# Class 12 RD Sharma Solutions – Chapter 2 Functions – Exercise 2.3

**Question 1. Find fog and gof, if**

**(i) f (x) = e**^{x},

^{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) = xClearly, 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) = x**^{2}, g(x) = cos x

^{2}, 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) = cos^{2}xClearly, the range of f is a subset of the domain of g.

⇒ fog: R→R

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

= g (x

^{2})

(gof)(x) = cos x^{2}

**(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}

^{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}+ 1Clearly, 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**^{−1}x, g(x) = x^{2}

^{−1}x, g(x) = x

^{2}

**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(x

^{2})

(fog)(x) = sin^{−1}(x^{2})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^{−1}x)^{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 + 1Now 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 + 4Clearly, 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 x**^{2}

^{2}

**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(sinx

^{2})

(fog)(x) = cClearly, the range of f is a subset of the domain of g.

⇒ fog: R→ R

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

= g(c)

(gof)(x) = sinc^{2}

**(ix) f(x) = x**^{2}+ 2 and

^{2}+ 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(x

^{2}+ 2)

**Question 2. Let f(x) = x**^{2} + x + 1 and g(x) = sin x. Show that fog ≠ gof.

^{2}+ x + 1 and g(x) = sin x. Show that fog ≠ gof.

**Solution:**

Given f(x) = x

^{2}+ x + 1 and g(x) = sin xNow we have to prove fog ≠ gof

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

= f(sin x)

(fog)(x) = sin

^{2}x + sin x + 1 …..(1)And (gof)(x) = g (f (x))

= g (x

^{2}+ 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) = x**^{2} + 1 be two real functions, then describe each of the following functions:

^{2}+ 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 (x

^{2}+ 1)= 2 (x

^{2 }+ 1) + 5=2x

^{2 }+ 2 + 5

= 2x^{2}+7

**(ii) gof**

**Solution:**

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

= g (2x +5)

= (2x + 5)

^{2 }+ 1

= 4x^{2}+ 20x + 26

**(iii) fof**

**Solution:**

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

= f (2x +5)

= 2 (2x + 5) + 5

= 4x + 10 + 5

= 4x + 15

**(iv) f ^{2}(x)**

**Solution:**

f

^{2}(x) = f(x) x f(x)= (2x + 5)(2x + 5)

= (2x + 5)

^{2}

= 4x^{2}+ 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. If****and****are 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. If****and g(x) = x**^{2} + 1 be two real functions, find fog and gof.

^{2}+ 1 be two real functions, find fog and gof.

**Solution:**

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

= f(x

^{2}+ 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) f**^{2}

^{2}

**Solution:**

f

^{2}(x) = f(x).f(x)=

f^{2(}x) = x – 2

**Question 12. Let****find 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

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