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

### (i) f (x) = ex,

Solution:

Let f: R â†’ (0, âˆž); and g: (0, âˆž) â†’ R

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

So, fog: (0, âˆž) â†’ 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.

â‡’ fog: Râ†’ R

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

= g(ex)

(gof)(x) = x

### (ii) f (x) = x2, g(x) = cos x

Solution:

f: Râ†’ [0, âˆž) ; g: Râ†’[âˆ’1, 1]

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

â‡’ Domain (fog) = {x: x âˆˆ domain of g and g (x) âˆˆ domain of f}

â‡’ Domain (fog) = x: x âˆˆ R and cos x âˆˆ R}

â‡’ Domain of (fog) = R

(fog): Râ†’ R

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

= f(cosx)

(fog)(x) = cos2x

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

â‡’ fog: Râ†’R

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

= g (x2)

(gof)(x) = cos x2

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

Solution:

f: R â†’ (0, âˆž) ; g : Râ†’[âˆ’1, 1]

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

â‡’ fog: Râ†’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.

â‡’ fog : Râ†’ R

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

= g (|x|)

(gof)(x) = sin |x|

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

Solution:

f: Râ†’R ; g: R â†’ [ 1, âˆž)

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

â‡’ fog: Râ†’R

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

= f(ex)

(fog)(x) = ex + 1

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

â‡’ fog: Râ†’R

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

= g(x+1)

(gof)(x) = ex+1

### (v) f (x) = sinâˆ’1x, g(x) = x2

Solution:

f: [âˆ’1,1]â†’ [(-Ï€)/2 ,Ï€/2]; g : R â†’ [0, âˆž)

Domain (fog) = {x: x âˆˆ R and x âˆˆ [âˆ’1, 1]}

So, Domain of (fog) = [âˆ’1, 1]

fog: [âˆ’1,1] â†’ R

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

= f(x2)

(fog)(x) = sinâˆ’1(x2)

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

fog: [âˆ’1, 1] â†’ R

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

= g (sinâˆ’1 x)

(gof)(x) = (sinâˆ’1x)2

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

Solution:

f: Râ†’R ; g: Râ†’[âˆ’1, 1]

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

Set of the domain of f.

â‡’ fog: Râ†’ 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.

â‡’ fog: R â†’ 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: Râ†’R ; g: R â†’ R

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

â‡’ fog: Râ†’ 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.

â‡’ fog: R â†’ 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 â†’ {c} ; g: Râ†’ [ 0, 1 ]

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

fog: Râ†’R

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

= f(sinx2)

(fog)(x) = c

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

â‡’ fog: Râ†’ R

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

= g(c)

(gof)(x) = sinc2

### (ix) f(x) = x2+ 2 and

Solution:

f: R â†’ [2, âˆž)

For domain of g: 1âˆ’ x â‰  0

â‡’ x â‰  1

â‡’ Domain of g = R âˆ’ {1}

=

Range of g = R âˆ’ {1}

So, g: R âˆ’ {1} â†’ R âˆ’ {1}

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

â‡’ fog: R âˆ’ {1} â†’ R

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

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

â‡’ gof: Râ†’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 â‰  gof.

Solution:

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

Now we have to prove fog â‰  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 â‰  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), âˆ€x âˆˆ 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.

â‡’ f: R â†’ R and g: R â†’ R.

So, fog: R â†’ R and gof: R â†’ 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: Râ†’ [âˆ’1, 1] and g: Râ†’ R

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

gof: Râ†’ 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 â†’ R

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

= f(2x)

= sin(2x)

Clearly, fog â‰  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

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 â€“ 2

### Question 12. Letfind fof.

Solution:

Range of f = [0,3]

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

### Question 13. If f, g : R â†’ R be two functions defined as f(x) = |x| + x and g(x) = |x|- x, âˆ€ xâˆˆR. Then find fog and gof. Hence find fog(â€“3), fog(5) and gof (â€“2).

Solution:

It is given that, f(x) = |x| + x and g(x) = |x| -x, âˆ€x âˆˆ R

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

= ||x| âˆ’ x| + (|x| âˆ’ x)

gof =  g (f(x)) = |f(x)| âˆ’ f (x)

= ||x| + x| âˆ’ (|x| + x)

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

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

fog (5) = 0, as fog = 0 for x â‰¥ 0

gof(âˆ’2) = 0, as gof = 0 for x < 0

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