# Class 12 RD Sharma Solutions – Chapter 2 Functions – Exercise 2.2

• Last Updated : 02 Feb, 2021

### Question 1(i). Find g o f and f o g when f: R -> R and g: R -> R is defined by f(x) = 2x + 3 and g(x) = x2 + 5

Solution:

f: R -> R and g: R -> R

Therefore, f o g: R -> R and g o f: R -> R

Now, f(x) = 2x + 3 and g(x) = x2 + 5

g o f(x) = g(2x + 3) =(2x + 3)2 + 5

=> g o f(x) = 4x2 + 12x + 14

f o g(x) = f(g(x)) = f(x2 + 5) = 2(x2 + 5) + 3

=> f o g(x) = 2x2 + 13

### Question 1(ii). Find g o f and f o g when f: R -> R and g: R -> R is defined by f(x) = 2x + x2 and g(x) = x3

Solution:

f(x) = 2x + x2 and g(x) = x3

g o f(x) = g(f(x)) = g(2x + x2)

g o f(x) =(2x + x2)3

f o g(x) = f(g(x)) = f(x3)

f o g(x) = 2x3 + x6

### Question 1(iii). Find g o f and f o g when f: R -> R and g: R -> R is defined by f(x) = x2 + 8 and g(x) = 3x3 + 1

Solution:

f(x) = x2 + 8 and g(x) = 3x3 + 1

Thus, g o f(x) = g [f(x)]

=> g o f(x) = g [x2 + 8]

=> g o f(x) = 3 [x2 + 8]3 + 1

Similarly, f o g(x) = f [g(x)]

=> f o g(x) = f [3x3 + 1]

=> f o g(x) = [3x3 + 1]2 + 8

=> f o g(x) = [9x6 + 1 + 6x3] + 8

=> f o g(x) = 9x6 + 6x3 + 9

### Question 1(iv). Find g o f and f o g when f: R -> R and g: R -> R is defined by f(x) = x and g(x) = | x |

Solution:

f(x) = x and g(x) = | x |

Now, g o f = g(f(x)) = g(x)

g o f(x) = | x |

g o f(x) = | x |

and, f o g(x) = f(g(x)) = f(x)

f o g(x) = | x |

### Question 1(v). Find g o f and f o g when f: R -> R and g: R -> R is defined by f(x) = x2 + 2x – 3 and g(x) = 3x – 4

Solution:

f(x) = x2 + 2x – 3 and g(x) = 3x – 4

Now, g o f(x) = g(f(x)) = g(x2 + 2x – 3)

g o f(x) = 3(x2 + 2x – 3) -4

g o f(x) = 3x2 + 6x – 13

and, f o g(x) = f(g(x)) = f(3x – 4)

f o g(x) =(3x – 4)2 + 2(3x – 4) – 3

= gx2 + 16 – 24x + 6x – 8 – 3

f o g(x) = 9x2 – 18x + 5

### Question 1(vi). Find g o f and f o g when f: R -> R and g: R -> R is defined by f(x) = 8x3 and g(x) = x1/3

Solution:

f(x) = 8x3 and g(x) = x1/3

Now, g o f(x) = g(f(x)) = g(8x3)

=(8x,3),1/3

g o f(x) = 2x

and, f o g(x) = f(g(x)) = f(x1/3)

= 8(x1/3)3

f o g(x) = 8x

### Question 2. Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3), (4, 9), (5, 9)} Show that g o f and f o g are both defined. Also, find f o g and g o f.

Solution:

Let f = {(3, 1),(9, 3),(12, 4)} and

g = {(1, 3),(3, 3),(4, 9),(5, 9)

Now,

range of f = {1, 3, 4}

domain of f = {3, 9, 12}

range of g = {3, 9}

domain of g = {1, 3, 4, 5}

since, range of f ⊂ domain of g

Therefore g o f in well defined.

range of g ⊂ domain of g

g o f in well defined.

Now g o f = {(3, 3),(9, 3),(12, 9)}

f o g = {(1, 1),(3, 1),(4, 3),(5, 3)}

### Question 3. Let f = {(1, -1), (4, -2), (9, -3), (16, 4)} and g = {(-1, -2), (-2, -4), (-3, -6), (4, 8)} Show that g o f is defined while f o g is not defined. Also, find g o f.

Solution:

We have,

f = {(1, -1),(4, -2),(9, -3),(16, 4)} and

g = {(-1, -2),(-2, -4),(-3, -6),(4, 8)}

Now,

Domain of f = {1, 4, 9, 16}

Range of f = {-1, -2, -3, 4}

Domain of g = {-1, -2, -3, 4}

Range of g = {-2, -4, -6, 8}

Clearly range of f = domain of g

Therefore, g o f is defined.

but, range of g != domain of f

Therefore, f o g is not defined.

Now,

g o f(1) = g(-1) = -2

g o f(4) = g(-2) = -4

g o f(g) = g(-3) = -6

g o f(16) = g(4) = 8

Therefore, g o f = {(1, -2),(4, -4),(9, -6),(16, 8)}

### Question 4. Let A = {a, b, c}, B = {u, v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as: f = {(a, v), (b, u), (c, w)},g = {(u, b), (v, a), (w, x)}. Show that f and g both are bijections and find f o g and g o f.

Solution:

A = {a, b, c}, B = {u, v, w} and

f = A -> B and g: B -> A defined by

f = {(a, v),(b, u),(c, w)} and

g = {(u, b) .(v, a),(w, c)}

For both f and g, different elements of domain have different images

Therefore, f and g are one-one

Again for each element in co-domain of f and g, there in a preimage in domain

Therefore, f and g are onto

So f and f are bijectives,

Now,

g o f = {(a, a),(b, b),(c, c)} and

f o g = {(u, u),(v, v),(w, w)}

### Question 5. Find f o g(2) and g o f(1) when: f: R -> R ; f(x) = x2 + 8 and g: R -> R ; g(x) = 3 x3 + 1.

Solution:

We have,

f: R -> R given by f(x) = x2 + 8 and

g: R -> R given by g(x) = 3x3 + 1

Therefore,

f o g(x) = f(g(x)) = f(3x3 + 1)

=(3x3 + 1)2 + 8

Therefore, f o g(2) =(3 * 8 + 1)2 + 8 = 625 + 8 = 633

Again

g o f(x) = g(f(x)) = g(x2 + 8)

= 3(x2 + 8)3 + 1

g o f(1) = 3(1 + 8)3 + 1 = 2188

### Question 6. Let R+ be the set of all non-negative real numbers . if: R+ -> R+ and g: R+ -> R+ are defined as f(x) = x2 and g(x) = +√x, find f o g and g o f . Are they equal functions ?

Solution:

We have, f: R+ -> R+ given by

f(x) = x2

g: R+ -> R+ given by

g(x) = √x

f o g(x) = f(g(x)) = f(√x) =(√x)2 = x

Also,

g o f(x) = g(f(x)) = g(x2 ) = √x2 = x

Thus,

f o g(x) = g o f(x)

### Question 7. Let f: R -> R and g: R -> R be defined by f(x) = x2 and g(x) = x + 1. Show that f o g != g o f.

Solution:

We have f: R -> R and g: R -> R are two functions defined by f(x) = x2 and g(x) = x + 1

Now,

f o g(x) = f(g(x)) = f(x + 1) =(x + 1)2

f o g(x) = x2 + 2x + 1 —> eq(i)

g o f(x) = g(f(x)) = g(f(x)) = g(x,2) = x2 + 1 —->(ii)

from(i),(ii)

f o g != g o f

### Question 8. Let f: R -> R and g: R -> R be defined by f(x) = x + 1 and g(x) = x – 1. Show that f o g = g o f = IR.

Solution:

Let f: R -> R and g: R -> R are defined as .

f(x) = x +1 and g(x) = x – 1

Now,

f o g(x) = f(g(x)) = f(x – 1) = x – 1 + 1

= x = IR —>(i)

Again,

f o g(x) = f(g(x)) = g(x + 1) = x + 1 – 1

= x = IR —>(ii)

from i and ii

f o g = g o f = IR

### Question 9. Verify associativity for the following three mappings: f: N -> Z0(the set of non-zero integers), g: Z0 -> Q and h: Q -> R given by f(x) = 2x, g(x) = 1 / x and h(x) = ex

Solution:

We have f: N -> Z0, g: Z0 ->Q and

h: Q -> R

Also, f(x) = 2x, g(x) = 1 / x and h(x) = ex

Now, f: N -> Z0 and h o g: Z0 -> R

(h og) of: N -> R

also, g o f: N -> Q and h: Q -> R

h o(g o f): N -> R

Thus,(h o g) o f and h o(g o f) exist and are function from N to set R.

Finally,(h o g) o f(x) =(h o g)(f(x)) =(h o g)(2x)

= h(1 / 2x)

= e1/2x

Now, h o(g o f)(x) = h o(g(2x)) = h(1 / 2x)

= e1/2x

Associativity verified.

### Question 10. Consider f: N -> N, g: N -> N and h:N -> R as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z for all x, y, z ∈ N . Show that h o(g o f) =(h o g) o f .

Solution:

We have,

h o(g o f)(x) = h(g o f(x)) = h(g(f(x)))

= h(g(2x)) h(3(2x) + 4)

= h(6x + 4) = sin(6x + 4) x ∈ N

((h o g) o f)(x) =(h o g)(f(x)) =(h o g)(2x)

= h(g(2x)) == h(3(2x) + 4)

= h(6x + 4) = sin(6x + 4) x ∈ N

This shows, h o(g o f) =(h o g) o f

### Question 11. Give examples of two functions f: N -> N and g: N -> N, such that g o f is onto but f is not onto.

Solution:

Define f: N -> N by, f(x) = x + 1

And, g: N -> N by,

g(x) = x – 1 if x > 1

1 if x = 1

first show that f is not onto.

for this, consider element 1 in co – domain N . It is clear that this element is not an image of any of the elements in domain N .

Therefore, f is not onto.

Now, g o f: N -> N is defined.

### Question 12. Give examples of two functions f: N -> Z and g: Z -> Z, such that g o f is injective but g is not injective.

Solution:

Define f: N -> Z as f(x) = x and g: z -> z as g(x) = | x | .

We first show that g is not injective.

It can be observed that:

g(-1) = | -1 | = 1

g(1) = | 1 | = 1

Therefore, g(-1) = g(1), but -1 != 1 .

Therefore, g is not injective.

Now, g o f: N -> Z is defined as g o f(x) = g(f(x)) = g(x) = | x |.

let x, y E N such that g o f(x) = g o f(y) .

=> | x | = | y |

Since x and y ∈ N, both are positive.

| x | = | y | => x = y

Hence, g o f is injective

### Question 13. If f: A -> B and g: B -> C are one-one functions, show that g o f is a one – one function.

Solution:

We have f: A -> B and g: B -> C are one – one functions

Now we have to prove: g o f: A -> C in one – one

let x, y ∈ A such that

g o f(x) = g o f(y)

=> g(f(x)) = g(f(y))

=> f(x) = f(y)

x = y

Therefore, g o f is one – one function.

### Question 14. If f: A -> B and g: B -> C are onto functions, show that g o f is a onto function.

Solution:

We have, f: A -> B and g: B -> C are onto functions

Now, we need to prove: g o f: A -> C in onto

let y ∈ C, then,

g o f(x) = y

g(f(x)) = y –>(i)

Since g is onto, for each element in c, then exists a preimage in B.

g(x) = y —->(ii)

from(i) and(ii)

f(x) = ∞

Since f is onto, for each element in B there exists a preimage in A

f(x) = ∞ —>(iii)

From(ii) and(iii) we can conclude that for each y ∈ c there exists a pre image in A

Such that g o f(x) = y

Therefore, g o f is onto

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