# NCERT Solutions for Class 12 Maths Exercise 1.3

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## NCERT Solutions for Class 12 Maths Exercise 1.3

### Relations and Functions Solutions

#### Q 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.

Answer)

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)}.

Answer)

To prove:

#### Q 3) Find gof and fog, if

(i) f(x) = | x | and g(x) = | 5x – 2 |

Answer)

#### Q 4) If , show that f o f(x) = x, for all . What is the inverse of f?

Answer) It is given that

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

#### Q 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)}

Answer)

(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.

#### Q 6) Show that f : [–1, 1] → R, given by f (x) = 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.,

Answer) 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

#### Q 7) Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Answer)

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

#### Q 8) Consider f: R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by, where R+ is the set of all non-negative real numbers.

Answer)

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

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

Answer)

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, ∞).

We now have:

#### Q 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) = IY(y) = fog2(y). Use one-one ness of f).

Answer)

Let fX → Y be an invertible function.

Also, suppose f has two inverses (say g1 and g2)
Then, for all y ∈Y, we have:

Hence, f has a unique inverse.

#### Q 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.

Answer)

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:

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:

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.

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

Answer)

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.

#### Q 13) If f: R → R be given by, then fof(x) is(A)(B) x3 (C) x (D) (3 − x3)

Answer) fR → R is given as

The correct answer is C.

#### Q 14) Let be a function defined as . The inverse of f is map g: Range Given by

Answer) It is given that

Let y be an arbitrary element of Range f.

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

The correct answer is B.

That’s it. These were the solutions of NCERT Class 12 Maths Exercise 1.3. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.