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, Df = R
and Dg = R
So f and g have same domain

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

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

(iii) $$\frac{1}{2}$$ ∈ Df
and 14 ∈ Dg
∴ 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 ∈ Df
∴ f (t) – f (- 2) = (t2 + 7) – ((- 2)2 + 7)
= t2 + 7 – 11
= t2 – 4

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

Question 2.
If f (x) = ex 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) = ex
and g (x) = log x
Df = R,
Dg = R+
[since log x is defined if x > 0]
Since f and g have different domains.
Let D = Df ∩ Dg
= R ∩ R+
= R+ ≠ Φ

(i) Then (f – g) (x) = f (x) – g (x)
= ex 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)
= ex . 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 Df = R = Dg.
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 Df = R = Dg.
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 Df : f (x) must be a real number
⇒ $$\sqrt{x-2}$$ must be a real number
⇒ x – 2 ≥ 0
⇒ x ≥ 2
∴ Df = [2, ∞)
For Dg : 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
∴ Dg = (- ∞, – 1] ∪ [1, ∞)
So f and g have different domains

Let D = Df ∩ Dg = [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)]