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 for x < 0