In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 Relations and Functions Ex 1.3 Pdf, These solutions are solved by subject experts from the latest MP Board books.

## MP Board Class 12th Maths Solutions 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:

gof = {(1, 3),(3, 1), (4, 3)}

Question 2.

Let f, g and h be functions from R to R. Show that

i. (f + g)oh = foh + goh

ii. (f. g)oh = (foh) . (goh)

Solution:

i. (f + g)oh(x) = (f + g)h(x)) = f(h(x)) + g(h(x)) = (foh)(x) + (goh)(x)

∴ (f + g)oh = foh + goh

ii. (f. g)oh(x) = (f.g)(h(x))

= f(h(x)).g(h(x))

= (foh)(x). (goh)(x)

= [(foh).(goh)](x)

∴ (f. g)oh = (foh) . (goh)

Question 3.

Find gof and fog, if

i. f(x) = |x| and g(x) = |5x – 2|

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

(March 2014, SAY 2015, March 2016)

Solution:

i. (gof)(x) = g(f(x)) = g(|x|) = |5|x| – 2|

(fog)(x) = f(g(x)) = f(|5x – 2|)

= ||5x – 2|| = |5x – 2|

ii. (gof)(x) = g(f(x)) = g(8x³) = (8x³ )A^{1/3} = 2x

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

Question 4.

If f(x) = \(\frac{(4 x+3)}{(6 x-4)}\), x ≠ 3, show that fof (x) = x, for all x ≠ \(\frac{2}{3}\). What is the inverse of f?

Solution:

Another method:

Without using the result fof(x) = x, we can directly find the inverse off.

Question 5.

State with reason whether following functions have inverse

i. f : {1,2, 3,4} → {10} with

f = {(1, 10), (2, 10), (3, 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. The elements 1,2,3 and 4 have the same image.

∴ g is not one-one.

Hence g has no inverse.

ii. The image of the elements 5 and 7 is the same

∴ g is not one-one

Hence g has no inverse.

iii.

From the arrow diagram, it is clear that h is one-one and onto.

Hence h has 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.

Solution:

Let x_{1}, x_{2} ∈ [-1, 1]

f(x_{1}) = f(x_{2}) ⇒ \(\frac{x_{1}}{2+x_{1}}\) = \(\frac{x_{2}}{2+x_{2}}\)

⇒ x_{1}(2 + x_{2}) = x_{2}(2 + x_{1})

⇒ 2x_{1} + x_{1}x_{2} = 2x_{2} + x_{1}x_{2}

⇒ 2x_{1} = 2x_{2}

⇒ x_{1} = x_{2}

∴ f is one-one

Let y = \(\frac{x}{x+2}\) ⇒ y(x + 2)= x

⇒ xy – x = – 2y

⇒ x (1 – y) = 2y

⇒ x = \(\frac{2y}{1-y}\), y ≠ 1

∴ f^{-1}(x) = \(\frac{2x}{1-x}\)

Question 7.

Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Solution:

f(x) = 4x + 3, x ∈ R

Domain of f = R and range of f = R

One-one

Let x_{1}, x_{2} ∈ R

f(x_{1}) = f(x_{2}) ⇒ 4x_{1} + 3 = 4x_{2} + 3

⇒ 4x_{1} = 4x_{2}

⇒ x_{1} = x_{2}

∴ f is one-one.

Onto

Let y ∈ range of f such that y = f(x)

i.e., corresponding to every y ∈ R, there exists a real number \(\frac{y-3}{4}\) such that f(\(\frac{y-3}{4}\)) = y

∴ f is onto

Hence f is a bijection and f^{-1} exists

To find f^{-1}

Let y = f(x)

Then x = \(\frac{y-3}{4}\) from (1)

∴ Inverse of/is given by f^{-1}(x) = \(\frac{x-3}{4}\)

Question 8.

Consider f : R_{+} → [4, ∞) given by f(x) = x^{2} + 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:

Let x_{1}, x_{2} ∈ R_{+} such that f(x_{1}) = f(x_{2})

⇒ x²_{1} + 4 = x²_{2} + 4

⇒ x²_{1} = x²_{2} ⇒ (x²_{1} – x²_{2}) = 0

⇒ (x_{1} – x_{2}) (x_{1} + x_{2}) = 0

⇒ x_{1} – x_{2} = 0 since x_{1}, x_{2} ∈ R

⇒ x_{1} = x_{2}

∴ f is one-one.

Let y ∈ [4, ∞) such that y = f(x)

⇒ y = x^{2} + 4 ⇒ x^{2} = y – 4

⇒ x = \(\sqrt{y – 4}\) since x ∈ R_{+}

i.e., x = \(\sqrt{y – 4}\) ∈ R_{+}+ …. (1)

f(x) = f(\(\sqrt{y – 4}\)) = (\(\sqrt{y – 4}\))² + 4

= y – 4 + 4 = y

∴ f is onto

Since f is one-one and onto, f is invertible.

To find f^{-1}

Let y = x^{2} + 4

⇒ x = \(\sqrt{y – 4}\) from(1)

∴ f^{-1}(y) = \(\sqrt{y – 4}\)

Question 9.

Consider f : R_{+} → [- 5, ∞) given by f(x) = 9x^{2} + 6x – 5. Show that f is invertible with f^{-1}(y) = \(\left(\frac{(\sqrt{y+6})-1}{3}\right)\)

Solution:

Let y = 9x^{2} + 6x – 5 = (3x + 1)²

⇒ (3x + 1)² = y + 6

⇒ 3x + 1 = \(\sqrt{y + 6}\) ⇒ 3x = \(\sqrt{y + 6}\) – 1

⇒ x = \(\frac{\sqrt{y+6}-1}{3}\)

Hence gof = fog = Identity function. Hence f is invertible and

f^{-1}(y) = \(\left(\frac{(\sqrt{y+6})-1}{3}\right)\)

Question 10.

Let f : X → Y be an invertible function. Show that f has unique inverse.

Solution:

Let g_{1} and g_{2} be two inverses of f.

Then for all y ∈ Y

(fog_{1})y = y and (fog_{2})y = y

⇒ (fog_{1})y = (fog_{2})(y)

⇒ f(g_{1}(y)) = f(g_{2}(y))

⇒ g_{1}(y) = g_{2}(y) since f is one-one

⇒ g_{1} = g_{2}

Hence the inverse of f is unique.

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:

f = {(1, a), (2, b), (3, c)}

f^{-1} = {(a, 1), (b, 2), (c, 3))

Clearly f is bijective

(f^{-1})^{-1} = {(1, a), (2, b), (3, c)} = f

Question 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:

f : X → Y is invertible. Then f^{-1} : Y → X

Let x ∈ X , then y = f(x) ∈ Y such that f^{-1} (y) = x

Now (f^{-1}of)(x) = f^{-1}(f(x)) = f^{-1}(y) = x

and (fof^{-1})(y) = f(f^{-1}(y)) = f(x) = y

i.e., f^{-1} of and fof^{-1} are identity functions.

Hence they are inverse of each other.

i.e., the inverse of f^{-1} is f.

i.e., (f^{-1})^{-1} = f.

Question 13.

If f : R → R be given by f(x) = (3 – x^{3})^{\(\frac { 1 }{ 3 }\)}, then fof (x) is

a. x^{\(\frac { 1 }{ 3 }\)}

b. x³

c. x

d. (3 – x³)

Solution:

c. x

Question 14.

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

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

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

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

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

Solution:

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

Let \(\frac { 4x }{ 3x+4 }\) = y ⇒ 4x = 3xy + 4y

⇒ 4x – 3xy = 4y ⇒ x = \(\frac { 4y }{ 4-3y }\)

∴ g(y) = \(\frac { 4y }{ 4-3y }\).