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Class 12 RD Sharma Solutions – Chapter 2 Functions – Exercise 2.3
  • Last Updated : 08 May, 2021

Question 1. Find fog and gof, if

(i) f (x) = ex,g (x) = \log_ex

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))

= f(\log_ex)



= e^{log_ex}

(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(ex)

= log_ee^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) = ex

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(ex)

(fog)(x) = ex + 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) = ex+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 andg(x) = 1 - \frac{1}{1 - x}

Solution:

f: R → [2, ∞)

For domain of g: 1− x ≠ 0

⇒ x ≠ 1

⇒ Domain of g = R − {1}

g(x) = 1 - \frac{1}{1 - x}

=\frac{(1 - x - 1)}{(1 - x)}

= \frac{-x}{(1 - x)}



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))

= f (\frac{-x}{1 - x})

(fog)(x) = (\frac{-x}{1 - x})^2 + 2

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

⇒ gof: R→R

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

= g(x2 + 2)

(gof)(x) = \frac{x^2 + 2}{x^2 + 1}

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. Iff(x) = \sqrt{1 - x} andg(x) = log_ex are two real functions, find fog and gof.

Solution:

Clearly the domain of f and g are R.

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

= f(log_ex)

fog(x)= \sqrt{1 - log_ex}

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

= g(\sqrt{1 - x})

(gof)(x)= log_e\sqrt{1 - x}

Question 9. If f(x) = tan x andg(x) = \sqrt{1 - x^2} , find fog and gof.

Solution:

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

= f(\sqrt{1 - x^2})

fog(x)= tan\sqrt{1 - x^2}

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

= g(tan x)

(gof)(x) = \sqrt{1 - tan^2x}

Question 10. Iff(x) = \sqrt{x + 3} and g(x) = x2 + 1 be two real functions, find fog and gof.

Solution:



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

= f(x2 + 1)

fog(x) = \sqrt{x^2 + 4}

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

= g(\sqrt{x + 3})

= (\sqrt{x + 3})^2 + 1

(gof)(x) = x + 4

Question 11. Let f be a real function given byf(x) = \sqrt{x - 2} . Find:

(i) fof

Solution:

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

= f(\sqrt{x - 2})

fof(x) = \sqrt{\sqrt{x - 2} - 2}

(ii) fofof

Solution:

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

= f(\sqrt{x - 2})

Thus,fof(x) = \sqrt{\sqrt{x - 2} - 2}

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

= fof(\sqrt{x - 2})

fofof(x) = \sqrt{\sqrt{\sqrt{x - 2} - 2} - 2}

(iii) f2

Solution:

f2(x) = f(x).f(x)



=(\sqrt{x - 2})^2

f2(x) = x – 2

Question 12. Letf(x)= \begin{cases}1+x,0\lex\le2\\3-x\ \ \ ,2\lex\le3\end{cases} find fof.

Solution:

Range of f = [0,3]

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

= f{\begin{cases}1+x,0\lex\le2\\3-x\ \ \ ,2\lex\le3\end{cases}}

fof(x) = \begin{cases}2+x\ \ ,0\lex\le1\\2-x\ \ \ \ \ ,1\lex\le2\\4-x\ \ \ \ \ ,2\lex\le3\end{cases}




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