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 = IRange 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 = IR
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 = IR+
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 = IR 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) = Iy(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 = IX and fog = IY, 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 = IX and hog = IY, 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 = IX and fog – IY.
Here, f-1 = g.
Now, gof = IX and fog = IY
⇒ f-1 = IX and fof-1 = IY
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).