# RD Sharma Solutions for Class 12 Maths Exercise 2.3 Chapter 2 Function (Updated for 2021-22)

RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3: In Chapter 2, Exercise 2.3 RD Sharma Class 12 Maths Solutions, you will solve problems concerning the composition of real functions. Class 12 is a turning point in a student’s life. It mainly prepares students to make important decisions about their future education and aspirations. You can refer to RD Sharma Solutions Class 12 Maths Chapter 2 Functions Exercise 2.3 free PDF for a better understanding of the topics taught in this exercise.

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## Download RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3 PDF

RD Sharma Solutions for Class 12 Maths Chapter 2 Exercise 2.3

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#### Exercise 2.3 Page No: 2.54

1. Find fog and gof, if

(i) f (x) = ex, g (x) = loge x

(ii) f (x) = x2, g (x) = cos x

(iii) f (x) = |x|, g (x) = sin x

(iv) f (x) = x+1, g(x) = ex

(v) f (x) = sin−1 x, g(x) = x2

(vi) f (x) = x+1, g (x) = sin x

(vii) f(x)= x + 1, g (x) = 2x + 3

(viii) f(x) = c, c ∈ R, g(x) = sin x2

(ix) f(x) = x2 + 2 , g (x) = 1 − 1/ (1-x)

Solution:

(i) Given f (x) = ex, g(x) = loge x

Let f: R → (0, ∞); and g: (0, ∞) → R

Now we have to calculate fog,

Clearly, the range of g is a subset of the domain of f.

fog: ( 0, ∞) → R

(fog) (x) = f (g (x))

= f (loge x)

= loge ex

= x

Now we have to calculate gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→ R

(gof) (x) = g (f (x))

= g (ex)

= loge ex

= x

(ii) f (x) = x2, g(x) = cos x

f: R→ [0, ∞) ; g: R→[−1, 1]

Now we have to calculate fog,

Clearly, the range of g is not a subset of the domain of f.

⇒ Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}

⇒ Domain (fog) = x: x ∈ R and cos x ∈ R}

⇒ Domain of (fog) = R

(fog): R→ R

(fog) (x) = f (g (x))

= f (cos x)

= cos2 x

Now we have to calculate gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→R

(gof) (x) = g (f (x))

= g (x2)

= cos x2

(iii) Given f (x) = |x|, g(x) = sin x

f: R → (0, ∞) ; g : R→[−1, 1]

Now we have to calculate fog,

Clearly, the range of g is a subset of the domain of f.

⇒ fog: R→R

(fog) (x) = f (g (x))

= f (sin x)

= |sin x|

Now we have to calculate gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog : R→ R

(gof) (x) = g (f (x))

= g (|x|)

= sin |x|

(iv) Given f (x) = x + 1, g(x) = ex

f: R→R ; g: R → [ 1, ∞)

Now we have calculate fog:

Clearly, range of g is a subset of domain of f.

⇒ fog: R→R

(fog) (x) = f (g (x))

= f (ex)

= e+ 1

Now we have to compute gof,

Clearly, range of f is a subset of domain of g.

⇒ fog: R→R

(gof) (x) = g (f (x))

= g (x+1)

= ex+1

(v) Given f (x) = sin −1 x, g(x) = x2

f: [−1,1]→ [(-π)/2 ,π/2]; g : R → [0, ∞)

Now we have to compute fog:

Clearly, the range of g is not a subset of the domain of f.

Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}

Domain (fog) = {x: x ∈ R and x2 ∈ [−1, 1]}

Domain (fog) = {x: x ∈ R and x ∈ [−1, 1]}

Domain of (fog) = [−1, 1]

fog: [−1,1] → R

(fog) (x) = f (g (x))

= f (x2)

= sin−1 (x2)

Now we have to compute gof:

Clearly, the range of f is a subset of the domain of g.

fog: [−1, 1] → R

(gof) (x) = g (f (x))

= g (sin−1 x)

= (sin−1 x)2

(vi) Given f(x) = x+1, g(x) = sin x

f: R→R ; g: R→[−1, 1]

Now we have to compute fog

Clearly, the range of g is a subset of the domain of f.

Set of the domain of f.

⇒ fog: R→ R

(fog) (x) = f (g (x))

= f (sin x)

= sin x + 1

Now we have to compute gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R → R

(gof) (x) = g (f (x))

= g (x+1)

= sin (x+1)

(vii) Given f (x) = x+1, g (x) = 2x + 3

f: R→R ; g: R → R

Now we have to compute fog

Clearly, the range of g is a subset of the domain of f.

⇒ fog: R→ R

(fog) (x) = f (g (x))

= f (2x+3)

= 2x + 3 + 1

= 2x + 4

Now we have to compute gof

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R → R

(gof) (x) = g (f (x))

= g (x+1)

= 2 (x + 1) + 3

= 2x + 5

(viii) Given f (x) = c, g (x) = sin x2

f: R → {c} ; g: R→ [ 0, 1 ]

Now we have to compute fog

Clearly, the range of g is a subset of the domain of f.

fog: R→R

(fog) (x) = f (g (x))

= f (sin x2)

= c

Now we have to compute gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→ R

(gof) (x) = g (f (x))

= g (c)

= sin c2

(ix) Given f (x) = x2+ 2 and g (x) = 1 – 1 / (1 – x)

f: R → [ 2, ∞ )

For domain of g: 1− x ≠ 0

⇒ x ≠ 1

⇒ Domain of g = R − {1}

g (x )= 1 – [1/(1 – x)] = (1 – x – 1)/ (1 – x) = -x/(1 – x)

For range of g

y = (- x)/ (1 – x)

⇒ y – x y = − x

⇒ y = x y − x

⇒ y = x (y−1)

⇒ x = y/(y – 1)

Range of g = R − {1}

So, g: R − {1} → R − {1}

Now we have to compute fog

Clearly, the range of g is a subset of the domain of f.

⇒ fog: R − {1} → R

(fog) (x) = f (g (x))

= f (-x/ (1 – x))

= ((-x)/ (1 – x))2 + 2

= (x2 + 2x2 + 2 – 4x) / (1 – x)2

= (3x– 4x + 2)/ (1 – x)2

Now we have to compute gof

Clearly, the range of f is a subset of the domain of g.

⇒ gof: R→R

(gof) (x) = g (f (x))

= g (x2 + 2)

= 1 – 1 / (1 – (x2 + 2))

= – 1/ (1 – (x2 + 2))

= (x2 + 2)/ (x2 + 1)

2. Let f(x) = x2 + x + 1 and g(x) = sin x. Show that fog ≠ gof.

Solution:

Given f(x) = x2 + x + 1 and g(x) = sin x

Now we have to prove fog ≠ gof

(fog) (x) = f (g (x))

= f (sin x)

= sinx + sin x + 1

And (gof) (x) = g (f (x))

= g (x2+ x + 1)

= sin (x2+ x + 1)

So, fog ≠ gof.

3. If f(x) = |x|, prove that fof = f.

Solution:

Given f(x) = |x|,

Now we have to prove that fof = f.

Consider (fof) (x) = f (f (x))

= f (|x|)

= ||x||

= |x|

= f (x)

So,

(fof) (x) = f (x), ∀x ∈ R

Hence, fof = f

4. If f(x) = 2x + 5 and g(x) = x2 + 1 be two real functions, then describe each of the following functions:
(i) fog
(ii) gof
(iii) fof
(iv) f2
Also, show that fof ≠ f2

Solution:

f(x) and g(x) are polynomials.

⇒ f: R → R and g: R → R.

So, fog: R → R and gof: R → R.

(i) (fog) (x) = f (g (x))

= f (x2 + 1)

= 2 (x+ 1) + 5

=2x2 + 2 + 5

= 2x2 +7

(ii) (gof) (x) = g (f (x))

= g (2x +5)

= (2x + 5)2 + 1

= 4x2 + 20x + 26

(iii) (fof) (x) = f (f (x))

= f (2x +5)

= 2 (2x + 5) + 5

= 4x + 10 + 5

= 4x + 15

(iv) f2 (x) = f (x) x f (x)

= (2x + 5) (2x + 5)

= (2x + 5)2

= 4x2 + 20x +25

Hence, from (iii) and (iv) clearly fof ≠ f2

5. If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?

Solution:

Given f(x) = sin x and g(x) = 2x

We know that

f: R→ [−1, 1] and g: R→ R

Clearly, the range of f is a subset of the domain of g.

gof: R→ R

(gof) (x) = g (f (x))

= g (sin x)

= 2 sin x

Clearly, the range of g is a subset of the domain of f.

fog: R → R

So, (fog) (x) = f (g (x))

= f (2x)

= sin (2x)

Clearly, fog ≠ gof

Hence they are not equal functions.

6. Let f, g, h be real functions given by f(x) = sin x, g (x) = 2x and h (x) = cos x. Prove that fog = go (f h).

Solution:

Given that f(x) = sin x, g (x) = 2x and h (x) = cos x

We know that f: R→ [−1, 1] and g: R→ R

Clearly, the range of g is a subset of the domain of f.

fog: R → R

Now, (f h) (x) = f (x) h (x) = (sin x) (cos x) = ½ sin (2x)

Domain of f h is R.

Since range of sin x is [-1, 1], −1 ≤ sin 2x ≤ 1

⇒ -1/2 ≤ sin x/2 ≤ 1/2

Range of f h = [-1/2, 1/2]

So, (f h): R → [(-1)/2, 1/2]

Clearly, range of f h is a subset of g.

⇒ go (f h): R → R

⇒ Domains of fog and go (f h) are the same.

So, (fog) (x) = f (g (x))

= f (2x)

= sin (2x)

And (go (f h)) (x) = g ((f(x). h(x))

= g (sin x cos x)

= 2sin x cos x

= sin (2x)

⇒ (fog) (x) = (go (f h)) (x), ∀x ∈ R

Hence, fog = go (f h)

## RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3: 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 2.3. 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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