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) = x^{2} + 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) = x^{2} + 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) = 3^{2} + 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) = (t^{2} + 7) – ((- 2)^{2} + 7)

= t^{2} + 7 – 11

= t^{2} – 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)

= e^{1} – 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)

= e^{1} log 1

= e × 0 = 0

(iii) \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{e^x}{\log x}\)

with domain D_{1}

where D_{1} = D except g (x) = 0

i.e. log x = 0

⇒ x = 1

D_{1} = 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) = x^{2} + 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) = x^{2} + 1

Clearly D_{f} = R = D_{g}.

So f and g have same domains

(i) (f + g) (x) = f (x) + g (x)

= 2x + 1 + x^{2} + 1

= x^{2} + 2x + 2 ∀ x ∈ R

(ii) (f – g) (x) = f (x) – g (x)

= 2x + 1 – x^{2} – 1

= 2x – x^{2} ∀ x ∈ R

(iii) (fg) (x) = f(x) g (x)

= (2x + 1) (x^{2} + 1) ∀ x ∈ R

(iv) \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\)

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

with Domain D_{1}

D_{1} = R except f (x) = 0

i.e. x^{2} + 1 = 0,

which is impossible ∀ x ∈ R

since x^{2} + 1 ≥ 1 ∀ x ∈ R

⇒ D_{1} = R.

Question 4.

If f (x) = x^{3} + 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) 2g^{2} – 3f.

Solution:

Given f (x) = x^{3} + 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)

= x^{3} + x + 2 ∀ x ∈ R

(ii) (g – f) (x) = g (x) – f (x)

= x + 1 – x^{3} – 1

= x – x^{3} ∀ x ∈ R

(iii) (fg) (x) = f (x) . g (x)

= (x^{3} + 1) (x + 1)

= x^{4} + x^{3} + 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) (2g^{2} – 3f) (x) = 2 (g (x))^{2} – 3 f (x)

= 2 (x + 1)^{2} – 3 (x^{3} + 1)

= 2 (x^{2} + 2x + 1) – 3 (x^{3} + 1)

= 2x^{2} + 4x – 3x^{3} – 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 f^{2} + √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

⇒ x^{2} – 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 D_{1}

where D_{1} [2, ∞) except g (x) = 0

i.e. \(\sqrt{x^2-1}\) = 0

⇒ x = ± 1

⇒ D_{1} = [2, ∞) = D

(vi) (2f^{2} + √3 g) (x) = 2 f^{2} (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 D_{2}

where D_{2} = [2, ∞) except f (x) = 0

i.e. \(\sqrt{x-2}\) = 0

⇒ x = 2

i.e. D_{2} = (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 D_{3}

where D_{3} = [2, ∞) except f (x) = 0

i.e. \(\sqrt{x-2}\) = 0

⇒ x = 2

⇒ D_{3} = (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 (\(\frac{1}{x}\)) = x^{3} – \(\frac{1}{x^3}+\frac{1}{x^3}\) – x^{3} = 0.

Question 7.

If f (x) = x + \(\frac{1}{x}\), prove that (f(x))^{3} = f (x^{3}) + 3 f (\(\frac{1}{x}\)).

Solution:

Given f (x) = x + \(\frac{1}{x}\)

L.H.S. = (f(x))^{3}

= (x + \(\frac{1}{x}\))^{3}

= x^{3} + \(\frac{1}{x^3}\) + 3 × x × \(\frac{1}{x}\) (x + \(\frac{1}{x}\))

= x^{3} + \(\frac{1}{x^3}\) + 3 (x + \(\frac{1}{x}\))

∴ L.H.S. = R.H.S.

Thus,

(f(x))^{3} = f (x^{3}) + 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)]