ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.7

The availability of ISC Mathematics Class 11 Solutions Chapter 2 Relations and Functions Ex 2.7 encourages students to tackle difficult exercises.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.7

Question 1.
If f and g are real functions defined by f (x) = x² + 7 and g (x) = 3x + 5, then find the values of
(i) f (3) + g (- 5)
(ii) f (- 2) + g (- 1)
(iii) f (\(\frac{1}{2}\)) × g (14)
(iv) f (t) – f (- 2)
(v) \(\frac{f(t)-f(5)}{t-5}\), t ≠ 5
Solution:
Given f(x) = x² + 7
and g(x) = 3x + 5
Thus, D_f = R
and D_g = R
So f and g have same domain

(i) Now 3 ∈ D
∴ f (3) = 3² + 7 = 16
– 5 ∈ D_g
∴ f (- 5) = 3 × (- 5) + 5 = – 10
∴ f (3) + g (- 5) = 16 – 10 = 6

(ii) – 2 ∈ D_f
∴ f (- 2) = (- 2)² + 7 = 11
and – 1 ∈D_g
∴ g (- 1) = 3 (- 1) + 5 = 2
∴ f (- 2) + g (- 1) = 11 + 2 = 13

(iii) \(\frac{1}{2}\) ∈ D_f
and 14 ∈ D_g
∴ f (\(\frac{1}{2}\)) = (\(\frac{1}{2}\))² + 7
= \(\frac{1}{4}\) + 7
= \(\frac{29}{4}\)
g (14) = 3 × 14 + 5 = 47
Thus, f (\(\frac{1}{2}\)) × g (14) = \(\frac{29}{4}\) × 47 = \(\frac{1363}{4}\)

(iv) Now, t, – 2 ∈ D_f
∴ f (t) – f (- 2) = (t² + 7) – ((- 2)² + 7)
= t² + 7 – 11
= t² – 4

(v) Since t ≠ 5 ∈ D_f
∴ \(\frac{f(t)-f(5)}{t-5}=\frac{\left(t^2+7\right)-\left(5^2+7\right)}{t-5}\)
= \(\frac{t^2-25}{t-5}\)
= t + 5, t ≠ 5.

Question 2.
If f (x) = e^x and g (x) = log x, then find :
(i) (f – g) (1)
(ii) (f g) (1)
(iii) \(\left(\frac{f}{g}\right)\) (3)
Solution:
Given f (x) = e^x
and g (x) = log x
D_f = R,
D_g = R⁺
[since log x is defined if x > 0]
Since f and g have different domains.
Let D = D_f ∩ D_g
= R ∩ R⁺
= R⁺ ≠ Φ

(i) Then (f – g) (x) = f (x) – g (x)
= e^x log x with domain R⁺
(f – g) 1 = f (1) – g (1)
= e¹ – log 1
= e – 0 = e,
since I ∈ R⁺

(ii) (fg) (x) = f (x) g (x)
= e^x . log x
with domain R
Since 1 ∈ R⁺,
(fg) (1) = f (1) g (1)
= e¹ log 1
= e × 0 = 0

(iii) \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{e^x}{\log x}\)
with domain D₁
where D₁ = D except g (x) = 0
i.e. log x = 0
⇒ x = 1
D₁ = R – {1}
∴ \(\left(\frac{f}{g}\right)(3)=\frac{f(3)}{g(3)}=\frac{e^3}{\log 3}\)

Short answer questions (3 to 4) :

Question 3.
If f and g arc two real valued functions defined by f(x) = 2x + 1 and g(x) = x² + 1, then find the following functions ;
(i) f + g
(ii) f – g
(iii) fg
(iv) \(\frac{f}{g}\)
Solution:
Given f (x) = 2x + 1
and g x) = x² + 1
Clearly D_f = R = D_g.
So f and g have same domains

(i) (f + g) (x) = f (x) + g (x)
= 2x + 1 + x² + 1
= x² + 2x + 2 ∀ x ∈ R

(ii) (f – g) (x) = f (x) – g (x)
= 2x + 1 – x² – 1
= 2x – x² ∀ x ∈ R

(iii) (fg) (x) = f(x) g (x)
= (2x + 1) (x² + 1) ∀ x ∈ R

(iv) \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\)
= \(\frac{2 x+1}{x^2+1}\)
with Domain D₁
D₁ = R except f (x) = 0
i.e. x² + 1 = 0,
which is impossible ∀ x ∈ R
since x² + 1 ≥ 1 ∀ x ∈ R
⇒ D₁ = R.

Question 4.
If f (x) = x³ + 1 and g (x) = x + 1 be two real functions, then find the following functions :
(i) f + g
(iî) g – f
(iii) fg
(iv) \(\frac{f}{g}\)
(v) 2g² – 3f.
Solution:
Given f (x) = x³ + 1
and g (x) = x + 1
Now D_f = R = D_g.
So f and g have same domains

(i) (f + g) (x) = f (x) + g (x)
= x³ + x + 2 ∀ x ∈ R

(ii) (g – f) (x) = g (x) – f (x)
= x + 1 – x³ – 1
= x – x³ ∀ x ∈ R

(iii) (fg) (x) = f (x) . g (x)
= (x³ + 1) (x + 1)
= x⁴ + x³ + x + 1, ∀ x ∈ R

(iv) \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^3+1}{x+1}\)
with domain D
where D = R except g (x) = 0 i.e. x + 1 = 0
⇒ D = R – {- 1}

(v) (2g² – 3f) (x) = 2 (g (x))² – 3 f (x)
= 2 (x + 1)² – 3 (x³ + 1)
= 2 (x² + 2x + 1) – 3 (x³ + 1)
= 2x² + 4x – 3x³ – 1 ∀ x ∈ R.

Long answer questions (5 to 8) :

Question 5.
If f (x) = \(\sqrt{x-2}\) and g (x) =\(\sqrt{x^2-1}\) be two real valued functions, then find the following functions :
(i) f + g
(ii) g – f
(iii) fg
(iv) 3f – 2g
(v) \(\frac{f}{g}\)
(vi) 2 f² + √3g
(vii) \(\frac{1}{f}\)
(viii) \(\frac{g}{f}\)
Solution:
Given f (x) = \(\sqrt{x-2}\)
and g (x) =\(\sqrt{x^2-1}\)
For D_f : f (x) must be a real number
⇒ \(\sqrt{x-2}\) must be a real number
⇒ x – 2 ≥ 0
⇒ x ≥ 2
∴ D_f = [2, ∞)
For D_g : g (x) must be a real number
⇒ \(\sqrt{x^2-1}\) must be a real number
⇒ x² – 1 ≥ 0
⇒ |x| ≥ 1
⇒ x ≥ 1
or x ≤ – 1
∴ D_g = (- ∞, – 1] ∪ [1, ∞)
So f and g have different domains

Let D = D_f ∩ D_g = [2, ∞)

(i) (f + g) (x) = f (x) + g (x)
= \(\sqrt{x-2}+\sqrt{x^2-1}\) ∀ x ≥ 2

(ii) (g – f) (x) = g (x) – f (x)
= \(\sqrt{x^2-1}-\sqrt{x-2}\) ∀ x ≥ 2

(iii) (fg) (x) = f (x) g (x)
= \(\sqrt{x-2} \sqrt{x^2-1}\)
= \(\sqrt{\left(x^2-1\right)(x-2)}\) ∀ x ≥ 2

(iv) (3f – 2g) (x) = 3 f (x) – 2 g (x)
= \(3 \sqrt{x-2}-2 \sqrt{x^2-1}\) ∀ x ≥ 2

(v) \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x-2}}{\sqrt{x^2-1}}\)
with domain D₁
where D₁ [2, ∞) except g (x) = 0
i.e. \(\sqrt{x^2-1}\) = 0
⇒ x = ± 1
⇒ D₁ = [2, ∞) = D

(vi) (2f² + √3 g) (x) = 2 f² (x) + √3 g (x)
= 2 (f(x))² + 6 g (x)
= \(2(\sqrt{x-2})^2+\sqrt{3} \sqrt{x^2-1}\)
= 2(x – 2) + \(\sqrt{3} \sqrt{x^2-1}\) ∀ x ∈ D

(vii) \(\left(\frac{1}{f}\right)(x)=\frac{1}{f(x)}=\frac{1}{\sqrt{x-2}}\)
with domain D₂
where D₂ = [2, ∞) except f (x) = 0
i.e. \(\sqrt{x-2}\) = 0
⇒ x = 2
i.e. D₂ = (2, ∞)

(viii) \(\left(\frac{g}{f}\right)(x)=\frac{g(x)}{f(x)}\)
= \(\frac{\sqrt{x^2-1}}{\sqrt{x-2}}\)
= \(\frac{\sqrt{x^2-1}}{\sqrt{x-2}}\) with domain D₃
where D₃ = [2, ∞) except f (x) = 0
i.e. \(\sqrt{x-2}\) = 0
⇒ x = 2
⇒ D₃ = (2, ∞).

Question 6.
If f (x) = \(x^3-\frac{1}{x^3}\), prove that f (x) + f (\(\frac{1}{x}\)) = 0.
Solution:
Given f (x) = \(x^3-\frac{1}{x^3}\)
∴ \(f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^3-\frac{1}{\left(\frac{1}{x}\right)^3}\)
= \(\frac{1}{x^3}-x^3\)
∴ f (x) + f (\(\frac{1}{x}\)) = x³ – \(\frac{1}{x^3}+\frac{1}{x^3}\) – x³ = 0.

Question 7.
If f (x) = x + \(\frac{1}{x}\), prove that (f(x))³ = f (x³) + 3 f (\(\frac{1}{x}\)).
Solution:
Given f (x) = x + \(\frac{1}{x}\)
L.H.S. = (f(x))³
= (x + \(\frac{1}{x}\))³
= x³ + \(\frac{1}{x^3}\) + 3 × x × \(\frac{1}{x}\) (x + \(\frac{1}{x}\))
= x³ + \(\frac{1}{x^3}\) + 3 (x + \(\frac{1}{x}\))
∴ L.H.S. = R.H.S.
Thus,
(f(x))³ = f (x³) + 3 f (\(\frac{1}{x}\))

Question 8.
If y = f(x) = \(\frac{6 x-5}{5 x-6}\), prove that f (y) = x, x ≠ \(\frac{6}{5}\).
Solution:
Given y = f(x)
= \(\frac{6 x-5}{5 x-6}\)
⇒ y(5x – 6) = 6x – 5
⇒ 5xy – 6y = 6x – 5
⇒ 5xy – 6x = 6y – 5
⇒ x (5y – 6) = 6y – 5
⇒ x = \(\frac{6 y-5}{5 y-6}\) = f (y) [using eqn. (1)]