# NCERT solution class 12 chapter 1 Relations and Functions exercise 1.3 mathematics part 1

## EXERCISE 1.3

#### Question 1:

Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by = {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}. Write down gof.

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as

= {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}. #### Question 2:

Let fg and h be functions from to R. Show that To prove:     #### Question 3:

Find goand fog, if

(i) (ii) (i)  (ii)  #### Question 4:

If , show that f f(x) = x, for all . What is the inverse of f?

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

#### 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)}

(i) 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

is not one-one.

Hence, function does not have an inverse.

(ii) 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.

is not one-one,

Hence, function g does not have an inverse.

(iii) 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 is one-one. Find the inverse of the function f: [−1, 1] → Range f.

(Hint: For y ∈Range fy = , for some x in [−1, 1], i.e., )

f: [−1, 1] → R is given as Let f(x) = f(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 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: Now, let us define g: Range f → [−1, 1] as gof = and fo  f−1 = g

⇒ #### Question 7:

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

fR → R is given by,

f(x) = 4x + 3

One-one:

Let f(x) = f(y). ∴ f is a one-one function.

Onto:

For y ∈ R, let y = 4x + 3. Therefore, for any ∈ R, there exists such that ∴ f is onto.

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

Let us define gR→ R by .  Hence, f is invertible and the inverse of f is given by #### Question 8:

Consider fR→ [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given by , where R+ is the set of all non-negative real numbers.

fR+ → [4, ∞) is given as f(x) = x2 + 4.

One-one:

Let f(x) = f(y). ∴ f is a one-one function.

Onto:

For y ∈ [4, ∞), let y = x2 + 4. Therefore, for any ∈ R, there exists such that .

∴ f is onto.

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

Let us define g: [4, ∞) → Rby,   Hence, f is invertible and the inverse of f is given by #### Question 9:

Consider fR+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with .

fR+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.

Let y be an arbitrary element of [−5, ∞).

Let y = 9x2 + 6− 5. f is onto, thereby range f = [−5, ∞).

Let us define g: [−5, ∞) → R+ as We now have:   and Hence, f is invertible and the inverse of f is given by #### Question 10:

Let fX → 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) = IY(y) = fog2(y). Use one-one ness of f).

Let fX → Y be an invertible function.

Also, suppose f has two inverses (say ).

Then, for all y ∈Y, we have: Hence, f has a unique inverse.

#### Question 11:

Consider f: {1, 2, 3} → {abc} given by f(1) = af(2) = b and f(3) = c. Find f−1 and show that (f−1)−1 = f.

Function f: {1, 2, 3} → {abc} is given by,

f(1) = af(2) = b, and f(3) = c

If we define g: {abc} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:  and , where X = {1, 2, 3} and Y= {abc}.

Thus, the inverse of exists and f−1 = g.

f−1: {abc} → {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} → {abc} as

h(1) = ah(2) = bh(3) = c, then we have:  , where X = {1, 2, 3} and Y = {abc}.

Thus, the inverse of exists and g−1 = h ⇒ (f−1)−1 = h.

It can be noted that h = f.

Hence, (f−1)−1 = f.

#### Question 12:

Let fX → Y be an invertible function. Show that the inverse of f−1 is f, i.e.,

(f−1)−1 = f.

Let fX → Y be an invertible function.

Then, there exists a function gY → X such that gof = IXand fo= IY.

Here, f−1 = g.

Now, gof = IXand fo= IY

⇒ f−1of = IXand fof−1= IY

Hence, f−1Y → X is invertible and f is the inverse of f−1

i.e., (f−1)−1 = f.

#### Question 13:

If f→ be given by , then fof(x) is

(A) (B) x3 (C) x (D) (3 − x3)

fR → R is given as . #### Question 14:

Let be a function defined as . The inverse of f is map g: Range (A) (B) (C) (D) It is given that Let y be an arbitrary element of Range f.

Then, there exists x ∈ such that  Let us define g: Range as Now,    Thus, g is the inverse of f i.e., f−1 = g.

Hence, the inverse of f is the map g: Range , which is given by  