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)]