RD Sharma Solutions for Class 12 Maths Exercise 2.3 Chapter 2 Function (Updated for 2024)

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.

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

Students’ exam preparation is accelerated by solving these problems in a simple manner. RD Sharma Solutions mostly aids in revision and increases student confidence before appearing for the board exam.

Access answers to RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3

Exercise 2.3 Page No: 2.54

1. Find fog and gof, if

(i) f (x) = e^x, g (x) = log_e x

(ii) f (x) = x², g (x) = cos x

(iii) f (x) = |x|, g (x) = sin x

(iv) f (x) = x+1, g(x) = e^x

(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) = e^x, g(x) = log_e 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 (log_e x)

= log_e e^x

= 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 (e^x)

= log_e e^x

= 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) = e^x

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 (e^x)

= e^x + 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)

= e^x+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

= (3x² – 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) 

= sin² x + 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 (x² + 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
  • The 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.

FAQs on RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.3