At cathode : Na⁺ (melt) + e^– → Na(s)
At anode : Cl^–(melt) → Cl(g) + e^–

(ii) Manufacture of NaOH : If an aqueous solution of NaCl is electrolysed, Cl₂ will be obtained at the anode but at the cathode, H₂ will be obtained instead of Na. This is because the standard reduction potential of Na (E^⊖ = – 2.71 V) is more negative than that of H₂O (E^⊖ = – 0.83 V). Hence, H₂O will get preference to get reduced at the cathode and as a result, H₂ is evolved.
NaCl(aq) → Na⁺(aq) + Cl^– (aq)
H₂O ⇌ H⁺(aq) + OH^–(aq)
At cathode : 2H₂O(l) + 2e^– → H₂(g) + 2OH^– (aq)
At anode : Cl^– (melt) → Cl(g) + e^–
2Cl (g) → Cl₂(g)
H₂ gas is obtained at cathode; chlorine gas at anode and NaOH is formed in the solution.
Na⁺(aq) + OH^–(aq) → NaOH (aq)

Question 25.
What is the role of graphite rod in the electrometallurgy of aluminium?
Answer:
In the electrometallurgy of aluminium, a fused mixture of purified alumina (Al₂O₃), cryolite (Na₃AlF₆) and fluorspar (CaF₂) is electrolysed. In this electrolysis, graphite is used as the anode and graphite-lined iron is used as the cathode. During the electrolysis, A1 is liberated at the cathode, while CO and CO₂ are liberated at the anode, according to the following equation :
At cathode: Al³⁺(melt) + 3e^– → Al(l)
At anode: C(s) + O^2- (melt) → CO(g) + 2e^–
C(s) + 2O^2- (melt) → CO₂(g) + 4e^–
If a metal is used instead of graphite as the anode, then 02will be liberated. This will not only oxidise the metal of the electrode, but also convert some of the A1 liberated at the cathode back into Al₂O₃. Hence, graphite is used for preventing the formation of O₂ at the anode.

Question 26.
Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
Answer:
(i) Zone refining : This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process.

(ii) Electrolytic refining : Electrolytic refining is the process of refining impure metals by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form of ions. The impurities present in the impure metal gets collected below the anode. This is known as anode mud.
At anode: M → Mⁿ⁺ + ne^–
At cathode: Mⁿ⁺ + ne^– → M

(iii) Vapour phase refining : Vapour phase refining is the process of refining metal by converting it into its volatile compound and then, decomposing it to obtain a pure metal. To carry out this process,

•  the metal should form a volatile compound with an available reagent, and
• the volatile compound should be easily decomposable so that the metal can be easily recovered.
  Nickel, zirconium, and titanium are refined using this method.

Question 27.
Predict conditions under which Al might be expected to reduce MgO. (Hint: See Intext Question 4)
Answer:
The equations for the formation of two oxides are :
\(\frac{4}{3}\)Al(s) + O₂(g) → \(\frac{2}{3}\)Al₂O₃(s)
2Mg(s) + O₂(g) → 2MgO(s)
If we observe the plots for the formation of the two oxides on the Ellingham diagram, we find the two curves intersect each other at a certain point. The corresponding value of △_fG^⊖ becomes zero for the reduction of MgO by aluminium metal.
2MgO(s) + \(\frac{4}{3}\)Al(s) ⇌ 2Mg(s) + \(\frac{2}{3}\)Al₂O₃(s)
This means that the reduction of MgO by A1 metal cannot occur below this temperature (1665 K). Instead, Mg can reduce Al₂O₃ to Al below 1665 K.
Aluminium metal (Al) can reduce MgO to Mg above 1665 K
because △_fG^⊖ for Al₂O₃ is less as compared to that of MgO.

Chemistry Guide for Class 12 PSEB General Principles and Processes of Isolation of Elements Textbook Questions and Answers

Question 1.
Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method?
Answer:
If the ore or the gangue can be attracted by the magnetic field, then the ore can be concentrated by the process of magnetic separation. The ores of iron such as haematite (Fe₂O₃), magnetite (Fe₃O₄), siderite (FeCO₃) and iron pyrites (FeS₂) can be separated by the process of magnetic separation.

Question 2.
What is the significance of leaching in the extraction of aluminium?
Answer:
In the extraction of aluminium, the significance of leaching is to concentrate pure alumina (Al₂O₃) from bauxite ore. Bauxite usually contains silica, iron oxide, and titanium oxide as impurities. In the process of leaching, alumina is concentrated by digesting the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar. Under these conditions, alumina (Al₂O₃) dissolves as sodium meta-aluminate and silica (SiO₂) dissolves as sodium silicate leaving the impurities behind.

The impurities are then filtered and the solution is neutralised by passing CO₂ gas. In this process, hydrated Al₂O₃ gets precipitated and sodium silicate remains in the solution. Precipitation is induced by seeding the solution with freshly prepared samples of hydrated Al₂O₃.
2Na[Al(OH)₄](aq) + CO₂(g) → Al₂O₃∙xH₂O(S) + 2NaHCO₃(aq)
Hydrated alumina
Hydrated alumina Al₂O₃∙xH₂O is filtered, dried, and heated to give back pure alumina (Al₂O₃).

Question 3.
The reaction,
Cr₂O₃ + 2Al > Al₂O₃ + 2Cr (△_fG^⊖ = -421kJ) is thermodynamically feasible as is apparent from the Gibbs energy value.
Why does it not take place at room temperature?
Answer:
The change in Gibbs energy is related to the equilibrium constant, K as
△G = – RT in K
At room temperature, all reactants and products of the given reaction are in the solid state. As a result, equilibrium does not exist between the reactants and th e prod ac ts lienee, the reaction does not take place at room temperature.
However, at a higher temperature, chromium melts and the reaction takes place.
We also know that according to the equation
△G = △H – T△S,
Increasing the temperature increases die value of T△S, making the value of △G more and more negative. Therefore, the reaction becomes more and more feasible as the temperature is increased.

Question 4.
Is it true that under certain conditions. Mg can reduce Al₂O₃ and Al can reduce MgO? What are those conditions?
Answer:
If we look at the Ellingbam diagram wo observe that the plots for Al and Mg cross each other at 1350°C (1623k) Below this temperature Mg can reduce Al₂O₃ and above this temperature., Al can reduce MgO.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1

Direction (1-10):
Find the principal values of the following.
Question 1.
sin^-1 (- \(\frac{1}{2}\))
Solution.
Let sin^-1 (- \(\frac{1}{2}\)) = y
Then, sin y = – \(\frac{1}{2}\)
= – sin (\(\left(\frac{\pi}{6}\right)\))
= sin (- \(\left(\frac{\pi}{6}\right)\))
We know that the range of the principal value of sin^-1 y is
[latex]-\frac{\pi}{2}, \frac{\pi}{2}[/latex] and sin[- \(\left(\frac{\pi}{6}\right)\)] = – \(\frac{1}{2}\)
Therefore, the principal value of sin^-1 (- \(\frac{1}{2}\)) is – \(\left(\frac{\pi}{6}\right)\).

Question 2.
cos^-1 (\(\frac{\sqrt{3}}{2}\))
Solution.
Let cos^-1 (\(\frac{\sqrt{3}}{2}\)) = y.
Then, cos y = \(\frac{\sqrt{3}}{2}\) = cos (\(\left(\frac{\pi}{6}\right)\))
We know that the range of the principal value of cos^-1 y is [0, π] and cos (\(\left(\frac{\pi}{6}\right)\)) = \(\frac{\sqrt{3}}{2}\).
Therefore, the principal value of cos^-1 (\(\frac{\sqrt{3}}{2}\)) is \(\left(\frac{\pi}{6}\right)\).

Question 3.
cosec^-1 (2)
Solution.
Let cosec^-1 (2) = y. Then, cosec y = 2 = cosec (\(\left(\frac{\pi}{6}\right)\))
We know that the range of the principal value of cosec^-1 y is (\(-\frac{\pi}{2}, \frac{\pi}{2}\)) – {0} and cosec (\(\left(\frac{\pi}{6}\right)\)) = 2.
Therefore, the principal value of cosec^-1 (2) is \(\left(\frac{\pi}{6}\right)\).

Question 4.
tan^-1 (- √3)
Solution.
Let tan^-1 (- √3) = y.
Then, tan y = – √3 = – tan \(\left(\frac{\pi}{3}\right)\) = tan (- \(\left(\frac{\pi}{3}\right)\))
We know that the range of the principal value of tan^-1 y is (\(-\frac{\pi}{2}, \frac{\pi}{2}\)) and tan(- \(\left(\frac{\pi}{2}\right)\)) is – √3
Therefore, the principal value of tan^-1 (- √3) is – \(\frac{\pi}{3}\).

Question 5.
cos^-1 (- \(\frac{1}{2}\))
Solution.
Let cos^-1 (- \(\frac{1}{2}\)) = y. Then,
cos y = – \(\frac{1}{2}\) = – cos (\(\left(\frac{\pi}{3}\right)\))
= cos (π – \(\frac{\pi}{3}\)) = cos \(\left(\frac{2 \pi}{3}\right)\)
We know that the range of the principal value of cos^-1 y is [0, π] and cos \(\left(\frac{2 \pi}{3}\right)\) = – \(\frac{1}{2}\).
Therefore, the principal value of cos^-1 (- \(\frac{1}{2}\)) is \(\left(\frac{2 \pi}{3}\right)\).

Question 6.
tan^-1 (- 1)
Solution.
Let tan^-1 (- 1) = y.
Then, tan y = – 1 = – tan (\(\frac{\pi}{4}\)) = tan (- \(\frac{\pi}{4}\))
We know that the range of the principal value of tan^-1 y is (\(-\frac{\pi}{2}, \frac{\pi}{2}\)) and tan (- \(\frac{\pi}{4}\)) = – 1.
Therefore, the principal value of tan^-1 (- 1) is (- \(\frac{\pi}{4}\))

Question 7.
sec^-1 (\(\frac{2}{\sqrt{3}}\))
Solution.
Let sec^-1 (\(\frac{2}{\sqrt{3}}\)) = y.
Then, sec y = \(\frac{2}{\sqrt{3}}\) = sec (\(\frac{\pi}{6}\))
We know that the range of the principal value of sec^-1 y is [0, π] – {\(\frac{\pi}{2}\)} and sec (\(\frac{\pi}{6}\)) = \(\frac{2}{\sqrt{3}}\).
Therefore, the principal value of sec^-1 (\(\frac{2}{\sqrt{3}}\)) is \(\frac{\pi}{6}\).

Question 8.
cot^-1 (√3)
Solution.
Let cot^-1 (√3) = y. Then, cot y = √3 = cot (\(\frac{\pi}{6}\))
We know that the range of the principal value of cot^-1 y is (0, π) and cot (\(\frac{\pi}{6}\)) = √3
Therefore, the principal value of cot^-1 (√3) is \(\frac{\pi}{6}\).

Question 9.
cos^-1 (- \(\frac{1}{\sqrt{2}}\))
Solution.
Let cos^-1 (- \(\frac{1}{\sqrt{2}}\)) = y. Then,
cos y = – \(\frac{1}{\sqrt{2}}\) = – cos (\(\frac{\pi}{4}\))
= cos(\(\pi-\frac{\pi}{4}\)) = cos(\(\frac{3 \pi}{4}\))
We know that the range of the principal value of cos^-1 y is [0, π] and cos (\(\frac{3 \pi}{4}\)) = – \(\frac{1}{\sqrt{2}}\)
Therefore, the principal value of cos^-1 (- \(\frac{1}{\sqrt{2}}\)) is \(\frac{3 \pi}{4}\).

Question 10.
cosec^-1 (√2)
Solution.
Let cosec^-1 (√2) = y.
Then, cosec y = – √2 = – cosec (\(\frac{\pi}{4}\)) = cosec (- \(\frac{\pi}{4}\))
We know that the range of the principal value of cosec^-1 y is
[\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0} and cosec (- \(\frac{\pi}{4}\)) = – √2.
Therefore, the principal value of cosec^-1 (- √2) is – \(\frac{\pi}{4}\).

DirectIon (11 – 14): Find the value of the following.

Question 11.
tan^-1 (1) + cos^-1 (- \(\frac{1}{2}\)) + sin^-1 (- \(\frac{1}{2}\))
Solution.
Let tan^-1 (1) = x. Then, tan x = 1 = tan \(\frac{\pi}{4}\)
∴ tan^-1 (1) = \(\frac{\pi}{4}\)
Let cos^-1 (- \(\frac{1}{2}\)) = y.
Then, cos y = – \(\frac{1}{2}\)
= – cos (\(\frac{\pi}{3}\))
= cos (π – \(\frac{\pi}{3}\))
= cos \(\frac{2 \pi}{3}\)
∴ cos^-1 (- \(\frac{1}{2}\)) = \(\frac{2 \pi}{3}\)
Let sin^-1 (- \(\frac{1}{2}\)) = z.
Then, sin z = – \(\frac{1}{2}\)
= – sin (\(\frac{\pi}{6}\))
= sin (- \(\frac{\pi}{6}\))
∴ sin^-1 (- \(\frac{1}{2}\)) = – \(\frac{\pi}{6}\)
∴ tan^-1 (1) + cos^-1 (- \(\frac{1}{2}\)) + sin^-1 (- \(\frac{1}{2}\)) = \(\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}\)
= \(\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}\).

Question 12.
cos^-1 (\(\frac{1}{2}\)) + 2 sin^-1 \(\frac{1}{2}\)
Solution.
Let cos^-1 (\(\frac{1}{2}\)) = x.
Then, cos x = \(\frac{1}{2}\) = cos (\(\frac{\pi}{3}\)).
∴ cos^-1 (\(\frac{1}{2}\)) = \(\frac{\pi}{3}\)
Let sin^-1 (\(\frac{1}{2}\)) = y.
Then, sin y = \(\frac{1}{2}\) = sin (\(\frac{\pi}{6}\))
∴ sin^-1 (\(\frac{1}{2}\)) = \(\frac{\pi}{6}\)
∴ cos^-1 (\(\frac{1}{2}\)) + 2 sin^-1 (\(\frac{1}{2}\)) = \(\frac{\pi}{3}+\frac{2 \pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}\)

Question 13.
If sin^-1 x = y, then
(A) 0 ≤ y ≤ K
(B) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
(C) 0 < y < π
(D) \(-\frac{\pi}{2}<y<\frac{\pi}{2}\)
Solution.
It is given that sin^-1 x = y.
We know that the range of the principal value branch of sin^-1 is [latex]-\frac{\pi}{2}, \frac{\pi}{2}[/latex]
Therefore, \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\).
Hence, the correct option is (B).

Question 14.
tan^-1 √3 – sec^-1 (- 2) is equal to
(A) π
(B) – \(\frac{\pi}{3}\)
(C) \(\frac{\pi}{3}\)
(D) \(\frac{2 \pi}{6}\)
Solution.
Let tan^-1 √3 = x.
Then, tan x = √3 = tan \(\frac{\pi}{3}\)
We know that the range of the principal value of tan^-1 x is (\(-\frac{\pi}{2}, \frac{\pi}{2}\))
∴ tan^-1 √3 = \(\frac{\pi}{3}\)
Let sec^-1 (- 2) = y.
Then, sec y = – 2 = – sec (\(\frac{\pi}{3}\))
= sec (π – \(\frac{\pi}{3}\)) = sec(\(\frac{2 \pi}{3}\))
Now, tan^-1 (√3) – sec^-1 (- 2) = \(\frac{\pi}{3}\) – \(\frac{2 \pi}{3}\)
= – \(\frac{\pi}{3}\)
Hence, correct option is (B).

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 5 Surface Chemistry

Very Short Answer Type Questions

Question 1.
Define desorption.
Answer:
The process of removal of an adsorbed substance from a surface on which it is adsorbed is called desorption.

Question 2.
What is the effect of temperature on chemisorption?
Answer:
Chemisorption initially increases then decreases with rise in temperature. The initial increase is due to the fact that heat supplied acts as activation energy. The decrease afterwards is due to the exothermic nature of adsorption equilibrium.

Question 3.
What is the role of diffusion in heterogeneous catalysis?
Answer:
The gaseous molecules diffuses on to the surface of the solid catalyst and get adsorbed. After the required chemical changes, the products diffuse away from the surface of the catalyst leaving the surface free for more reactant molecules to get adsorbed and undergo reaction.

Question 4.
What is the type of charge on Agl colloidal sol formed when AgNO₃ solution is added to KI solution?
Answer:
Negatively charged sol, Agl/I^– is formed when AgNO₃ solution is added to KI solution.

Question 5.
What causes Brownian movement in a colloidal solution?
Answer:
Unbalanced bombardment of the particles of dispersed phase by molecules of dispersion medium causes Brownian motion. This stabilises the sol.

Question 6.
Based on the type of dispersed phase, what type of colloid is micelles?
Answer:
Associated colloids

Question 7.
Name the temperature above which the formation of micelles takes place.
Answer:
Kraft temperature.

Question 8.
How do emulsifying agents stabilise the emulsion?
Answer:
The emulsifying agent forms an interfacial layer between suspended particles and the dispersion medium thereby stabilising the emulsion.

Question 9.
Write the dispersed phase and dispersion medium of butter.
Answer:
Dispersed phase — Liquid
Dispersion medium — Solid.

Question 10.
Write the main reason for the stability of colloidal sols.
Answer:
All the particles of colloidal sol carry the same charge so they keep on repelling each other and do not aggregate together to form bigger particles.

Question 11.
How is Brownian movement responsible for the stability of sols?
Answer:
The Brownian movement has a stirring effect, which does not allow the particles to settle down.

Short Answer Type Questions

Question 1.
Differentiate among a homogeneous solution, a suspension and a colloidal solution, giving a suitable example of each.
Answer:

Property	Homogeneous solution	Colloidal solution	Suspension
(i) Particle size	Less than 1 nm	Between 1 nm to 1000 nm	More than 1000 nm
(ii) Separation by
ordinary filtration	Not possible	Not possible	Not possible
ultra filtration	Not possible	Possible	Possible
(iii) Settling of particles	Do not settle	Settle only on coagulation	Settle under gravity
(iv) Appearance	Transparent	Opaque	Translucent
(v) Example	Glucose dissolved in water	Smoke, milk, gold sol	Sand in water

Question 2.
Classify colloids where the dispersion medium is water. State their characteristics and write an example of each of these classes.
Answer:
These are of two types
(i) Hydrophilic
Stability: More stable as the stability is due to charge and water envelope surrounding the sol particles.
Nature: Reversible
Examples: Starch, gum etc.

(ii) Hydrophobic
Stability: Less stable as the stability is due to charge only.
Nature: Irreversible
Examples: Metal hydroxide like Fe(OH)₃ and metal sulphide like As₂S₃.

Question 3.
Explain the cleansing action of soap. Why do soaps not work in hard water?
Answer:
The cleansing action of soap such as sodium stearate is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles. Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus, soap helps in emulsification and washing away of oils and fats.

Hard water contains calcium and magnesium salts. In hard water, soap gets precipitated as calcium and magnesium soap which being insoluble stick to the clothes as gummy mass. Therefore, soaps do not work in hard water.

Question 4.
Adsorption of a gas on the surface of solid is generally accompanied by a decrease in entropy still it is a spontaneous process. Why?
Answer:
According to the equation
△G = △H – T△S
For a process to be spontaneous, △G should be negative. Even though △S is negative here, △G is negative because reaction is highly exothermic, i.e., △H is negative.

Question 5.
Define the following terms:
(i) Brownian movement,
(ii) Peptization.
Answer:
(i) Brownian movement : The motion of the colloidal particles in a zig zag path due to unbalanced bombardment by the particles of dispersion medium is called Brownian movement.

(ii) Peptization : The process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of a small amount of suitable electrolyte is called peptization. During peptization, the precipitate absorbs one of the ions of the electrolyte on its surface. This causes development of positive or negative charge on precipitates, which ultimately break up into particles of colloidal dimension.

Question 6.
(i) Write the expression for Freundlich’s equation to describe the behaviour of adsorption from solution.
(ii) What causes charge on sol particles?
(iii) Name the promoter used in the haber’s process for the manufacture of ammonia.
Answer:
(i) \(\frac{x}{m}\) = KC^{\(\frac{1}{n}\)}
(ii) The charge on the sol particles is due to :

• electron capture by sol particles during electro dispersion.
• preferential anolsorption of ions from solution.
•  formulation of electrical double layer.

(iii) Molybdenum acts in a promoter for iron.

Long Answer Type Questions

Question 1.
Consider the adsorption isotherms given alongside and interpret the variation in the extent of adsorption (xlm) when

(a) (i) temperature increases at constant pressure.
(ii) pressure increases at constant temperature.
(b) Name the catalyst and the promoter used in Haber’s process for manufacture of ammonia.
Answer:
(a) (i) At constant pressure, extent of adsorption \(\left(\frac{x}{m}\right)\) decreases with increase in temperature as adsorption is an exothermic process.

(ii) At constant temperature, first adsorption \(\left(\frac{x}{m}\right)\) increases with increase in pressure up to a particular pressure and then it
At low pressure, \(\frac{x}{m}\) = kp m
At intermediate range of pressure, \(\frac{x}{m}\) = kp^1/n (n > 1)
At high pressure, \(\frac{x}{m}\) = k (independent of pressure)

(b) Finely divided iron is used as a catalyst and molybdenum is used as promoter.

Question 2.
Explain the following observations:
(i) Sun looks red at the time of setting.
(ii) Cottrell’s smoke precipitator is fitted at the mouth of the chimney used in factories.
(iii) Physical adsorption is multilayered while chemical adsorption is monolayered.
Answer:
(i) At the time of setting, the sun is at horizon. The light emitted by the sun has to travel a relatively longer distance through the atmosphere. As a result, blue part of light is scattered away by the particulate in the atmosphere causing red part to be visible.

(ii) Cottrell’s smoke precipitator, neutralises the charge on unburnt carbon particles, coming out of chimney and they get precipitated and settle down at the floor of the chamber.

(iii) Physical adsorption involves van der Waals’ forces, so any number of layers may be formed one over the other on the surface of the adsorbent. Chemical adsorption takes place as a result of the reaction between adsorbent and adsorbate. When the surface of adsorbent is covered with one layer, no further reaction can take place.

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 5 Surface Chemistry Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry

PSEB 12th Class Chemistry Guide Surface Chemistry InText Questions and Answers

Question 1.
Distinguish between the meaning of the terms adsorption and absorption.
Give one example of each.
Answer:

Adsorption	Absorption
1. It is the surface phenomenon.	It is the bulk phenomenon.
2. It is the phenomenon as a result of which the species of one substance gets concentrated mainly on the surface of another substance.	It is the phenomenon as a result of which one substance gets distributed uniformly throughout the total volume of another substance.
3. Adsorption is fast in the beginning then slows down due to non­availability of the surface.	Absorption proceeds at uniform rate.
4. The concentration on the surface of the adsorbent is different from that in the bulk. e.g., Water vapours on silica gel.	The concentration is same throughout the material. e.g., Water vapours are absorbed by anhydrous CaCl2.

Question 2.
What is the difference between physisorption and chemisorption?
Answer:

Physisorption	Chemisorption
1. In this type of adsorption, the adsorbate is attached to the surface of the adsorbent with weak van der Waal’s forces of attraction.	In this type of adsorption, strong chemical bonds are formed between the adsorbate and the surface of the adsorbent.
2. No new compound is formed in the process.	New compounds are formed at the surface of the adsorbent.
3. It is generally found to be reversible in nature.	It is usually irreversible in nature.
4. Enthalpy of adsorption is low as weak van der Waal’s forces of attraction are involved. The values lie in the range of 20-40 kJ mol-1.	Enthalpy of adsorption is high as chemical bonds are formed. The values lie in the range of 40-400 kJ mol-1.
5. It is favoured by low temperature conditions.	It is favoured by high temperature conditions.
6. It is an example of multi-layer adsorption	It is an example of mono-layer adsorption.

Question 3.
Give reason why a finely divided substance is more effective as an adsorbent.
Answer:
Adsorption is a surface phenomenon. Therefore, adsorption is directly proportional to the surface area. A finely divided substance has a large surface area. Both physisorption and chemisorption increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent.

Question 4.
What are the factors which influence the adsorption of a gas on a solid?
Answer:
There are various factors that affect the rate of adsorption of a gas on a solid surface.

1. Nature of the gas : Easily liquefiable gases such as NH₃, HCl etc. are adsorbed to a great extent in comparison to gases such as H₂, O₂ etc. This is because van der Waal’s forces are stronger in easily liquefiable gases.
2. Surface area of the solid : The greater the surface area of the adsorbent, the greater is the adsorption of a gas on the solid surface.
3. Effect of pressure : Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.
4. Effect of temperature : Adsorption is an exothermic process. Thus, in accordance with Le-Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 5.
What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
Answer:

The plot between the extent of absorption \(\left(\frac{x}{m}\right)\) against the pressure of gas (P) at constant temperature (T) is called the adsorption isotherm.

Freundlich adsorption isotherm : Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by the unit mass of solid adsorbent and pressure at a specific temperature.
From the given plot it is clear that at pressure P_s, \(\frac{x}{m}\) reaches the maximum value. Ps is called the saturation pressure. Three cases arise from the graph now :
Case I-At low pressure
The plot is straight and sloping, indicating that the pressure is directly proportional to \(\frac{x}{m}\) i.e., \(\frac{x}{m}\) ∝ P.
\(\frac{x}{m}\) = kP (k is a constant)

Case II-At high pressure
When pressure exceeds, the saturated pressure, \(\frac{x}{m}\) becomes independent of P values.
\(\frac{x}{m}\) ∝ P^o
\(\frac{x}{m}\) = kP^o

Case III-At intermediate pressure
At intermediate pressure, \(\frac{x}{m}\) depends on P raised to the powers between 0 and 1. This relationship is known as the Freundlich adsorption isotherm.
\(\frac{x}{m}\) ∝ P^{\(\frac{1}{n}\)}
\(\frac{x}{m}\) = kP^1/n n > 1
Now, taking log
log\(\frac{x}{m}\) = log k + \(\frac{1}{n}\)logP
On plotting the graph between log \(\left(\frac{x}{m}\right)\) and log P, a straight line is obtained with the slope equal to \(\frac{1}{n}\) and intercept equal to log k.

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Answer:
By activating an adsorbent, we tend to increase the adsorbing power of the adsorbent. Some ways to activate an adsorbent are:

1. By increasing the surface area of the adsorbent. This can be done by breaking it into smaller pieces or powdering it.
2. Some specific treatments can also lead to the activation of the adsorbent. For example, wood charcoal is activated by heating it between 650 K and 1330 K in vacuum or air. It expels all the gases absorbed or adsorbed and thus, creates a space for adsorption of gases.

Question 7.
What role does adsorption play in heterogeneous catalysis?
Answer:
Heterogeneous catalysis : A catalytic process in which the catalyst and the reactants are present in different phases is known as a heterogeneous catalysis. This heterogeneous catalytic action can be explained in terms of the adsorption theory. The mechanism of catalysis involves the following steps:

1. Adsorption of reactant molecules on the catalyst surface.
2. Occurrence of a chemical reaction through the formation of an intermediate.
3. Desorption of products from the catalyst surface.
4. Diffusion of products away from the catalyst surface.

In this process, the reactants are usually present in the gaseous state and the catalyst is present in the solid state. Gaseous molecules are then adsorbed on the surface of the catalyst. As the concentration of reactants on the surface of the catalyst increases, the rate of reaction also increases. In such reactions, the products have very less affinity for the catalyst and are quickly desorbed, thereby making the surface free for other reactants.

Question 8.
Why is adsorption always exothermic?
Answer:
Adsorption is always exothermic. This statement can be explained in two ways:
(i) Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.

(ii) AH of adsorption is always negative. When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., ∆S is negative. Now for a process to be spontaneous, ∆G should be negative.
∴ ∆G – ∆H – T∆S
Since, ∆S is negative, ∆H has to be negative to make ∆G negative. Hence, adsorption is always exothermic.

Question 9.
How are the colloidal solutions classified on the basis of physical stjates of the dispersed phase and dispersion medium?
Answer:
One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

Dispersed phase	Dispersion medium	Type of colloid	Example
Solid	Solid	Solid Sol	Gemstones, glasses
Solid	Liquid	Sol	Paints, cell fluids
Solid	Gas	Aerosol	Smoke, dust
Liquid	Solid	Gel	Cheese, butter
Liquid	Liquid	Emulsion	Milk, hair cream
Liquid	Gas	Aerosol	Fog, mist, cloud
Gas	Solid	Solid Sol	Pumice stone, foam rubber
Gas	Liquid	Foam	Froth, soap lather

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Answer:
Effect of pressure on adsorption : At constant temperature, the extent of adsorption of a gas (x / m) on a solid increases with pressure. A graph between x / m and the pressure p of a gas at constant temperature is called adsorption isotherm.

(i) At lower range of pressure, x / m is directly proportional tothe applied pressure.
\(\frac{x}{m}\) ∝ p¹ or \(\frac{x}{m}\) = kp

(ii) At high pressure range, the extent of adsorption of a gas (x / m) is independent of the applied pressure, i.e.,
\(\frac{x}{m}\) ∝ p^o or \(\frac{x}{m}\) = k

(iii) At intermediate pressure range, the value of x / m is proportional to a fractional power of pressure, i. e.,
\(\frac{x}{m}\) ∝ p^1/n or \(\frac{x}{m}\) = kp^1/n
where 1 / n is a fraction. Its value may be between 0 and 1.
log\(\left(\frac{x}{m}\right)\) = log k + \(\frac{1}{n}\) log p

Effect of temperature on adsorption : Adsorption is generally temperature dependent. Mostly adsorption processes are exothermic and hence adsorption decreases with increasing temperature. However, for an endothermic adsorption process, adsorption increases with increase in temperature.

Question 11.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Answer:
(i) Lyophilic sols : Colloidal sols directly formed by mixing substances in a suitable dispersion medium are called lyophilic sols. These sols are reversible in nature i.e., if two constituents of the sol are separated by any means (such as evaporation), then the sol can be prepared again by simply mixing the dispersion medium with the dispersion phase and shaking the mixture, e.g., gum, gelatin, starch, rubber etc.

(ii) Lyophobic sols : When substances such as metals and their sulphides etc. are mixed with the dispersion medium, they do not form colloidal sols. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are irreversible in nature, e.g., gold sol, AS₂O₃ etc.

Now, the stability of hydrophilic sols depends on two things—the presence of a charge and the salvation of colloidal particles. On the other hand, the stability of hydrophobic sols is only because of the presence of a charge. Therefore, the latter are much less stable than the former. If the charge of hydrophobic sols is removed (by addition of electrolytes), then the particles present in them come closer and form aggregates, leading to precipitation.

Question 12.
What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Answer:
(i) In multi-molecular colloids, the colloidal particles are an aggregate of atoms or small molecules with a diameter of less than 1 nm. The molecules in the aggregate are held together by van der Waal’s forces of attraction. Examples of such colloids include gold sol and sulphur sol.

(ii) In macro-molecular colloids, the colloidal particles are large molecules . having colloidal dimensions. These particles have a high molecular mass. When these particles are dissolved in a liquid, sol is obtained. For example: starch, nylon, cellulose, etc.

(iii) Certain substances tend to behave like normal electrolytes at lower concentrations. However, at higher concentrations, these substances behave as colloidal solutions due to the formation of aggregated particles. Such colloids are called associated colloids.

Question 13.
What are enzymes? Write in brief the mechanism of enzyme catalysis.
Answer:
Enzymes are basically protein molecules of high molecular masses. These form colloidal solutions when dissolved in water. These are complex, nitrogenous organic compounds produced by living plants and animals. Enzymes are also called “biochemical catalysts’.

On the surface of the enzymes, various cavities are present with characteristic shapes. These cavities possess active groups such as—NH₂, —COOH, etc. The reactant molecules having a complementary shape fit into the cavities just like a key fits into a lock. This leads to the formation of an activated complex. This complex then decomposes to give the product.

1. Binding of enzyme to substrate (reactant) to form activated complex.
   E + S → ES*
2. Decomposition of the activated complex to form product.
   ES* → E + P

Question 14.
How are colloids classified on the basis of
(i) physical states of components
(ii) nature of dispersion medium and
(iii) interaction between dispersed phase and dispersion medium?
Answer:
(i) One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

Dispersed phase	Dispersion medium	Type of colloid	Example
Solid	Solid	Solid Sol	Gemstones, glasses
Solid	Liquid	Sol	Paints, cell fluids
Solid	Gas	Aerosol	Smoke, dust
Liquid	Solid	Gel	Cheese, butter
Liquid	Liquid	Emulsion	Milk, hair cream
Liquid	Gas	Aerosol	Fog, mist, cloud
Gas	Solid	Solid Sol	Pumice stone, foam rubber
Gas	Liquid	Foam	Froth, soap lather

(ii) On the basis of the nature of dispersion medium, colloids can be divided as:

Dispersion medium	Name of sol
Water	Aquasol or hydrosol
Alcohol	Alcosol
Benzene	Benzosol
Gases	Aerosol

(iii) On the basis of the nature of the interaction between the dispersed phase and dispersion medium, the colloids can be classified as lyophilic (solvent attracting) and lyophobic (solvent repelling).

Question 15.
Explain what is observed
(i) when a beam of light is passed through a colloidal sol.
(ii) an electrolyte, NaCl is added to hydrated ferric oxide sol.
(iii) electric current is passed through a colloidal sol?
Answer:
(i) When a beam of light is passed through a colloidal solution, then scattering of light is observed. This is known as the Tyndall effect. This scattering of light illuminates the path of the beam in the colloidal solution.

(ii) When NaCl is added to hydrated ferric oxide sol, it dissociates to give Na⁺ and Cl^– ions. Particles of ferric oxide sol are positively charged. Thus, they get coagulated in the presence of negatively charged Cl^– ions.

(iii) The colloidal particles are charged and carry either a positive or negative charge. The dispersion medium carries an equal and opposite charge. This makes the whole system neutral. Under the influence of an electric current, the colloidal particles move towards the oppositely charged electrode. When they come in contact with the electrode, they lose their charge and coagulate.

Question 16.
What are emulsions? What are their different types? Give example of each type.
Answer:
The colloidal solution in which both the dispersed phase and dispersion medium are liquids is called an emulsion. There are two types of emulsions:
(a) Oil in water type : Here, oil is the dispersed phase while water is the dispersion medium. For example: milk, vanishing cream, etc.

(b) Water in oil type : Here, water is the dispersed phase while oil is the dispersion medium. For example: cold cream, butter, etc.

Question 17.
What is demulsification? Name two demulsifiers.
Answer:
The process of decomposition of an emulsion into its constituent liquids is called demulsification. Examples of demulsifiers are surfactants, ethylene oxide, etc.

Question 18.
Action of soap is due to emulsification and micelle formation. Comment.
Answer:
The cleansing action of soap is due to emulsification and micelle formation. Soaps are basically sodium and potassium salts of long chain fatty acids, R-COO^– Na⁺. The end of the molecule to which the sodium is attached is polar in nature, while the alkyl-end is non-polar. Thus, a soap molecule contains a hydrophilic (polar) and a hydrophobic (non-polar) part.

When soap is added to water containing dirt, the soap molecules surround the dirt particles in such a manner that their hydrophobic parts get attached to the dirt molecule and the hydrophilic parts point away from the dirt molecule. This is known as micelle formation. Thus, we can say that the polar group dissolves in water while the non-polar group dissolves in the dirt particle. Now, as these micelles are negatively charged, they do not coalesce and a stable emulsion is formed.

Question 19.
Give four examples of heterogeneous catalysis.
Answer:
Examples of heterogeneous catalysis
(i) Oxidation of sulphur dioxide to form sulphur trioxide. In this reaction, Pt acts as a catalyst.

(ii) Formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron.

This process is called the Haber’s process.

(iii) Oswald’s process: Oxidation of ammonia to nitric oxide in the presence of platinum.

(iv) Hydrogenation of vegetable oils in the presence of Ni.

Question 20.
What do you mean by activity and selectivity of catalysts?
Answer:
(a) Activity of a catalyst : The activity of a catalyst is its ability to increase the rate of a particular reaction. Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants on the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

(b) Selectivity of the catalyst : The ability of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H₂ and CO.

Question 21.
Describe some features of catalysis by zeolites.
Answer:
1. Zeolites are widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerisation. They are also used for removing permanent hardness of water,
e.g., ZSM-5 is a catalyst used in petroleum industry

2. Zeolites are shape selective catalysts having honey comb like structure.
3. They are microporous aluminosilicates with Al—O—Si framework and general formula M x / n [(AlO₂)_x (SiO₂)_y] ∙ mH₂O
4. The reactions taking place in zeolites depend upon the size and shape of the reactant and product molecules as well as upon the pores and cavities of the zeolites.

Question 22.
What is shape selective catalysis?
Answer:
A catalytic reaction which depends upon the pore structure of the catalyst and on the size of the reactant and the product molecules is called shape-selective catalysis. For example, catalysis by zeolites is a shape-selective catalysis. The pore size present in the zeolites ranges from 260-740 pm. Thus, molecules having a pore size more than this cannot enter the zeolite and undergo the reaction.

Question 23.
Explain the following terms:
(i) Electrophoresis
(ii) Coagulation
(iii) Dialysis
(iv) Tyndall effect.
Answer:
(i) Electrophoresis : The movement of colloidal particles under the influence of an applied electric field is known as electrophoresis. Positively charged particles move to the cathode, while negatively charged particles move towards the anode. As the particles reach oppositely charged electrodes, they become neutral and get coagulated.

(ii) Coagulation : The process of settling down of colloidal particles i.e., conversion of a colloid into a precipitate is called coagulation.

(iii) Dialysis : The process of removing a dissolved substance from a colloidal solution by the means of diffusion through a membrane is known as dialysis. This process is based on the principle that ions and small molecules can pass through animal membranes unlike colloidal particles.

(iv) Tyndall effect : When a beam of light is allowed to pass through a colloidal solution, it becomes visible like a column of light. This is known as the Tyndall effect. This phenomenon takes place as particles of colloidal dimensions scatter light in all directions.

Question 24.
Give four uses of emulsions.
Answer:
Uses of emulsions

1. Cleansing action of soaps is based on the formation of emulsions.
2. Digestion of fats in intestines takes place by the process of emulsification.
3. Antiseptics and disinfectants when added to water form emulsions.
4. The process of emulsification is used to make medicines.

Question 25.
What are micelles? Give an example of a micellers system.
Answer:
The aggregate of colloidal particles which have both hydrophobic and hydrophilic parts are called micelles. These are formed above a particular temperature called Krafts temperature (T_k)and above certain concentrations, called Critical Miceller Concentration (CMC).

These molecules are arranged radially with the hydrocarbon or non-polar part towards the centre and the polar part towards the periphery, e.g., soap solution in water is an example of micelles system.

Question 26.
Explain the terms with suitable examples:
(i) Alcosol
(ii) Aerosol
(iii) Hydrosol
Answer:
(i) Alcosol : A colloidal solution having alcohol as the dispersion medium and a solid substance as the dispersed phase is called an alcosol.
For example: colloidal sol of cellulose nitrate in ethyl alcohol is an alcosol.

(ii) Aerosol : A colloidal solution having a gas as the dispersion medium and a solid as the dispersed phase is called an aerosol. For example: fog, mist, cloud, etc.

(iii) Hydrosol: A colloidal solution having water as the dispersion medium and a solid as the dispersed phase is called a hydrosol. For example: starch sol or gold sol etc.

Question 27.
Comment on the statement that “colloid is not a substance but a state of substance”.
Answer:
Common salt (a typical crystalloid in an aqueous medium) behaves as a colloid in a benzene medium. Hence, we can say that a colloidal substance does not represent a separate class of substances. When the size of the solute particle lies between 1 nm and 1000 nm, it behaves as a colloid.

Hence, we can say that colloid is not a substance but a state of the substance which is dependent on the size of the particle. A colloidal state is intermediate between a true solution and a suspension.

Chemistry Guide for Class 12 PSEB Surface Chemistry Textbook Questions and Answers

Question 1.
Write any two characteristics of chemisorption.
Answer:

1. Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate.
2. Like physisorption, chemisorption also increases with an increase in the surface area of the adsorbent.

Question 2.
Why does physisorption decrease with the increase of temperature?
Answer:
Physisorption is exothermic in nature. Therefore, in accordance with Le-Chateliere’s principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature.

Question 3.
Why are powdered substances more effective adsorbents than their crystalline forms?
Answer:
Powdered substances are more effective adsorbents than their crystalline forms because when a substance is powdered, its surface area increases and physisorption is directly proportional to the surface area of the adsorbent.

Question 4.
In Haber’s process, hydrogen is obtained by reacting methane with steam in presence of NiO as catalyst. The process is known as steam reforming.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
Answer:
Carbon monoxide acts as a poison for catalyst iron and promoter molybdenum in Haber’s process, i.e., the efficiency of catalyst and promoter is decreased. It also combines with Fe to form iron carbonyl, Fe(CO)₅ which interfere in the production of ammonia. Hence, CO must be removed from the reaction mixture.

Question 5.
Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
Answer:
The chemical equation for ester hydrolysis can be represented as:
Ester + Water → Acid + Alcohol
The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts.

Question 6.
What is the role of desorption in the process of catalysis?
Answer:
The role of desorption in the process of catalysis is to make the surface of the solid catalyst free for the fresh adsorption of the reactants on the surface.

Question 7.
What modification can you suggest in the Hardy-Schulze law?
Answer:
Hardy-Schulze law states that ‘the greater the valence of the flocculating ion added, the greater is its power to cause precipitation.’ This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy-Schulze law can be stated as ‘the greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.’

Question 8.
Why is it essential to wash the precipitate with water before estimating it quantitatively?
Answer:
When a substance gets precipitated, some ions that combine to form the precipitate get adsorbed on the surface of the precipitate. Therefore, it becomes important to wash the precipitate before estimating it quantitatively in order to remove these adsorbed ions or other such impurities.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Miscellaneous Exercise Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise

Question 1.
Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that gof = fog = I_R.
Solution.
It is given that f: R → R is defined as f(x) = 10x + 7.
One – one :
Let f(x) = f(y), where x, y ∈ R.
⇒ 10x + 7 = 10y + 7
⇒ x = y
∴ f is a one-one function.

Onto :
For y ∈ R, let y = 10x + 7.
⇒ x = \(\frac{y-7}{10}\) ∈ R
Therefore, for any y ∈ R, there exists x = \(\frac{y-7}{10}\) ∈ R such that
f(x) = f(\(\frac{y-7}{10}\))
= 10(\(\frac{y-7}{10}\)) + 7
= y – 7 + 7 = y
∴ f is onto.
Therefore f is one-one and onto.
Thus f is an invertible function.
Let us define g : R → R as g(y) = \(\frac{y-7}{10}\)
Now, we have
gof(x) = g(f(x)) = g(10x + 7)
= \(\frac{(10 x+7)-7}{10}=\frac{10 x}{10}\) = x
And, fog(y) = f(g(y))
= f(\(\frac{y-7}{10}\))
= 10(\(\frac{y-7}{10}\)) + 7
= y – 7 + 7 = y
∴ gof = I_R and fog = I_R
Hence, the required functiong:R → R is defined as g(y) = \(\frac{y-7}{10}\)

Question 2.
Let f: W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.
Solution.
It is given that
f: W → W is defined as f(n) =

One-one :
Let f(n) = f(m)
It can be observed that if n is odd and m is even, then we will have
n – 1 = m + 1
⇒ n – m = 2
However, this is impossible.
Similarly, the possibility of n being even and m being odd can also ignored under a similar argument.
∴ Both n and m must be either odd or even.
Now, if both n and m are odd, then we have
f(n) = f(m) ⇒ n – 1 = m – 1 ⇒ n = m
Again, if both n and m are even , then we have
f(n) = f(m) ⇒ n + 1 = m+1 ⇒ n = m
∴ f is one – one.

Onto :
It is clear that any odd number 2r + 1 in co-domain W is the image of 2r in domain W and any even number 2r in co-domain W is the image of 2r + 1 in domain W.
∴ f is onto.
Hence, f is an invertible function.
Let us define g : W → W as

g(m) =

Now, when n is odd
gof(n) = g(f(n)) = g(n – 1) = n – 1 + 1 = n
and, when n is even
gof(n) = g(f(n)) = g(n + 1) = n + 1 – 1 = n
Similarly, when m is odd
fog(m) = f(g(m)) = f(m – 1) = m – 1 + 1 = m
and when m is even
fog(m) = f(g(m)) = f(m + 1) = m + 1 – 1 = m
∴ gof = I_W and fog = I_W
Thus, f is invertible and the inverse of f is given by f^-1 = g, which is the same as f.
Hence, the inverse of f is itself.

Question 3.
If f: R → R is defined by f(x) = x² – 3x + 2, find f(f(x)).
Solution.
It is given that f: R → R is defined as f(x) = x² – 3x + 2.
f(f(x)) = f(x² – 3x + 2)
= (x² – 3x + 2)² – 3(x² -3x + 2) + 2
= x⁴ + 9x² + 4 – 6x³ – 12x + 4x² – 3x² + 9x – 6 + 2
= x⁴ – 6x³ + 10x² – 3x.

Question 4.
Show that the function f: R → {x ∈ R: – 1 < x < 1} defined by f(x) = \(\frac{x}{1+|x|}\) ∈ R is one-one and onto function.
Solution.
It is given that f: R → {x ∈ R: – 1 < x < 1} is defined as f(x) = \(\frac{x}{1+|x|}\), x ∈ R.
Suppose f(x) = f(y), where x,y ∈ R ⇒ \(\frac{x}{1+|x|}=\frac{y}{1+|y|}\)
It can be observed that if x is positive and y is negative, then we have \(\frac{x}{1+x}=\frac{y}{1-y}\)
⇒ 2xy = x – y
Since x is positive and y is negative, then x > y ⇒ x – y > 0
But, 2xy is negative.
Then, 2xy ≠ x – y.
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out.
∴ x and y have to be either positive or negative.
When x and y are both positive, we have x y
f(x) = f(y)
⇒ \(\frac{x}{1+x}=\frac{y}{1+y}\)
⇒ x + xy = y + xy
⇒ x = y
When x and y are both negative, we have
f(x) = f(y)
⇒ \(\frac{x}{1-x}=\frac{y}{1-y}\)
⇒ x – xy = y – yx
⇒ x = y
∴ f is one-one.
Now, let y ∈ R such that – 1 < y < 1.
If y is negative, then there exists x = \(\frac{y}{1+y}\) ∈ R such that
f(x) = f(\(\frac{y}{1+y}\))
= \(\frac{\left(\frac{y}{1+y}\right)}{1+\left|\frac{y}{1+y}\right|}=\frac{\frac{y}{1+y}}{1+\left(\frac{-y}{1+y}\right)}=\frac{y}{1+y-y}\) = y
If y is positive, then there exists x = \(\frac{y}{1-y}\) ∈ R such that
f(x) = \(f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left(\frac{y}{1-y}\right)}\)

= \(\frac{\frac{y}{1-y}}{1+\left(\frac{y}{1-y}\right)}=\frac{y}{1-y+y}\) = y
∴ f is onto.
Hence, f is one-one and onto.

Question 5.
Show that the function f: R → R given by f(x) = x³ in injective.
Solution.
f: R → R is given as f(x) = x³.
Suppose f(x) = f(y), where x, y ∈ R.
⇒ x³ = y³ …………(i)
Now, we need to show that x = y
Suppose x * y, their cubes will also not be equal.
⇒ x³ ≠ y³
However, this will be a contradiction to Eq. (i).
∴ x = y
Hence, f is injective.

c

Question 6.
Give examples of two functions f: N → Z and g: Z → Z such that gof is injective but g is not injective.
[Hint: consider f(x) x and g(x) = |x|].
Solution.
Define f: N → Z as f(x) – x and g: Z → Z as g(x) =|x|
We first show that g is not injective.
It can be observed that
g(- 1) = |- 1|= 1; g(1) = |1|= 1
∴ g(- 1) = g(1), but – 1 ≠ 1.
∴ g is not injective.
Now, gof: N → Z is defined as gof(x) = g(f(x)) = g(x) =|x|.
Let x, y ∈ N such that gof(x) – gof(y).
⇒ |x| = |y|
Since x and y ∈ N, both are positive.
∴ |x |= |y |=> x = y
Hence, gof is injective.

Question 7.
Give examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.
[Hint: consider f(x) = x + 1 and g(x) = i]
Solution.
Define f: N → N by f(x) = x +1
and, g: N → N by g(x) =
We first show that f is not onto.
For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N.
∴ f is not onto.
Now, gof: N → N is defined by,
gof(x) = g(f(x)) = g(x + 1) = (x + 1) – 1 = x [∵ x ∈ N ⇒ (x + 1) > 1]
Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y.
Hence, gof is onto.

Question 8.
Given a non-empty set X, consider P(X) which is the set of all subsets of X.
Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if A c B. Is R an equivalence relation on P(X)? Justify your answer.
Solution.
Since every set is a subset of itself, ARA for all A ∈ P(X).
∴ R is reflexive.
Let ARB ⇒ A ⊂ B.
This cannot be implied to B ⊂ A.
For instance, if A = {1, 2} and B = {1, 2,3}, then it cannot be implied that B is related to A.
∴ R is not symmetric.
Further, if ARB and BRC, then A c B and B c C.
⇒ A ⊂ C
⇒ ARC
R is transitive.
Hence, R is not an equivalence relation since it is not symmetric.

Question 9.
Given a non-empty set X, consider the binary operation *: P(X) × P(X) P(X) given by A * B = A ∩ B ∀ A, B in P(X) where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation*.
Solution.
It is given that * : P(X) × P(X) → P(X) is defined as A * B = A ∩ B ∀ A, B ∈ P(X).
We know that A * X = A ∩ X = A = X ∩ A ∀ A ∈ P(X).
⇒ A * X = A = X * A ∀ A ∈ P (X)
Thus, X is the identity element for the given binary operation*.
Now, an element A ∈ P(X) is invertible if there exists B ∈ P(x) such that
A * B = X = B * A. (As X is the identity element)
i.e., A ∩ B = X = B ∩ A
This case in possible only when A = X = B.
Thus, X is the only invertible element in P(X) with respect to the given operation*.
Hence, the given result is proved.

Question 10.
Find the number of all onto functions from the set {1, 2, 3, n} to itself.
Solution.
Onto functions from the set {1, 2, 3, …, n} to itself is simply a permutation on n symbols 1, 2, …, n.
Thus, the total number of onto maps from {1, 2, 3,…, n} to itself is the same as the total number of permutations on symbols 1, 2,…, n, which is n!.

Question 11.
Let S = {a, b, c} and T = {1,2, 3}. Find F^-1 of the following functions F from S to T, if it exists.
(i) F = {(o, 3), (6, 2), (c, 1)}
(ii) F = {(a, 2), (6, 1), (c, 1)}
Solution.
Given, S = {a, b, c}, and T = {1, 2, 3}
F: S → T is defined as :
F = {(a, 3), (b, 2), (c, 1)}
⇒ f(a) = 3, F(b) = 2, F(c) = 1
Therefore, F^-1 : T → S is given by
F^-1 = {(3, a), (2, b), (1, c)}

(ii) F: S → T is defined as
F = {(a, 2), (b, 1), (c, 1)}
Since F (b) = F (c) = 1, F is not one-one.
Hence, F is not invertible i. e., F^-1 does not exist.

Question 12.
Consider the binary operations * : R × R → R and o: R × R → R defined as a * b = | a – b| and a o b = a, ∀ a, b ∈ R. Show that * is commutative hut not associative, o is associative but not commutative. Further, show that ∀ a, b, c ∈ R, a* (b o c) = (a * b) o (a * c). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
Solution.
It is given that *: R × R R and o: R × R → R is defined as a * b = |a – b| and a o b = a ∀ a, b ∈ R.
For a, b ∈ R, we have
a * b = |a – b|
b * a = |b – a| = |- (a – b)|= |a – b|
∴ a * b = b * a
Therefore, the operation * is commutative..
It can be observed that
(1 * 2) * 3 = (|1 – 2|) * 3 = 1 * 3 = |1 – 3|= 2
1 * (2 * 3) = 1 * (|2 – 3|) = 1 * 1 =|1 – 1 |= 0
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) (where 1, 2, 3 ∈ R)
Therefore, the operation * is not associative.
Now, consider the operation o
It can be observed that 1 o 2 = 1 and 2 o 1 = 2.
∴ 1 o 2 ≠ 2 o 1 where 1, 2 ∈ R
Therefore, the operation o is not commutative.
Let a, b, c ∈ R. Then, we have
(a o b) o c = a o c = a
a o (b o c) = a o b = a
⇒ (a o b) o c = a o (b o c)
Therefore, the operation o is associative.
Now, let a, b, c ∈ R, then we have
a * (b o c) = a * b = |a – b|
(a * b) o (a * c) = (|a – b|) o (|a – c|) = |a – b|
Hence a * (b o c) = (a * b) o (a * c)
Now, 1 o(2 * 3) = 1 o (|2 – 3|) = 1 o 1 = 1
(1 o 2) * (1 o 3) = 1 * 1 = |1 – 1|= 0
1 o (2 * 3) ≠ (1 o 2) * (1 o 3)
where 1, 2, 3 ∈ R Therefore, the operation o does not distribute over *.

Question 13.
Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B – (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A^-1 = A.
[Hint: (A – Φ) ∪ (Φ – A) = A and (A – A) ∪ (A – A) = A * A = Φ]
Solution.
It is given that *: P(X) × P(X) → P(X) is defined as
A * B = (A – B) ∪ (B – A) ∀ A, B, ∈ P(X).
Let A ∈ P(X). Then, we have
A * (Φ) = (A – Φ) ∪ (Φ – A) = A ∪ Φ = A
Φ * A = (Φ – A) ∪ (A – Φ) = Φ ∪ A = A
A * Φ = A = Φ * A ∀ A ∈ P(X)
Thus, Φ is the identity element for the given operation *.
Now, an element A s P(X) will be invertible if there exists B ∈ P(X) such that
A * B = Φ = B * A. (As Φ is the identity element)
Now, we observed that
A * A = (A – A) ∪ (A – A) = Φ ∪ Φ = Φ ∀ A ∈ P(X).
Hence, all the elements A of P(X) are invertible with A^-1 = A.

Question 14.
Define a binary operation * on the set {0, 1, 2, 3, 4, 5) as
a * b =
Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.
Solution.
(i) e is the identity element if a * e = e * a = a
a * 0 = a + 0, 0 * a = 0 + a = a
⇒ a * 0 = 0 * a = a
∴ 0 is the identity of the operation.

(ii) b is the inverse of a if a * b = b * a = e
Now a * (6 – a) = a + (6 – a) – 6 = 0
(6 – a) * a = (6 – a) + a – 6 = 0
Hence, each element of a of the set is invertible with inverse.

Question 15.
Let A = {-1, 0, 1, 2}, B = {-4,-2, 0,2} and f, g: A → B be functions defined by f(x) = x² – x, x ∈ A and g(x) = 2|x – \(\frac{1}{2}\)| – 1, x ∈ A. Are f and g equal? Justify your answer.
[Hint: One may not be that two functions f: A → B and g: A → B
such that f(a) = g(a) ∀ a ∈ A, are called equal functions.]
Solution.
It is given that A = {- 1,0,1, 2}, B = {- 4, – 2, 0, 2).
Also, it is given that f, g: A → B are defined by f(x) = x² – x, x ∈ A and
g(x) = 2 |x – \(\frac{1}{2}\)| – 1, x ∈ A
It is observed that
f(- 1) = (- 1)² – (- 1) = 1 + 1 = 2
and g(- 1) = 2|(- 1) – \(\frac{1}{2}\)| – 1
= 2(\(\frac{3}{2}\)) – 1 = 3 – 1 = 2
⇒ f(- 1) = g(- 1)

⇒ f(0) = (0)² – 0 = 0
and g(0) = 2|0 – \(\frac{1}{2}\)|
= 2(\(\frac{3}{2}\)) – 1 = 1 – 1 = 0

⇒ f(0) = g(0)
f(1) = (1)² – 1 = 1 – 1 = 0
and g(1)= 2|1 – \(\frac{1}{2}\)|
= 2(\(\frac{1}{2}\)) – 1 = 1 – 1 = o

⇒ f(1) = g(1)
f(2) = (2)² – 2 = 4 – 2 = 2
and g(2) = 2 |2 – \(\frac{1}{2}\)|
= 2(\(\frac{3}{2}\)) – 1 = 3 – 1 = 2
⇒ f(2) = g(2)
∴ f(a) = g(a) ∀ a ∈ A
Hence, the functions f and g are equal.

Question 16.
Let A = {1, 2, 3}. Then, number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive, is
(A) 1
(B) 2
(C) 3
(D) 4
Solution.
This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.
Relation R is symmetric since (1, 2), (2 ,1) ∈ R and (1, 3), (3, 1) ∈ R.
But relationR is not transitive as (3, 1), (1, 2) ∈ R but (3, 2) ∈ R.
Now, if we add any one of the two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.
Hence, the total number of desired relation is one.
Thus, the correct option is (A).

Question 17.
Let A = {1, 2, 3}. Then, number of equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C)3
(D) 4
Solution.
It is given that A = {1, 2, 3}
The smallest equivalence relation containing (1, 2) is given by,
R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four pairs i. e., (2, 3), (3, 2), (1, 3), and (3, 1).
If we add any one pair [say (2, 3)] to R₁ then for symmetry we must add (3, 2). Also, for transitivity, we are required to add (1, 3) and (3,1). Hence, the only equivalence relation (bigger than R₁) is the universal relation.
This shows that the total number of equivalence relations containing (1, 2) is two. The correct option is (B).

Question 18.
Let f: R → R be the signum function defined as
f(x) =
and g: R → R be the greatest integer function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?
Solution.
It is given that
f: R → R is defined as f(x) =
Also, g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x .
Now, let x ∈ (0, 1]
Then, we have
[x] = 1, if x = 1 and [x] = 0 if 0 < x < 1. ∴ fog(x) = f (g(x)) = f([x]) = gof(x) = g(f(x))= g(1) [∵ x > 0]
= [1] = 1 .
Thus, when x ∈ (0, 1), we have fog(x) = 0 and gof(x) = 1.
But fog (1) ≠ gof (1)
Hence, fog and gof do not coincide in (0,1].

Question 19.
Number of binary operations on the set {a, b} are (A) 10 (B) 16 (C) 20 (D) 8
Solution.
A binary operation * on {a, b} is a function from {a, b} × {a, b} → {a, b} i. e.,* is a function from {(a, a), (a, b), (b, a), (b, b)} → {a, b}.
Hence, the total number of binary operations on the set {a, b} is 2⁴ i.e. 16.
Thus, the correct option is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Ex 1.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.4

Question 1.
Determine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for this.
(i) On Z⁺, define * by a * b = a – b
(ii) On Z⁺, define * by a * b = ab
(iii) On R, define * by a * b = ab²
(iv) On Z⁺, define * by a * b = |a – b|
(v) On Z⁺, define * by a * b = a
Sol.
(i) On Z⁺, * is defined by a * b = a – b.
It is not a binary operation as the image of (1, 2) under * is
1 * 2 = 1 – 2 = – 1 ∉ Z⁺.

(ii) On Z⁺, * is defined by a * b = ab.
It is seen that for each a, b ∈ Z⁺, there is a unique element ab in Z⁺.
This means that * carries each pair (a, b) to a unique element a * b = ab in Z⁺. Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab².
It is seen that for each a, b ∈ R, there is a unique element ab² in R.
This means that * carries each pair (a, b) to a unique element a * b = ab² in R. Therefore, * is a binary operation.

(iv) On Z⁺, * is defined by a * b =|a – b|.
It is seen that for each a, b ∈ Z⁺, there is a unique element | a – b | in Z⁺. This means that * carries each pair (a, b) to a unique element a * b = |a – b|in Z⁺. Therefore, * is a binary operation.

(v) On Z⁺, * is defined by a * b = a.
It is seen that for each a, b ∈ Z⁺, there is a unique element a ∈ Z⁺. This means that * carries each pair (a, b) to a unique element a * b = a in Z⁺. Therefore, * is a binary operation.

Question 2.
For each operation * defined below, determine whether * is binary commutative or associative.
(i) On Z, define a* b = a – b
(ii) On Q, define a * b = ab + 1
(iii) On Q, define a* b = \(\frac{a b}{2}\)
(iv) On Z⁺, define a * b = 2^ab
(v) On Z⁺, define a * b = a^b
(vi) On R – {- 1},define a * b = \(\frac{a}{b+1}\)
Solution.
(i) On Z, operation * is defined as
(a) a * b = a – b
⇒ b * a = b – a
But a – b ≠ b – a
⇒ a * b ≠ b * a
∴ Defined operation is not commutative.

(b) a – (b – c) ≠ (a – b) – c
∴ Binary operation * as defined is not associative.

(ii) On Q, operation * is defined as a * b = ab +1
(a) ab + 1 = ba + 1, a * b = b * a
∴ Defined binary operation is commutative.

(b) a * (b * c) = a * (bc + 1) = a (bc + 1) + 1 = abc + a + 1
and (a * b)* c = (ab + 1) * c = (ab + 1)c + 1
= abc + c + 1
a * (b * c) ≠ (a * b) * c
∴ Binary operation defined is not associative.

(iii) (a) On Q, operation * is defined as a * b = \(\frac{ab}{2}\)
∴ a * b = b * a
∴ Operation binary defined is commutative.

(b) a * b = a * \(\frac{b c}{2}=\frac{a b c}{4}\)
and (a * b) * c = \(\frac{b c}{2}\) * c = \(\frac{a b c}{4}\)
⇒ Defined binary operation is associative.

(iv) On Z⁺, operation * is defined as a * b = 2^ab
(a) a * b = 2^ab, b * a = 2^ba = 2^ab
a * b = b * a
Binary operation defined is commutative.

(b) a * (b * c) = a * 2^ba = 2^{a . bc}
(a * b) * c = 2^ab * c = 2^2ab
Thus, (a * b) * c ≠ a * (b * c)
∴ Binary operation * as defined is not associative.

(v) On Z⁺, a * b = a^b
(a) b * a = b^a
∴ a^b = b^a
⇒ a * b ≠ b * a
* is not commutative.

(b) (a * b) * c = a^b * c
= (a^b)^c = a^bc
a * (b * c) = a * bc = a^bc.
This (a * b) * c ≠ (a * b * c)
∴ Operation * is not associative.

(vi) On Z⁺ operation * is defined as
a * b = \(\frac{a}{b+1}\), b ≠ – 1
∴ b * a = \(\frac{b}{a+1}\)
(a) a * b ≠ b * a
Binary operation defined is not commutative.

(b) (a * b) * c = \(a^{*}\left(\frac{b}{c+1}\right)=\frac{a}{\frac{b}{c+1}+1}=\frac{a(c+1)}{b+c+1}\)

(a * b) * c = \(\frac{a}{b+1} * c=\frac{\frac{a}{b+1}}{c+1}=\frac{a}{(b+1)(c+1)}\)

∴ a * (b * c) ≠ (a * b) * c
⇒ Binary operation defined above is not associative.

Question 3.
Consider the binary operation ^ on the set (1, 2, 3, 4, 5} defined by a ^ b = min {a, b}. Write the multiplication table of the operation ^.
Solution.
The binary operation ^ on the set {1, 2, 3, 4, 5} is defined as
a ^ b = min{a, b} for a, b ∈ {1, 2, 3, 4, 5}.
Thus, the operation table for the given operation ^ can be given as

Question 4.
Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.
(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Is * commutative?
(iii) Compute (2* 3) * (4* 5).
(Hint: use the following table) (i)

Solution.
(i) We have (2 * 3) *4 = 1 * 4 = 1
and 2 * (3 * 4) = 2 * 1 = 1

(ii) For every a, b ∈ (1, 2, 3, 4, 5}, we have a * b = b * a. Therefore, the operation * is commutative.
(iii) We have (2 * 3) = 1 and (4 * 5) = 1 .
∴ (2 * 3) * (4 * 5) = 1 * 1 = 1.

Question 5.
Let *’ be the binary operation on the set {1, 2, 3, 4, 5} is defined by a *’ b = H.C.F. of a and b. Is the operation *’ same as the operation * defined in Q. 4 above? Justify your answer.
Solution.
The binary operation *’ on the set {1, 2, 3, 4, 5} is defined as
a*’ b = HCF of a and b.
The operation table for the operation * can be given as :

We observe that the operation table for the operations * and *’ are the same.
Thus, the operation *’ is same as the operation *.

Question 6.
Let * be the binary operation on N given by a * b = L.C.M. of a and b.
(i) Find 5 * 7, 20 * 16
(ii) Is * commutative?
(iii) Is * associative?
(iv) Find the identity of * in N.
(v) Which elements of N are invertible for the operation *?
Solution.
The binary operation * defined as a * b = L.C.M. of a and b
(i) 5 * 7 = L.C.M. of 5 and 7 = 35
and 20 * 16 = L.C.M. of 20 and 16 = 80

(ii) a * b = L.C.M. of a and b
b * a = L.C.M. of b and a
⇒ a * b = b * a L.C.M. of a, b and b, a are equal
∴ Binary operation * is commutative.

(iii) a * (b * c) = L.C.M. of a, b, c
and (a * b)* c = L.C.M. of a, b, c
⇒ a * (b * c) = (a * b) * c
⇒ Binary operation * is associative.

(iv) Identity of * in N is 1
1 * a = a * 1 = a = L.C.M. of 1 and a.

(v) Let * : N × N → N defined as a * b = L.C.M. of (a, b)
For a = 1, b = 1, a * b = 1 = b * a. Otherwise a * b ≠ 1
∴ Binary operation * is not invertible.
⇒ 1 is invertible for operaiton *.

Question 7.
Is * defined on the set {1, 2, 3, 4, 5} by a * 6 = L.C.M. of a and 6 a binary operation? Justify your answer.
Solution.
The operation * on the set A = {1, 2, 3, 4, 5} is defined as a * b = L.C.M. of a and b.
Now, 2 * 3 = L.C.M. of 2 and 3 = 6.
But 6 does not belong to the given set.
Hence, the given operation * is not a binary operation.

Question 8.
Let * be the binary operation on N defined by a * 6 = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
Solution.
The binary operation * on N is defined as a * b = H.C.F. of a and b It is known that
H.C.F. of a and b = H.C.F. of b and a V a, b ∈ N.
∴ a * b = b * a
Thus, the operation * is commutative.
For a, b, c ∈ N, we have
(a * b) * c = (H.C.F. of a and b) * c = H.C.F. of a, b and c
a * (b * c) = a * (H.C.F. of b and c) = H.C.F. of a, b, and c
∴ (a * b) * c = a* (b * c)
Thus, the operation * is associative.
Now, an element e ∈ N will be the identity for the operation * if a * e = a = e * a for ∀ a ∈ N.
But this relation is not true for any a ∈ N.
Thus, the operation * does not have any identity in N.

Question 9.
Let * be a binary operation on the set Q of rational numbers as
(i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
(iv) a * b = (a – b)²
(v) a * b = \(\frac{ab}{4}\)
(vi) a * b = ab²
Find which of the binary operations are commutative and which are associative.
Solution.
Operation is on the set Q.
(i) defined as a * b = a – b
(a) Now b * a = b – a But a – b *b – a
∴ a * b * b * a
∴ Operation * is not commutative.

(b) a* (b * c) = a * (b – c) = a – (b – c) = a – b + c
(a * b) * c = (a – b) * c = a – b – c
Thus, a * (b * c) ^ (a * b) * c
∴ The operation * as defined is not associative.

(ii) (a) a * b = a² + b²
b * a = b² + a² = a² + b²
∴ a * b = b * a
∴ This binary operation is commutative.

(b) a * (b * c) = a * (b² + c²)
= a² + (b² + c²)²
(a * b) * c = (a² + b²) * c = (a² + b²)² + c²
⇒ a * (b * c) * (a * b) * c
∴ The operation * given is not associative.

(iii) Operation * is defined as
a * b = a + ab
(a) b* a = b + ba
∴ a * b ≠ b * a
∴ This operation is not commutative.

(b) a * (b * c) = a * (b + bc)
= a + a(b + bc)
= a + ab + abc
(a* b) * c = (a + ab) *c = a + ab + (a + ab) . c
= a + ab + ac + abc
⇒ a* (b* c)& (a* b)* c
⇒ The binary operation is not associative.

(iv) The binary operation is defined as a * b = (a – b)²
(a) b * a = (b – a)² = (a – b)²
⇒ a * b = b * a
∴ This binary operation * is commutative.

(b) a * (b * c) = a * (b – c)²
= [a – (b – c)²]²
(a * b) * c = (a – b)² * c = [(a – b)² – c]²
⇒ (a * b) * c ≠ a * (b * c)
Thus, the operation given is associative.

(v) Binary operation is * defined as
a * b = \(\frac{ab}{4}\)

(a) b * a = \(\frac{ba}{4}\) = \(\frac{ab}{4}\)
a* b^b* a
∴ The operation is not commutative.

(b) a * (b * c) = a * \(\frac{bc}{4}\)
= \(\frac{a}{4}\left(\frac{b c}{4}\right)=\frac{a b c}{16}\)
(a * b) * c = \(\frac{ab}{4}\) * c
= \(\frac{a b}{4} \cdot \frac{c}{4}=\frac{a b c}{16}\)
⇒ a * (b* c) = (a * b) * c
Thus, the operation given is associative.

(vi) Binary operation is defined as
a * b = ab²
(a) b * a = ba² ≠ ab²
∴ a * b ≠ b * a
∴ The operation is not commutative.

(b) a * (b * c) = a * bc²
= a(bc²)²
= ab²c⁴
(a * b)* c = ab² * c
= (ab²)c²
= ab²c²
∴ a * (b * c) ≠ (a * b) * c
∴ Binary operation * given is not associative.

Question 10.
Find which of the operations given above has identity.
Solution.
An element e ∈ Q will be the identity element for the operation * if
a * e = a = e * a, ∀ a ∈ Q
(i) a * b = a – b
lf a * e = a, a ≠ 0
⇒ a – e = a, a ≠ 0 ⇒ e = 0
Also, e * a = a
⇒ e – a = a ⇒ e = 2 a
e = 0 = 2a, a ≠ 0
But the identity is unique. Hence this operation has no identity.

(ii) a * b = a² + b²
If a * e = a, then a² + e² = a
For a = – 2, (- 2)² + e² = 4 + e² ≠ – 2
Hence, there is no identity element.

(iii) a * b = a + ab
If a * e = a
⇒ a + ae a
⇒ ae = 0
⇒ e = 0, a ≠ 0
Also a * e = a
⇒ e + ae = a
⇒ e = \(\frac{a}{a+1}\), a ≠ 1
∴ e = 0 = \(\frac{a}{a+1}\), a ≠ 0
But the identity in unique. Hence this operation has no identify.

(iv) a * b = (a – b)²
If a* e = a, then (a – e)² = a.
A square is always positive, so for a = – 2, (- 2 – e)² ≠ – 2.
Hence, there is no identity element.

(v) a * b – ab/ 4
If a * e = a, then ae / 4 = a.
Hence, e = 4 is the identity element.
∴ a * 4 = 4 * a = 4a/4 = a.

(vi) a * b = ab²
If a * e = a
⇒ ae² = a
⇒ e² = 1
⇒ e = ±1
But identity is unique. Hence this operation has no identity.
Therefore only part (v) has an identity element.

Question 11.
Let A = N × N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Find the identity element for * on A, if any.
Solution.
Given that A = N × N and * is a binary operation on A and is defined by (a, b) * (c, d) = (a + c,b + d.)
Let (a, b), (c, d) ∈ A
Then, a, b, c, d ∈ N
We have (a, b) * (c, d) = (a + c, b + d)
and (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
[Addition is commutative in the set of natural numbers]
∴ (a, b) * (c, d) = (c, d) * (a, b)
Therefore, the operation * is commutative.
Now, let (a, b), (c, d), (e, f) ∈ A
Then, a, b, c, d, e, f ∈ N
We have {(a, b) * (c, d)} * (e, f) = (a + c,b + d) * (e, f)
= (a+ c + e, b + d + f)
(a, b) * ((c, d) * (e, f)) = (a, b) * (c + e, d + f) = (a + c + e, b + d + f)
((a, b) * (c, d)) * (e, f) = (a, b) * ((c, d) * (e, f))
Therefore, the operation * is associative.
An element e = (e₁, e₂) ∈ A will be an identity element for the operation * if
a * e = a = e * a ∀ a = (a₁, a₂) ∈ A, i.e., (a₁ + e₁, a₂ + e₂)
= (a₁, a₂) = (e₁ + a₁; e₂ + a₂)
which is not true for any element in A.
Therefore, the operation * does not have any identity element.

Question 12.
State whether the following statements are true or false. Justify.
(i) For an arbitrary binaiy operation * on a set N, a * a = a ∀ a ∈ N.
(ii) If * is a commutative binary operation on N, then a* (b* c) = (c * b) * a
Solution.
(i) Define an operation * on IV as a * b – a + b ∀ a, b ∈ N
Then, in particular, for b = a = 3, we have 3 * 3 = 3 + 3 = 6 ≠ 3
Therefore, statement (i) is false.

(ii) R.H.S. = (c * b) * a
= (b * c) * a [* is commutative]
= a * (b * c) [Again, as * is commutative]
= L.H.S.
∴ a * (b * c) = (c * b) * a
Therefore, statement (ii) is true.

Question 13.
Consider a binary operation * on N defined as a * b = a3 +b3. Choose the correct answer.
(A) Is * both associative and commutative?
(B) Is * commutative but not associative?
(C) Is * associative but not commutative?
(D) Is * neither commutative nor associative?
Solution.
On N, the operation * is defined as a * b = a³ + b³.
For, a, b ∈ N, we have
a * b = a³ + b³
= b³ + a³ = b * a [Addition is commutative in N]
Therefore, the operation * is commutative.
It can be observed that
(1 * 2) * 3 = (1³ + 2³) * 3 = 9 * 3
= 9³ + 3³
= 729 + 27 = 756

1 * (2 * 3) = 1 * (2³ +3³)
= 1 * (8 + 27) = 1 * 35
= 1³ + 35³
= 1 + (35)³
= 1 + 42875 = 42876
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) where 1, 2, 3 ∈ N
Therefore, the operation * is not associative.
Hence, the operation * is commutative, but not associative.
Thus, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Ex 1.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3

Question 1.
Let f:{1, 3, 4} → {1, 2, 5} and g:{ 1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5,1)}. Write down gof.
Solution.
The functions f :{1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as f = {(1, 2), (3, 5), (4,1)} and g = {(1, 3), (2, 3), (5,1)}.
gof (1) = g(f(1)) = g(2) = 3 [∵ f(1) = 2 and g(2) = 3]
gof (3) = g(f(3)) = g(5) = 1 [∵ f(3) = 5 and g(5) = 1]
gof (4) = g(f(4)) = g(1) = 3 [∵ f(4) = 1 and g(1) = 3]
∴ gof = {(1,3), (3,1), (4, 3)}.

Question 2.
Let f, g and h be functions from R to R. Show that (f + g)oh = foh + goh (f . g)oh = (foh) (goh)
Solution.
To prove (f + g)oh = foh + goh Consider
((f + g)oh)(x) = (f + g)(h(x))
f(h(x)) + g(h(x)) = (foh)(x) + (goh)(x) = {{foh) + (goh)}(x)
((f + g)oh)(x) = {(foh) + (goh)} (x) ∀ x ∈ R
Hence, (f + g)oh = foh + goh.
To prove (f . g)oh = (foh) . (goh)
Consider
((f . g)oh) (x) = (f . g) (h(x)) = f(h(x)) . g(h(x))
= (foh)(x).(goh)(x)
= {(foh) . (goh)}(x)
∴ ((f . g)oh)(x) = {(foh) . (goh)}(x) ∀ x ∈ R
Hence, (f . g)oh = (/oh) . (goh)

Question 3.
Find gof and fog, if
(i) f(x) = |x| and g(x) = |5x – 2|
(ii) f(x) = 8x³ and g(x) = \(x^{\frac{1}{3}}\)
Solution.
(i) f(x) =|x| and g(x) = |5x – 2|
∴ (gof)(x) = g(f (x)) = g(| x |) =| 5| x | – 2 |
(fog(x)) = f(g(x)) = f(| 5x – 2 |) = | | 5x-2 || = |5x – 2|

(ii) f(x) = 8x³ and g(x) = \(x^{\frac{1}{3}}\)
∴ gof(x) = g(f(x))
= g(8x³)
= (8x³)^{\(\frac{1}{3}\)}
= 8x

(fog)(x) = f(g(x))
= f(\(x^{\frac{1}{3}}\))
= 8(\(x^{\frac{1}{3}}\))³
= 8x

Question 4.
If f(x) = \(\frac{(4 x+3)}{(6 x-4)}\), x ≠ \(\frac{2}{3}\) show that fof(x) = x for all x ≠ \(\frac{2}{3}\) What is the inverse of f?
Solution.
It is given that f(x) = \(\frac{(4 x+3)}{(6 x-4)}\), x ≠ \(\frac{2}{3}\)
(fof)(x) = f(f(x)) = f(\(\frac{(4 x+3)}{(6 x-4)}\))
= \(\frac{4\left(\frac{4 x+3}{6 x-4}\right)+3}{6\left(\frac{4 x+3}{6 x-4}\right)-4}\)
= \(\frac{16 x+12+18 x-12}{24 x+18-24 x+16}=\frac{34 x}{34}\) = x
Therefore, fof(x) = x, for all x ≠ \(\frac{2}{3}\).
⇒ fof = 1.
Hence, the given function f is invertible and the inverse of f is itself.

Question 5.
State with reason whether the following functions have inverse
(i) f: {1, 2, 3, 4} → {10} with  f = {(1, 10), <2,10), (8, 10), <4, 10)}
(ii) g: {5, 6, 7,8} → {1, 2, 3, 4,} with g = {(5, 4), (6,3), (7,4), (8, 2)}
(iii) h:{2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}.
Solution.
(i) Function f:{1, 2, 3, 4} {10} defined as
f = {(1,10), (2,10), (3,10), (4,10)}
From the given definition of f, we can see that f is a many-one function as:
f(1) = f(2) = f(3) = f(4) = 10
∴ f is not one-one.
Hence, function f does not have an inverse.

(ii) Function g:{5, 6, 7,8} → {1,2, 3, 4,} defined as g = {(5, 4), (6, 3), (7, 4), (8, 2)}
From the given definition of g, it is seen that g is a many-one function as : g(5) = g(7) = 4.
∴ g is not one-one,
Hence, function g does not have an inverse.

(iii) Function h:{2, 3, 4, 5,} → {7, 9,11,13} defined as h = {(2, 7), (3, 9), (4,11), (5,13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.
∴ Function h is one-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5} such that h(x) = y.
Thus, h is a one-one and onto function. Hence, h has an inverse.

Question 6.
Show that f: [- 1,1] → R, given by f(x) = \(\frac{x}{x+2}\) is one-one. Find the inverse of the function f: [- 1, 1] → Range f.
[Hint : For y ∈ R Range f, y = f(x) = \(\frac{x}{x+2}\), for some x in [- 1, 1] i.e., x = \(\frac{2 y}{1-y}\)]
Solution.
f: [- 1, 1] → R, is given as f(x) = \(\frac{x}{x+2}\)
Let f(ix) = f(y).
⇒ \(\frac{x}{x+2}=\frac{y}{y+2}\)
⇒ 2x = 2y
⇒ x = y
∴ f is one-one function.
It is clear that f: [- 1,1] Range f is onto.
∴ f: [- 1, 1] → Range f is one-one and onto and therefore, the inverse of the function :
f: [- 1, 1] → Range f exists.
Let g: Range f → [- 1, 1] be the inverse of f.
Let y be an arbitrary element of range f.
Since, f: [- 1, 1] → Range f is onto , we have
y = f(x) for some x ∈ [- 1, 1]
⇒ y = \(\frac{x}{x+2}\)
⇒ xy + 2y = x
⇒ x(1 – y) = 2y
⇒ x = \(\frac{2 y}{1-y}\), y ≠ 1
Now, let us define g: Range f → [- 1, 1] as g(y) = \(\frac{2 y}{1-y}\), y ≠ 1.
Now, (gof)(x) = g(f(x))
= g(\(\frac{x}{x+2}\)) = \(\frac{2\left(\frac{x}{x+2}\right)}{1-\frac{x}{x+2}}\)
= \(\frac{2 x}{x+2-x}=\frac{2 x}{2}\) = x

(fog)(y) = f(g(y))
= f(\(\frac{2 y}{1-y}\)) = \(\frac{\frac{2 y}{1-y}}{\frac{2 y}{1-y}+2}\)
= \(\frac{2 y}{2 y+2-2 y}=\frac{2 y}{2}\) = y
∴ gof = I_{[- 1, 1]} and fog = I_{Range f}
∴ f^-1 = g
⇒ f^-1(y) = \(\frac{2 y}{1-y}\), y ≠ 1.

Question 7.
Consider f:R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Solution.
Here, f: R → R is given by f(x) = 4x +3
Let x, y ∈ R, such that
f(x) = f(y)
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y
⇒ x = y
Therefore, f is a one-one function. .
Let y = 4x +3
⇒ There exists, x = \(\frac{y-3}{7}\) ∈ R, ∀ y ∈ R
Therefore for any y ∈ R, there exists x = \(\frac{y-3}{4}\) ∈ R such that
f(x) = f(\(\frac{y-3}{4}\)) = 4 (\(\frac{y-3}{4}\)) + 3 = y
Therefore, f is onto function.
Thus, f is one-one and onto and therefore, f^-1 exists.
Let us define g: R → R by g(x) = \(\frac{x-3}{4}\)
Now, (gof)(x) = g(f(x)) = g(4x + 3)
= \(\frac{(4 x+3)-3}{4}\) = x

(fog)(y) = f(g(y))
= \(f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)\) + 3
= y – 3 + 3 = y
Therefore, gof = fog = I_R
Hence, f is invertible and the inverse of f is given by
f^-1 (y) = g(y) = \(\frac{y-3}{4}\)

Question 8.
Consider f: R → [4, ∞) given by f(x) = x² + 4 Show that f is invertible with the inverse f^-1 of f given by f^-1(y) = \(\sqrt{y-4}\), where R is the set of all non-negative real numbers.
Solution.
Function f: R₊ → [4, ∞) is given as f(x) = x² + 4.

One-one :
Let f(x) = f(y).
⇒ x² + 4 = y² + 4
⇒ x² = y²
⇒ x = y [as x = y ∈ R₊]
∴ f is one-one function.

Onto :
For y ∈ [4, ∞), let y = x² + 4.
⇒ x² = y – 4 ≥ 0 [as y ≥ 4]
⇒ x = \(\sqrt{y-4}\) > 0
Therefore, for any y ∈ R, there exists x = \(\sqrt{y-4}\) ∈ R such that
f(x) = f(\(\sqrt{y-4}\))
= (\(\sqrt{y-4}\))² + 4
= y – 4 + 4 = y
∴ f is onto.
Thus, f is one-one and onto and therefore, f^-1 exists.
Let us define g:[4, ∞) → R₊ by,
g(y) = \(\sqrt{y-4}\)
Now, gof (x) = g(f(x)) = g(x² + 4)
= \(\sqrt{\left(x^{2}+4\right)-4}=\sqrt{x^{2}}\) = x

and, fog(y) = f(g(y))
= f(\(\sqrt{y-4}\))
= \((\sqrt{y-4})^{2}+4\)
= (y – 4) + 4 = y
∴ gof = fog = I_R+
Hence, f is invertible and the inverse of f is given by
f^-1 (y) = g(y) = \(\sqrt{y-4}\)

Question 9.
Consider f: R → [- 5, ∞) given by f(x) = 9x² + 6x – 5. Show that f is invertible with f^-1 (y) = \(\left(\frac{(\sqrt{y+6}-1}{3}\right)\)
Solution.
f: R₊ → [- 5, ∞) is given as f(x) = 9x² + 6x – 5
Let y be an arbitrary element of (- 5, ∞)
Let y = 9x² + 6x – 5
y = (3x + 1)² – 1 – 5
= (3x + 1)² – 6
⇒ (3x + 1)² = y + 6
⇒ 3x + 1 = \(\sqrt{y+6}\) [as y ≥ – 5 ⇒ y + 6 > 0]
⇒ x = \(\frac{\sqrt{y+6}-1}{3}\)
∴ f is onto, thereby range f = [- 5, ∞]
Let us define g: [- 5, ∞) → R₊ as g(y) = \(\frac{\sqrt{y+6}-1}{3}\)
We now have :
(gof)(x) = g(f(x)) = g(9x² + 6x – 5) = g((3x +1)² – 6)
= \(\frac{\sqrt{(3 x+1)^{2}-6+6}-1}{3}=\frac{3 x+1-1}{3}\) = x
and, (fog)(y) = f(g(y))
= \(f\left(\frac{\sqrt{y+6}-1}{3}\right)=\left[3\left(\frac{(\sqrt{y+6})-1}{3}\right)+1\right]^{2}-6\)
= \((\sqrt{y+6})^{2}\) – 6 = y + 6 – 6 = y
∴ gof = I_R and fog = I_{[ – 5, ∞]}
Hence f is invertible and inverse of f is given by
f^-1(y) = g(y) = \(\frac{\sqrt{y+6}-1}{3}\)

Question 10.
Let f: X → Y be an invertible function. Show that f has unique inverse.
[Hint: Suppose g₁ and g₂ are two inverses of f. Then for all y ∈ Y, fog₁ (y) = 1, (y) = fog₂ (y). Use one-one ness of f].
Solution.
Let f: X → Y be an invertible function.
Also, suppose f has two inverses (say g₁ and g₂).
Then, for all y ∈ Y, we have
fog₁ (y) = I_y(y) = fog₂(y)
⇒ f(g₁(y)) = f(g₂(y))
⇒ g₁(y) = g₂(y) [f is invertible ⇒ f is one-one, g is one-one]
⇒ g₁ = g₂
Hence, f has a unique inverse.

Question 11.
Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f^-1 and show that (f^-1)^-1 = f.
Solution.
Function f: {1,2, 3} → {a, b, c} is given by f(1) = a, f(2) = b and f(3) = c
If we define g :{a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we
(fog)(a) = f(g(a)) = f(1) = a
(fog)(b) = f(g(b) = f(2) = b
(fog)(c) = f(g(c)) = f(3) = c
(gof)(2) = g(f(2)) = g(b) = 2
(gof)(3) = g(f(3)) = g(c) = 3
∴ gof = I_X and fog = I_Y, where X = {1, 2, 3} and Y = {a, b, c}. Thus, the inverse of f exists and f^-1 = g.
∴ f^-1 : {a, b, c} → {1, 2, 3} is given by,
f^-1(a) = 1, f^-1(b) = 2, f^-1(c) = 3
Let us now find the inverse of f^-1 i.e., find the inverse of g.
If we define h:{ 1,2, 3} → {a, b, c} as h(1) = a, h(2) = b, h(3) = c, then we have
(goh)(1) = g(h(1l)) = g(a) = 1
(goh) (2) = g(h(2)) = g(b) = 2
(goh) (3) = g(h(3)) = g(c) = 3
and,(hog)(a) = h(g(a)) – h(1) = a
(hog) (b) = h(g(b)) = h(2) = b
(hog)(c) = h(g(c)) = h(3) = c
∴ goh = I_X and hog = I_Y, where X = {1, 2, 3} and Y = {a, b, c}.
Thus, the inverse of g exists and g^-1 = h
⇒ (f^-1)^-1 = h
It can be noted that h = f.
Hence, (f^-1)^-1 = f.

Question 12.
Let f: X → Y be an invertible function. Show that the inverse of f1 is f, i.e., (f ^-1)^-1 = f.
Solution.
Let f:X → Y be an invertible function.
Then, there exists a functiong:Y → X such that gof = I_X and fog – I_Y.
Here, f^-1 = g.
Now, gof = I_X and fog = I_Y
⇒ f^-1 = I_X and fof^-1 = I_Y
Hence, f^-1: Y → X is invertible and f is the inverse of f^-1 i-e., (f^-1)^-1 = f

Question 13.
If f : R → R be given by fix) = (3 – x³)^{\(\frac{1}{3}\)}, then fof(x) is
(A) x^{\(\frac{1}{3}\)}
(B) x³
(C) x
D) (3 – x³)
Solution.
Function f: R → R is given as f(x) = {3 – x³)^{\(\frac{1}{3}\)}; f(x) = (3 – x³)^{\(\frac{1}{3}\)}
∴ fof(x) = f(f(x)) = f(3 – x³)^{\(\frac{1}{3}\)}
= [3 – ((3 – x³)^{\(\frac{1}{3}\)})³]^{\(\frac{1}{3}\)}
= [3 – (3 – x³)]^{\(\frac{1}{3}\)}
= (x³)^{\(\frac{1}{3}\)} = x
∴ fof(x) = x
The correct answer is (C)

Question 14.
Let f: R – {- \(\frac{4}{3}\)} R be a function defined as f(x) = \(\frac{4 x}{3 x+4}\). The inverse of f is the map g: Range f → R given by
(A) g(y) = \(\frac{3 y}{3-4 y}\)

(B) g(y) = \(\frac{4 y}{4-3 y}\)

(C) g(y) = \(\frac{4 y}{3-4 y}\)

(D) g(y) = \(\frac{3 y}{4-3 y}\)
Solution.
Given that f : R – {- \(\frac{4}{3}\)} → R is a function defined as
f(x) = \(\frac{4 x}{3 x+4}\)
i.e., y = \(\frac{4 x}{3 x+4}\)
3 xy + 4y = 4x
4y = 4x – 3xy
4 y = x(4 – 3y)
x = \(\frac{4 y}{4-3 y}\)
∴ f^-1(y) = g(y) = \(\frac{4 y}{4-3 y}\)
The correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Ex 1.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2

Question 1.
Show that the function F: R → R, defined by f(x) = \(\frac{1}{x}\) is one-one and onto, where R, is the set of all non-zero real numbers. Is the result true, if the domain R. is replaced by N with co-domain being same as R?
Solution.
It is given that f: R. → R. is defined by f(x) = \(\frac{1}{x}\)
One-one :
f(x) = f(y)
⇒ \(\frac{1}{x}\) = \(\frac{1}{y}\)
⇒ x = y
∴ f is one-one.

Onto :
It is clear that for y ∈ R., there exists x = \(\frac{1}{y}\) ∈ R. (Exists as y ≠ 0) such that f(x) = \(\frac{1}{\left(\frac{1}{y}\right)}\) = y.
∴ f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function g :N →R, defined by
g(x) = \(\frac{1}{x}\)
We have,
g(x₁) = g(x₂)
⇒ \(\frac{1}{x_{1}}=\frac{1}{x_{2}}\)
x₁ = x₂
∴ g is one-one.
Further, it is clear that g is^not onto as for 1.2 ∈ R, there does not exist any x in N such that g(x) = \(\frac{1}{1.2}\).
Hence, function g is one-one but not onto.

Question 2.
Check the injectivity and surjectivity of the following functions
(i) f: N → N given by f(x) = x²
(ii) f: Z → Z given by f(x) = x²
(iii) f: R → R given by f(x) = x²
(vi) f: N → N given by f(x)) = x³
(v) f: Z → Z given by f(x) = x³
Solution.
(i) f: N → N is given by,
f(x) = x²
It is seen that for x, y ∈ N, f(x) = f(y)
⇒ x² = y²
⇒ x = y
∴ f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x² = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.

(ii) f: Z → Z is given by,
f(x) = x²
It is seen that f(- 1) = f(1) = 1, but = – 1 ≠ 1.
∴ f is not injective.
Now, – 2 ∈ Z. But, there does not exist any element x ∈ Z such that f(x) = x² = – 2
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

(iii) f: R → R is given by, f(x) = x²
It is seen that f(- 1) = f(1) = 1, but -1 ≠ 1.
∴ f is not injective.
Now, – 2 ∈ R. But , there does not exist any element x ∈ R such that f(x) = x² = – 2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

(iv) f : N → N given by,
f(x) = x³
It is seen that for x, y ∈ N, f(x) = f(y)
⇒ x³ = y³
⇒ x = y
∴ f is injective.
Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x³ = 2.
∴ f is not surjective
Hence, function f is injective but not surjective.

(v) f: Z → Z is given by, f(x) = x3
It is seen that for x, y ∈ Z, f(x) = f(y)
⇒ x³ = y³
⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x³ = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.

Question 3.
Prove that the greatest integer function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution.
f: R → R is given by,
f(x) = [x]
It is seen that /(1.2) = [1.2] = 1,
f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one-one.
Now, consider 0.7 ∈ R.
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.

Question 4.
Show that the modulus function f: R → R given by f(x) = |x|, is neither one-one nor onto, where x is x, if x is positive or 0 and |x| is – x, if x is negative.
Solution.
f: R → R is given by,
f(x) = |x| = {x, if x ≥ 0; – x if x < 0
It is seen that f(- 1) = |- 1| = 1, f(1) = |1| = 1.
∴ f(- 1) = f(1),but – 1 ≠ 1.
∴ f is not one-one.
Now, consider – 1 ∈ R.
It is known that f(x) = |x| is always non-negative,. Thus, there does not exist any element x in domain R such that f(x) = |x| = – 1.
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.

Question 5.
Show that the signum function f: R → R, given by

is neither one-one nor onto.
Solution.
f: R → R is given by,

It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴ f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or – 1) for the element – 2 in co-domain R, there does not exist any x in domain R such that f(x) = – 2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.

Question 6.
Let A = {1, 2, 3,}, B = {4 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Solution.
It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f(1) = 4, f(2) = 5, f(3) = 6
It is seen that the images of distinct elements of A under f are distinct. Hence, function f is one-one.

Question 7.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f: R → R defined as f(x) = 3 – 4x
(ii) f: R → R defined as f(x) = 1 + x³
Solution.
(i ) f: R → R is defined as f(x) = 3- 4x.
Let x₁, x₂ ∈ R such that f(x₁) = f(x₂)
⇒ 3 – 4x₁ = 3 – 4x₂
⇒ – 4x₁ = – 4x₁
⇒ x₁ = x₂
∴ f is one-one.
For any real number (y) in R, there exists \(\frac{3-y}{4}\) in R such that
f(\(\frac{3-y}{4}\)) = 3 – 4(\(\frac{3-y}{4}\)) = y
∴ f is onto.
Hence, f is bijective.

(ii) f: R → R is defined as f(x) = 1 + x²
Let x₁, x₂ ∈ R such that f(x₁) = f(x₂)
⇒ 1 + x₁² = 1 + x₂²
⇒ x₁² = ± x₂²
⇒ x₁ = x₂
⇒ f(x₁) = f(x₂) does not imply that x₁ = x₂.
For instance, f(1) = f(- 1) = 2
∴ f is not one-one.
Consider an element – 2 in co-domain R.
k is seen that f(x) = 1 + x² is positive for all x ∈ R.
Thus, there does not exist any x in domain R such that f(x) = – 2.
∴ f is not onto.
Hence, f is neither one-one nor onto.

Question 8.
Let A and B be sets. Show that f: A × B – B × A such that f (a, b) (b, a) is bijective function.
Solution.
f: A × B → B × A is defined as f(a, b) = (b, a).
Let(a₁, b₁), (a₂, b₂) ∈ A × B such that f(a₁, b₁) = f(a₂, b₂)
⇒ (b₁, a₁) = (b₂, a₂)
⇒ b₁ = b₂ and a₁ = a₂
⇒ (a₁, b₁) = (a₂, b₂)
∴ f is one – one.
Now, let (b, a) ∈ B × A be any element.
Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a). [by definition of f]
∴ f is onto.
Hence, f is bijective.

Question 9.
Let f: N → N be defined by

(n) =

State whether the function is bijective. Justify your answer.
Solution.

It can be observed that:
f(1) = \(\frac{1+1}{2}\) = 1 amnd f(2) = \(\frac{2}{2}\) = 1 [by definition of f]
∴ f(1) = f(2), where 1 ≠ 2.
∴ f is not one-one.
Consider a natural number (n) in co-domain N.

Case I: n is odd.
∴ n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1 ∈ N such that
f(4r + 1) = \(\frac{4 r+1+1}{2}\) = 2r+ 1

Case II : n is even,
∴ n – 2r for some r ∈ N. Then there exists 4r ∈ N such that 4r
f(4r) = \(\frac{4r}{2}\) = 2r.
∴ f is onto.
Hence, f is not a bijective function.

Question 10.
Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by f(x) = \(\left(\frac{x-2}{x-3}\right)\). Is f one-one and onto? Justify your answer.
Solution.
Here, A = R – {3}, B = R – {1}
and f: A → B is defined as f(x) = \(\left(\frac{x-2}{x-3}\right)\)
Let x, y ∈ A such that f(x) = f(y).
⇒ \(\frac{x-2}{x-3}=\frac{y-2}{y-3}\)
⇒ (x – 2) (y – 3) = (y – 2) (x – 3)
⇒ xy – 3x – 2y + 6 = xy – 3y – 2x + 6
⇒ – 3x – 2y = – 3y – 2x
⇒ 3x – 2x = 3y – 2y
⇒ x = y
∴ f is one-one.
Let y ∈ B = R – {1}. Then, y ≠ 1.
The function f is onto if there exists x ∈ A such that f(x) = y.
Now, f(x) = y
⇒ \(\frac{x-2}{x-3}\) = y
⇒ x – 2 = xy – 3y
⇒ x(1 – y) = – 3y + 2
⇒ x = \(\frac{2-3 y}{1-y}\) ∈ A

Thus, for any y B, there exists \(\frac{2-3 y}{1-y}\) ∈ A such that
f(\(\frac{2-3 y}{1-y}\)) = \(\frac{\left(\frac{2-3 y}{1-y}\right)-2}{\left(\frac{2-3 y}{1-y}\right)-3}\)

= \(\frac{2-3 y-2+2 y}{2-3 y-3+3 y}=\frac{-y}{-1}\) = y

∴ f is onto.
Hence, function f is one-one and onto.

Question 11.
Let f: R → R be defined as f(x) = x⁴. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Solution.
f : R → R is defined as f(x) = x4 Let x, yeR such that f(x) = f(y).
⇒ x⁴ = y⁴
⇒ x = ±y
∴ f(x₁) = f(x₂) does not imply that x₁ = x₂.
For instance,
f(1) = f(- 1) = 1
∴ f is not one-one.
Consider an element 2 in co-domain it. It is clear that there does not exist any x in domain R such that f(x) – 2 .
∴ f is not onto.
Hence, function f is neither one-one nor onto.
The correct answer is (D).

Question 12.
Let f: R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one not onto
(D) f is neither one-one nor onto
Solution.
f: R → R is defined as f(x) = 3x.
Let x, y ∈ R such that f(x) = f(y).
⇒ 3x – 3y
⇒ x = y .
∴ f is one-one.
Also any real number (y) in co-domain R, there exists \(\frac{y}{3}\) in R such that
f(\(\frac{y}{3}\)) = 3(\(\frac{y}{3}\)) = y
∴ f is onto.
Hence, function f is one-one and onto.
The correct answer is (A).