Class 12 RD Sharma Solutions - Chapter 2 Functions - Exercise 2.3

Question 1. Find fog and gof, if

(i) f (x) = e^x,

Solution:

Let f: R → (0, ∞); and g: (0, ∞) → R

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

So, fog: (0, ∞) → R and we know, (fog)(x) = f(g(x))

(fog)(x) = x

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

⇒ fog: R→ R

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

= g(e^x)

(gof)(x) = x

(ii) f (x) = x², g(x) = cos x

Solution:

f: R→ [0, ∞) ; g: R→[−1, 1]

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(cosx)

(fog)(x) = cos²x

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

⇒ fog: R→R

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

= g (x²)

(gof)(x) = cos x²

(iii) f(x) = |x|, g(x) = sin x

Solution:

f: R → (0, ∞) ; g : R→[−1, 1]

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

⇒ fog: R→R

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

= f (sin x)

(fog)(x) = |sin x|

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

⇒ fog : R→ R

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

= g (|x|)

(gof)(x) = sin |x|

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

Solution:

f: R→R ; g: R → [ 1, ∞)

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

⇒ fog: R→R

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

= f(e^x)

(fog)(x) = e^x + 1

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

⇒ fog: R→R

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

= g(x+1)

(gof)(x) = e^x+1

(v) f (x) = sin⁻¹x, g(x) = x²

Solution:

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

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

So, Domain of (fog) = [−1, 1]

fog: [−1,1] → R

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

= f(x²)

(fog)(x) = sin⁻¹(x²)

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)

(gof)(x) = (sin⁻¹x)²

(vi) f(x) = x+1, g(x) = sinx

Solution:

f: R→R ; g: R→[−1, 1]

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(sinx)

(fog)(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)

(gof)(x) = sin(x+1)

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

Solution:

f: R→R ; g: R → R

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

(fog)(x) = 2x + 4

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

(gof)(x) = 2x + 5

(viii) f (x) = c, g (x) = sin x²

Solution:

f: R → {c} ; g: R→ [ 0, 1 ]

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

fog: R→R

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

= f(sinx²)

(fog)(x) = c

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

⇒ fog: R→ R

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

= g(c)

(gof)(x) = sinc²

(ix) f(x) = x²+ 2 and

Solution:

f: R → [2, ∞)

For domain of g: 1− x ≠ 0

⇒ x ≠ 1

⇒ Domain of g = R − {1}

=

Range of g = R − {1}

So, g: R − {1} → R − {1}

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

⇒ fog: R − {1} → R

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

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

⇒ gof: R→R

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

= g(x² + 2)

Question 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)

(fog)(x) = sin²x + sin x + 1 .....(1)

And (gof)(x) = g (f (x))

= g (x²+ x + 1)

(gof)(x) = sin (x2+ x + 1) ....(2)

From (1) and (2), we get

fog ≠ gof.

Question 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.

Question 4. If f(x) = 2x + 5 and g(x) = x² + 1 be two real functions, then describe each of the following functions:

(i) fog

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

Solution:

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

= g (2x +5)

= (2x + 5)² + 1

= 4x² + 20x + 26

(iii) fof

Solution:

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

= f (2x +5)

= 2 (2x + 5) + 5

= 4x + 10 + 5

= 4x + 15

(iv) f²(x)

Solution:

f²(x) = f(x) x f(x)

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

= (2x + 5)²

= 4x² + 20x +25

Question 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.

Question 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(fh).

Solution:

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

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

= f(2x)

fog(x) = sin2x ....(1)

And (go (f h)) (x) = g ((f(x). h(x))

= g (sin x cos x)

= 2sin x cos x

= sin (2x) ....(2)

From (1) and (2), fog(x) = go(fh) (x).

Question 7. Let f be any real function and let g be a function given by g(x) = 2x. prove that: gof = f+f.

Solution:

We know, (gof)(x) = g(f(x))

= 2(f(x))

= f(x) + f(x)

= f + f.

Hence proved.

Question 8. Ifandare two real functions, find fog and gof.

Solution:

Clearly the domain of f and g are R.

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

fog(x)

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

(gof)(x)

Question 9. If f(x) = tan x and, find fog and gof.

Solution:

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

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

= g(tan x)

Question 10. Ifand g(x) = x² + 1 be two real functions, find fog and gof.

Solution:

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

= f(x² + 1)

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

(gof)(x) = x + 4

Question 11. Let f be a real function given by. Find:

(i) fof

Solution:

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

(ii) fofof

Solution:

We know, fof(x) = f(f(x))

Thus,

Now, fofof(x) = fof(f(x))

(iii) (fofof) (38)

Solution:

As obtained from the previous part, we have

So we get,

fofof (38) = 

= 0

(iv) f²

Solution:

f²(x) = f(x).f(x)

=

f²⁽x) = x – 2

Question 12. Letfind fof.

Solution:

Range of f = [0,3]

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

Question 13. If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x|- x, ∀ x∈R. Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).

Solution:

It is given that, f(x) = |x| + x and g(x) = |x| -x, ∀x ∈ R

fog = f(g(x)) = | g (x) | + g(x)

= ||x| − x| + (|x| − x)

gof =  g (f(x)) = |f(x)| − f (x)

= ||x| + x| − (|x| + x)

So, g (f(x)) = gof = 0

Now, fog(−3) =(4)(−3) = −12, as fog = 4x for x < 0

fog (5) = 0, as fog = 0 for x ≥ 0

gof(−2) = 0, as gof = 0

Summary

Exercise 2.3 in Chapter 2 of RD Sharma's Class 12 mathematics textbook likely focuses on the topic of functions, specifically dealing with the composition of functions. This exercise typically covers how to compose two or more functions, find the domain and range of composite functions, and solve problems involving function composition. It may also include topics such as finding the inverse of composite functions and understanding the properties of function composition, like associativity and non-commutativity.

Pratice Questions

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

2. Given f(x) = √(x + 2) and g(x) = x² - 1, determine the domain of (f ∘ g)(x).

3. If f(x) = 1/(x-1) and g(x) = x + 2, find (f ∘ g)⁻¹(x).

4. Let f(x) = 2x + 1 and g(x) = x² - 3. Solve the equation (f ∘ g)(x) = 11.

5. If f(x) = x³ and g(x) = ³√x, prove that (f ∘ g)(x) = x but (g ∘ f)(x) ≠ x.

6. Given f(x) = sin x and g(x) = cos x, find (f ∘ g ∘ f)(π/4).

7. If f(x) = x² - 2x + 3 and (f ∘ g)(x) = x² + 2x + 5, find g(x).

8. Let f(x) = 2x + 1 and g(x) = x - 3. Find a function h(x) such that (f ∘ h ∘ g)(x) = x.

9. If f(x) = 1/x and g(x) = x + 1, determine the range of (g ∘ f)(x).

10. Given f(x) = x² and g(x) = x + 1, find the value of x for which (f ∘ g)(x) = (g ∘ f)(x).