# RD Sharma Solutions for Class 12 Maths Exercise 2.2 Chapter 2 Function (Updated for 2021-22)

RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2: RD Sharma Solutions for Class 12, Maths Chapter 2, helps students who wish to achieve a good academic score in the board exam. RD Sharma Class 12 Maths Solutions are expertly designed to increase students’ confidence to understand the concepts covered in this chapter and how to solve problems in a short period of time.

## Download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2 PDF

RD Sharma Solutions for Class 12 Maths Chapter 2 Exercise 2.2

Our experts prepare these materials based on the Class 12 CBSE syllabus, keeping in mind the types of questions asked in the RD Sharma solution. Chapter 2 Functions explains the function and domains of functions and functions. It has four practices. Students can easily get answers to problems in RD Sharma Maths Solutions for Class 12 Chapter 2 Exercise 2.2 from this article.

### Exercise 2.2 Page No: 2.46

1. Find gof and fog when f: R → R and g : R → R is defined by

(i) f(x) = 2x + 3 and g(x) = x2 + 5.

(ii) f(x) = 2x + x2 and g(x) = x3

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

(iv) f (x) = x and g(x) = |x|

(v) f(x) = x2 + 2x − 3 and g(x) = 3x − 4

(vi) f(x) = 8x3 and g(x) = x1/3

Solution:

(i) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

Also given that f(x) = 2x + 3 and g(x) = x2 + 5

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

= g (2x +3)

= (2x + 3)2 + 5

= 4x2+ 9 + 12x +5

=4x2+ 12x + 14

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

= f (x2 + 5)

= 2 (x2 + 5) +3

= 2 x2+ 10 + 3

= 2x2 + 13

(ii) Given, f: R → R and g: R → R

so, gof: R → R and fog: R → R

f(x) = 2x + x2 and g(x) = x3

(gof) (x)= g (f (x))

= g (2x+x2)

= (2x+x2)3

Now, (fog) (x) = f (g (x))

= f (x3)

= 2 (x3) + (x3)2

= 2x+ x6

(iii) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

f(x) = x2 + 8 and g(x) = 3x3 + 1

(gof) (x) = g (f (x))

= g (x2 + 8)

= 3 (x2+8)3 + 1

Now, (fog) (x) = f (g (x))

= f (3x3 + 1)

= (3x3+1)2 + 8

= 9x6 + 6x+ 1 + 8

= 9x+ 6x+ 9

(iv) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

f(x) = x and g(x) = |x|

(gof) (x) = g (f (x))

= g (x)

= |x|

Now (fog) (x) = f (g (x))

= f (|x|)

= |x|

(v) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

f(x) = x2 + 2x − 3 and g(x) = 3x − 4

(gof) (x) = g (f(x))

= g (x+ 2x − 3)

= 3 (x+ 2x − 3) − 4

= 3x+ 6x − 9 − 4

= 3x+ 6x − 13

Now, (fog) (x) = f (g (x))

= f (3x − 4)

= (3x − 4)+ 2 (3x − 4) −3

= 9x+ 16 − 24x + 6x – 8 − 3

= 9x− 18x + 5

(vi) Given, f: R → R and g: R → R

So, gof: R → R and fog: R → R

f(x) = 8x3 and g(x) = x1/3

(gof) (x) = g (f (x))

= g (8x3)

= (8x3)1/3

= [(2x)3]1/3

= 2x

Now, (fog) (x) = f (g (x))

= f (x1/3)

= 8 (x1/3)3

= 8x

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

Solution:

Given f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}

f : {3, 9, 12} → {1, 3, 4} and g : {1, 3, 4, 5} → {3, 9}

Co-domain of f is a subset of the domain of g.

So, gof exists and gof: {3, 9, 12} → {3, 9}

(gof) (3) = g (f (3)) = g (1) = 3

(gof) (9) = g (f (9)) = g (3) = 3

(gof) (12) = g (f (12)) = g (4) = 9

⇒ gof = {(3, 3), (9, 3), (12, 9)}

Co-domain of g is a subset of the domain of f.

So, fog exists and fog: {1, 3, 4, 5} → {3, 9, 12}

(fog) (1) = f (g (1)) = f (3) = 1

(fog) (3) = f (g (3)) = f (3) = 1

(fog) (4) = f (g (4)) = f (9) = 3

(fog) (5) = f (g (5)) = f (9) = 3

⇒ fog = {(1, 1), (3, 1), (4, 3), (5, 3)}

3. Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.

Solution:

Given f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}

f: {1, 4, 9, 16} → {-1, -2, -3, 4} and g: {-1, -2, -3, 4} → {-2, -4, -6, 8}

Co-domain of f = domain of g

So, gof exists and gof: {1, 4, 9, 16} → {-2, -4, -6, 8}

(gof) (1) = g (f (1)) = g (−1) = −2

(gof) (4) = g (f (4)) = g (−2) = −4

(gof) (9) = g (f (9)) = g (−3) = −6

(gof) (16) = g (f (16)) = g (4) = 8

So, gof = {(1, −2), (4, −4), (9, −6), (16, 8)}

But the co-domain of g is not same as the domain of f.

So, fog does not exist.

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, c)}.
Show that f and g both are bijections and find fog and gof.

Solution:

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

Also given that A = {a, b, c}, B = {u, v, w}

Now we have to show f and g both are bijective.

Consider f = {(a, v), (b, u), (c, w)} and f: A → B

Injectivity of f: No two elements of A have the same image in B.

So, f is one-one.

Surjectivity of f: Co-domain of f = {u, v, w}

Range of f = {u, v, w}

Both are same.

So, f is onto.

Hence, f is a bijection.

Now consider g = {(u, b), (v, a), (w, c)} and g: B → A

Injectivity of g: No two elements of B have the same image in A.

So, g is one-one.

Surjectivity of g: Co-domain of g = {a, b, c}

Range of g = {a, b, c}

Both are the same.

So, g is onto.

Hence, g is a bijection.

Now we have to find fog,

we know that Co-domain of g is same as the domain of f.

So, fog exists and fog: {u v, w} → {u, v, w}

(fog) (u) = f (g (u)) = f (b) = u

(fog) (v) = f (g (v)) = f (a) = v

(fog) (w) = f (g (w)) = f (c) = w

So, fog = {(u, u), (v, v), (w, w)}

Now we have to find gof,

Co-domain of f is same as the domain of g.

So, fog exists and gof: {a, b, c} → {a, b, c}

(gof) (a) = g (f (a)) = g (v) = a

(gof) (b) = g (f (b)) = g (u) = b

(gof) (c) = g (f (c)) = g (w) = c

So, gof = {(a, a), (b, b), (c, c)}

5. Find fog (2) and gof (1) when f: R → R; f(x) = x2 + 8 and g: R → R; g(x) = 3x3 + 1.

Solution:

Given f: R → R; f(x) = x2 + 8 and g: R → R; g(x) = 3x3 + 1.

Consider (fog) (2) = f (g (2))

= f (3 × 2+ 1)

= f(3 × 8 + 1)

= f (25)

= 252 + 8

= 633

(gof) (1) = g (f (1))

= g (1+ 8)

= g (9)

= 3 × 9+ 1

= 2188

6. Let R+ be the set of all non-negative real numbers. If f: R+ → R+ and g : R+ → R+ are defined as f(x)=x2 and g(x)=+ √x, find fog and gof. Are they equal functions.

Solution:

Given f: R+ → R+ and g: R+ → R+

So, fog: R+ → R+ and gof: R+ → R+

Domains of fog and gof are the same.

Now we have to find fog and gof also we have to check whether they are equal or not,

Consider (fog) (x) = f (g (x))

= f (√x)

= √x2

= x

Now consider (gof) (x) = g (f (x))

= g (x2)

= √x2

= x

So, (fog) (x) = (gof) (x), ∀x ∈ R+

Hence, fog = gof

7. Let f: R → R and g: R → R be defined by f(x) = x2 and g(x) = x + 1. Show that fog ≠ gof.

Solution:

Given f: R → R and g: R → R.

So, the domains of f and g are the same.

Consider (fog) (x) = f (g (x))

= f (x + 1) = (x + 1)2

= x+ 1 + 2x

Again consider (gof) (x) = g (f (x))

= g (x2) = x+ 1

So, fog ≠ gof

## RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2: Important Topics

Let us have a look at some of the important concepts that are discussed in this chapter.

• Classification of functions
• Types of functions
• Constant function
• Identity function
• Modulus function
• Integer function
• Exponential function
• Logarithmic function
• Reciprocal function
• Square root function
• Operations on real functions
• Kinds of functions
• One-one function
• On-to function
• Many one function
• In to function
• Bijection
• Composition of functions
• Properties of the composition of functions
• Composition of real function
• Inverse of a function
• Inverse of an element
• Relation between graphs of a function and its inverse

We have provided complete details of RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.2. If you have any queries related to CBSE Class 12 Exam, feel free to ask us in the comment section below.

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