NCERT Solution Class 12th Maths chapter – 1 Relations and Functions Exercise 1.3

June 17, 2022

NCERT Solutions Class 12th Maths Chapter – 1 Relations and Functions Exercise 1.3

Textbook	NCERT
class	Class – 12th
Subject	Mathematics
Chapter	Chapter – 1
Chapter Name	Relations and Functions
grade	Class 12th Maths solution
Medium	English
Source	last doubt

NCERT Solutions Class 12th Maths Chapter – 1 Relations and Functions Exercise 1.3

?Chapter – 1?

✍ Relations and Functions✍

?Exercise – 1.3?

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) }

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 = (foh) .(goh)

3. (i) Find gof and fog, if f(x) = |x| and g(x) = |5x – 2|

Solution:

f(x)=|x| and g(x)=|5x-2|

∴ (gof) (x) = g g(f(x)) = g(|x|) = |5|x|-2|

(fog) (x) = F(g(x))= f |5x – 2|) = ||5x -2|=|5x-2

(ii) Find gof and fog, if f(x)=8x³ and g (x) = x^1/3

Solution:

f (x) = 8x³and g(x) = x^1/2

∴ (gof) (x) = g(f(x)) = g(8x³) = g(8x³)^{1/3 _{= 2x}}

(fog) (x) = f(g(x)) = f(x^1/3)= 8(x^1/3)^{3 _{= 8x}}

4. if f(x) = 4x+3/6x-4,x≠ 2/3 show that fof(x) = x, for all x ≠ 2/3 . What is the inverse of f?

Solution:

t is given that f(x)=4x+3/6x-4,x≠2/3

(fof) (x) = f(f(x))=f(4x+3/6x-4)

=4(4x+3/6x-4)+3/6(4x+36x-4)-4 = 16x+12+18x-12/24x+18-24x+16 = 34x/34 = x

Therefore fof(x) = x for all x≠2/3

=> fof = 1

Hence, the given function f is invertible and the inverse of f is f itself.

5. (i) State with reason whether following functions have inverse
f: {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}

Solution:

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) State with reason whether following functions have inverse g: {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}

Solution:

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) State with reason whether following functions have inverse h: {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Solution:

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.

6. Show that f: [−1, 1] → R, given by f(x) = x/x+2 is one-one. Find the inverse of the function f: [−1, 1] → Range f. (Hint: For y in Range f, y = f(x)=x/x+2 for some x in [-1, 1] ie x = 2 y/1-y

Solution:

`f: [−1, 1] → R is given as f(x)=x/x+2

Let f(x) = f(y).

⇒x/x+2 = y/y+2

=> xy + 2x = xy + 2y

=> 2x = 2y

=> x = y

∴ f is a 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=x/x+2

=> xy + 2y = x

=> x(1-y)= 2y

Relations and Functions — **7. Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.**

Solution:

f: R → R is given by,

f(x) = 4x + 3

One-one:

Let f(x) = f(y).

=> 4x + 3 = 4y + 3

=> 4x = 4y

=> x = y

∴ f is a one-one function.

Onto:

For y ∈ R, let y = 4x + 3.

⇒x = y-3/4 ∈R

Therefore, for any y ∈ R, there exists x = y-3/4 ∈R such that

f(x) = f(y-3/4) = 4(y-3/4)+3 = y

∴ f is onto.

Thus, f is one-one and onto and therefore, f⁻¹ exists.

Let us define g: R→ R by g(x)=y-3/4

Now, (gof) (x) = g(f(x))=g(4x+3)=(4x+3)-3/4=x

(fog)(y) = f(g(y)) = f(y-3/4) = 4(y-3/4)+3 = y-3+3 = y

∴gof=fog=IR

Hence, f is invertible and the inverse of f is given by

f-1=g(y)=y-3/4

Solution:

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 a one-one function.

Onto:

For y ∈ [4, ∞), let y = x² + 4.

⇒x²= y-4 ≥ 0 [as y ≥4]

⇒x = √y-4 ≥ 0

Therefore, for any y ∈ R, there exists x = √y-4 ∈R such that

f(x) = f(√y-4) = (√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,

Relations and Functions

9. Consider f: R₊ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with f ^{-1 _{(y) ( √y+6 – 1/3)}}

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 = √y+6 [as y≥5⇒y+6>0]

⇒x = √y+6-1/3

∴f is onto, thereby range f = [−5, ∞).

Let us define g: [−5, ∞) → R₊ as g(y) = √y+6-1/3

We now have:

(gof) (x) = g(f(x))=g(9x²+6x-5)

=g((3x+1)²-6)

Relations and Functions

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) = I_Y(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)) [f is invertible => f is one-one]

⇒g₁= g₂ [g is one- one]

Hence, f has a unique inverse.

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 have:

(fog)(a) = f(g(a)) = f(1) = a

(fog)(b) = f(g(b)) = f(2) = b

(fog) (c) = f(g(c)) = f(3) = c

And

(gof)(1) = g(f(1)) = g(a) = 1

(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(1)) = 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.

12. Let f: X → Y be an invertible function. Show that the inverse of f⁻¹ is f, i.e., (f⁻¹)⁻¹ = f.

Solution:

Let f: X → Y be an invertible function.

Then, there exists a function g: Y → X such that gof = I_Xand fog = I_Y.

Here, f⁻¹ = g.

Now, gof = I_Xand fog = I_Y

⇒ f⁻¹of = I_Xand 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.

13. If f: R → R be given by f(x)=(3-x3)^1/3 , then fof (x) is
(A) 1/_x3
(B) x3
(C) x
(D) (3 − x3)

Solution:

Relations and Functions

14. Let f:R -{-43}→R be a function defined as f(x)=4x/3x+4. The inverse of f is map g Range f→R-{-4/3}
(A) g(y)=3y/3-4y
(B) g(y)=4y/4-3y
(C) g(y)=4y/3-4y
(D) g(y)= 3y/4-3y

Solution:

t is given that f: R -{-4/3}→R is defined as f(x)=4x/3x-4

Let y be an arbitrary element of Range f.

Then, there exists x ∈ R-{-4/3} such that y=f(x)

⇒y=4x/3x+ 4

=>3xy + 4y = 4x Let us define g: Range asf→R-{-4/3} as g(y)=4y/4-3y

=> x(4 – 3y) = 4y

Relations and Functions

:. gof=I_R-(-4/3) and fog = I_” Range f”

Thus, g is the inverse of f i.e., f⁻¹ = g.

Hence, the inverse of f is the map `g: Range f -> R – {-4/3}` which is given by g(y)=4y/4-3y

The correct answer is B.