Class 12 RD Sharma Solutions - Chapter 2 Functions - Exercise 2.2

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) = x^1/3

Solution:

f(x) = 8x³ and g(x) = x^1/3

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(x^1/3)

= 8(x^1/3)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 = I_R.

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 = I_R --->(i)

Again,

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

= x = I_R --->(ii)

from i and ii

f o g = g o f = I_R

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) = e^x

Solution:

We have f: N -> Z₀, g: Z₀ ->Q and

h: Q -> R

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

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)

= e^1/2x

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

= e^1/2x

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

Summary:

Chapter 2, Exercise 2.2 of RD Sharma's Class 12 Mathematics textbook typically focuses on:

1. Algebra of functions (addition, subtraction, multiplication, division)

2. Composite functions and their properties

3. Identities and constant functions

4. Piecewise functions

5. Evaluation of functions at specific points

6. Domain and range of combined functions

Practice Questions:

1. If f(x) = x² - 3x + 2 and g(x) = 2x + 1, find (f + g)(x) and (f · g)(x).

2. Given f(x) = √x and g(x) = x + 4, find the domain of (f ∘ g)(x).

3. Let f(x) = |x - 1| and g(x) = x². Evaluate (g ∘ f)(2) and (f ∘ g)(2).

4. Define f(x) = {

x² if x < 0

x + 1 if x ≥ 0

Find f(-2), f(0), and f(3).

5. If f(x) = 1/(x - 2) and g(x) = x + 1, find (f/g)(x) and state its domain.

6. Prove that (f + g) ∘ h = (f ∘ h) + (g ∘ h) for any functions f, g, and h.

7. Given f(x) = x³ and g(x) = x + 2, find a function h such that f ∘ h = g.

8. Let f(x) = x² - 4 and g(x) = √x. Find the range of (g ∘ f)(x).

9. If f(x) = 2x + 3 and (f ∘ g)(x) = 2x² + 2x + 3, determine g(x).

10. Define f(x) = {

x if x is rational

1 - x if x is irrational

Is f continuous at x = 0? Justify your answer.