OP Malhotra Class 12 Maths Solutions Chapter 2 Functions Ex 2(b)

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S Chand Class 12 ICSE Maths Solutions Chapter 2 Functions Ex 2(b)

Question 1.
If f : N → R : f (x) = \(\frac { 2x-1 }{ 2 }\) and g : Q → R : g (x) = x + 2 be two functions, then find (fog)(\(\frac { -3 }{ 2 }\)).
Solution:
Given f : N → R defined by
f(x) = \(\frac { 2x-1 }{ 2 }\) ∀x∈N
and g : Q → R defined by
g (x) = x + 2 ∀x∈Q
\((\mathrm{fog})\left(\frac{-3}{2}\right)=f\left(g\left(\frac{-3}{2}\right)\right)\)
= f\(\left(\frac{-3}{2}+2\right)=f\left(\frac{1}{2}\right)\)
= \(\frac{2 \times \frac{1}{2}-1}{2}\) = 0

Question 2.
(i) If f : R → R and g : R → R are given f(x) = sin x and g(x) = 5x², then find gof (x).
(ii) If f(x) = 27x³ and g(x) = x^1/3, then find gof (x).
Solution:
(i) Given f : R→ R defined by f(x) = sin x and g : R → R defined by
g(x) = 5x²
Clearly R_f ⊂ Dg
∴ gof exists
Thus,
(gof) (x) = g(f(x)) = g (sin x) = 5
(sin x)² = 5 sin²x ∀x∈R

(ii) Given f(x) = 27x³ ; g (x) = x^1/3
∴ (gof) (x) = g(f(x) = g(27x³)
= (27x³)^1/3 = 3x

Question 3.
Find (gof)(3), (fog)(1) and(fof)(0) if
(i) f(x) = 3x – 2, g(x) = x²
(ii) f(x) = |x + 2|, g(x) = – x²
(iii) f(x) = x² – 1, g(x) = \(\sqrt{x}\)
Solution:
(i) Given f(x) = 3x – 2; g(x) = x²
(gof)(3) = g(f(3)) = g(3 x 3 – 2) = g(7)
= 7² = 49
(fog)(1) = f(g(1)) = f(1²) = f(x)
= 3 – 2 = 1
(fof) (0) = f(f(0)) = f(0 – 2) = f(- 2)
= 3(- 2) – 2 = – 8

(ii) Given f(x)=|x + 2|, g(x) = – x²
(gof)(3) = g(f(3)) = g(|3 + 2|) = g (5)
= – 5² = – 25
(fog)(1) = f(g(1)) = f(- 1²) = f(- 1)
= |- 1 + 2| = 1
(fof) (0) = ff(0)) = f(|0 + 2|)
= f(2) = |2 + 2| = 4

(iii) Given f (x) = x² – 1, g(x) = \(\sqrt{x}\)
(gof)(3) = g(f(3)) = g(3² – 1) = g (8)
= \(\sqrt{8}\) = 2\(\sqrt{2}\)
(fog) (1) = f(g(1)) = f(\(\sqrt{1}\)) = f(1))
= 1² – 1 = 0
(fof) (0) = f(g(1)) = f(0² – 1) = f(- 1) = (- 1²) – 1 = 0

Question 4.
If f(x) = x + 5 and g(x) = x² – 3, find the following:
(i) f(g(0))
(ii) g(f (0))
(iii) f(g(x))
(iv) g(f (x))
(v) g(f (x))
(vi) g(g (x))
(vii) f(f (-5))
(viii) g(g (2))
Solution:
Given f(x) = x + 5; g(x) = x² – 3
(i) f(g(0)) = f(0² – 3)
= f(-3) = – 3 + 5 = 2

(ii) g(f(0)) = g(0 + 5) = g(5)
= 5² – 3 = 22

(iii) f(g (x)) = f(x² – 3)
= x² – 3 + 5 = x² + 2

(iv) g(f (x)) = g(x + 5)
= (x + 5)² -3 = x² + 10x + 22

(v) f(f (x)) = f(x + 5)
= x + 5 + 5 = x + 10

(vi) g(g (x)) = g(x² – 3)
= (x² – 3)² – 3 = x⁴ – 6x² + 6

(vii) f(f(- 5)) = f(-5 + 5) = f(0)
= 0 + 5 = 5

(viii) g(g(2)) = g(2² – 3) = g(1)
= 1² – 3 = – 2

Question 5.
If u(x) = 4x – 5, v(x) = x² and f(x) = \(\frac { 1 }{ x }\) , find
(i) u (v (f (x)))
(ii) u(f(v(x)))
(iii) f(u(v(x)))
Solution:
Given u(x) = 4x – 5, v(x) = x and f(x) = \(\frac { 1 }{ x }\)
(i) u(v(f(x))) = u\(\left(v\left(\frac{1}{x}\right)\right)\)
= u\(\frac { 1 }{ x² }\) = \(\frac { 4 }{ x² }\) – 5

(ii) u(f(v(x))) = u(f(x²))
= u\(\left(\frac{1}{x^2}\right)=\frac{4}{x^2}\)

(iii) f(u(v(x))) = f(u(x²))
= f(4x² – 5) = \(\frac{1}{4 x^2-5}\)

Question 6.
Find the indicated values, where
g(t) = t² – 1 and f(x) = 1 + x
(i) g(f(0)) + f(g(0))
(ii) g(f(2) + 3)
Solution:
Given g(t) = t² – 1 and f(x) = 1 + x
∴ f(0) = 1 + 0 = 1 ;
g(0) = 0² – 1 = – 1
(i) Thus g (f (0)) + f (g (0))
= g(1) + f(-1) = (1² – 1) + (1 – 1) = 0 + 0 = 0

(ii) Now f(2) = 1 + 2 = 3
∴ g(f(2) + 3) = g(3 + 3) = g(6)
= 6² – 1 = 35

Question 7.
If f(x) = \(\frac{2 x+1}{3 x-2}\) then (fof)(2) is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Solution:

Question 8.
If f(x) = \(\frac{1-x}{1+x}\), then f{f(cos2θ) } is equal to
(a) cos 2θ
(b) tan 2θ
(c) sec 2θ
(d) cot 2θ
Solution:

Question 9.
Let [x] denote the greatest integer ≤ x. If f(x) = [x] and g(x) = |x|, then the value of \(f\left\{g\left(\frac{8}{5}\right)\right\}-g\left\{f\left(-\frac{8}{5}\right)\right\}\) is
(a) 2
(b) – 2
(c) 1
(d) 0
(d) – 1
Solution:
Given f(x) = [x] and g(x) = | x |
∴ g\(\left(\frac{8}{5}\right)=\left|\frac{8}{5}\right|=\frac{8}{5}\)
f\(\left(-\frac{8}{5}\right)=\left[\frac{-8}{5}\right]\) = [- 1, 6]
= [- 1 – 0.6] = [- 2 + 0.4] = – 2
f\(\left(g\left(\frac{8}{5}\right)\right)=f\left(\frac{8}{5}\right)=\left[\frac{8}{5}\right]\) = 1
and g\(\left\{f\left(\frac{-8}{5}\right)\right\}=g\left\{\left[\frac{-8}{5}\right]\right\}=g\{[-1.2]\}\)
= g(- 2) = |- 2| = 2
Thus, f\(\left\{g\left(\frac{-8}{5}\right)\right\}-g\left\{f\left(\frac{-8}{5}\right)\right\}\)
= 1 – 2 = – 1

Question 10.
If f : R → R and g : R → R are defined by f(x) = x – 3 and g(x) = x² +1, then the values of x for which g(f (x)) = 10 are
(a) 0, – 6
(b) 2, – 2
(c) 1, – 1
(d) 0, 6
(e) 0, 2
Solution:
Given f : R → R and g : R → R defined by f(x) = x- 3 and
g(x) = x² +1
Since R_f ⊂ D_g = R
∴ gof exists
Now |gof| (x) = g(f (x)) = g(x – 3)
= (x – 3)² + 1 = x² – 6x + 10 – 10, also
g(f(x)) = 10 (given)
Thus, x² – 6x + 10 = 10
⇒ x² – 6x = 0
⇒ x(x – 6) = 0
⇒ x = 0, 6

Question 11.
If f(x) = sin²x and the composite function g[f(x)] = | sinx |, then the function g(x) is equal to
(a) – \(\sqrt{x}\)
(b) \(\sqrt{x}\)
(c) \(\sqrt{x-1}\)
(d) \(\sqrt{x+1}\)
Solution:

Question 12.
If f : R → R and g : R → R are defined respectively as
f(x) = x² + 3x + 1 and g(x) = 2x – 3, find
(a) fog
(b) gof.
Solution:
Given f : R→ R defined by
f(x) = x² +3x + 1 and g : R → R defined by g (x) = 2x – 3
Since R_f ⊂ D_g
∴ gof exists.
and R_g ⊂ D_f
∴ fog exists (g(x))

(i) ∀x ∈ R, (fog) (x) = f(g(x)) = f(2x – 3) = (2x – 3)² + 3(2x – 3) + 1
= 4x² – 12x + 9 + 6x – 9 + 1
= 4x² – 6x + 1

(ii) x∈R, (gof) (x) = g(f(x))
= g(x² + 3x + 1)
= 2(x² + 3x + 1) – 3
= 2x² + 6x – 1

Question 13.
If f: R → R, f(x) = x²,
g : R→ R, g(x) = cos x ∀x ∈ R, find fog and gof and show that fog ≠ gof
Solution:
Given f : R → R defined by
f(x) = x² ∀x ∈ R
and g : R →R defined by
g(x) = cos x ∀x ∈ R
∀x ∈ R, (fog)(x) = f(g(x)) = f(cosx) = cos²x
(gof)(x) = g(f(x) = g(x²) = cos x²
Clearly (fog)(x) ≠ (gof)(x) ∀x ∈ R
⇒ fog ≠ gof

Question 14.
If the function f : R → R be defined as f(x) = \(\frac{3 x+4}{5 x-7}\left(x \neq \frac{7}{5}\right)\) and g : R → R be defined as g (x) = and g : R → R be defined as g(x) = \(\frac{7 x+4}{5 x-3}\left(x \neq \frac{3}{5}\right)\), show that (gof) (x) = (fog) (x).
Solution:
Given f : R → R defined as

Question 15.
If R → R is given by
f(x) = \(\left\{\begin{array}{r}
-1, \text { when } x \text { is a rational } \\
1 \text {, when } x \text { is irrational }
\end{array}\right.\)
then, (fof)(1 – \(\sqrt{3}\)) is equal to
(a) 1
(b) – 1
(c) \(\sqrt{3}\)
(d) 0
Solution:
Given f : R → R defined by
f(x) = \(\left\{\begin{array}{r}
-1, \text { when } x \text { is a rational } \\
1, \text { when } x \text { is irrational }
\end{array}\right.\)
Since (1 – \(\sqrt{3}\)) be an irrational number
∴ f(1 – \(\sqrt{3}\)) = 1
Now (fof)(1 – \(\sqrt{3}\)) = f{f(1 – \(\sqrt{3}\))}
= f(1) = – 1
[∵ 1 be a rational number ∴ f(x) = – 1]