OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Ex 2(e)

Effective OP Malhotra Class 11 Solutions Chapter 2 Relations and Functions Ex 2(e) can help bridge the gap between theory and application.

S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(e)

Question 1.
If the function f : N → N is defined by f (x) = \(\sqrt{x}\), then find \(\frac{f(25)}{f(15)+f(1)}\).
Solution:
Given a function f : N → N defined by f(x) = \(\sqrt{x}\)
f (25) = \(\sqrt{25}\) = 5; f (16) = \(\sqrt{16}\) = 4 ; f (1) = \(\sqrt{1}\) = 1
Thus \(\frac{f(25)}{f(16)+f(1)}=\frac{5}{4+1}\) = 1

Question 2.
If f(x) = \(\frac{x^3}{2}-\frac{x^2}{2}+x-16\), find f(\(\frac { 1 }{ 2 }\)).
Solution:

Question 3.
If f (x) = 7x⁴ – 2x³ – 8x – 5, find f (- 1).
Solution:
Given f (x) = 7x⁴ – 2x³ – 8x – 5
∴ f (- 1) = 7 (- 1)⁴ – 2 (- 1)³ – 8 (- 1) – 5 = 7 + 2 + 8 – 5 = 12

Question 4.
If f (x) = \(\left\{\begin{array}{l}
3 x-2 \text { when } x \leq 0 \\
x+1 \text { when } x>0
\end{array}\right.\), f (-1) and f (0).
Solution:
Given f (x) = \(\left\{\begin{array}{c}
3 x-2 ; x \leq 0 \\
x+1 ; x>0
\end{array}\right.\)
When x = 1 < 0 ∴ f (x) = 3x – 2 ⇒ f (-1 ) = – 3 – 2 = – 5
When x = 0 ∴ f (0) = 3 x 0 – 2 = – 2

Question 5.
If f (x) = log \(\left(\frac{1-x}{1+x}\right)\), show that f (a) + f (b) = f\(\left(\frac{a+b}{1+a b}\right)\).
Solution:

Question 6.
If f (x) = 2x\(\sqrt{1-x^2}\), then show that f \(\left(\sin \frac{x}{2}\right)\) = sin x.
Solution:
Given f (x) = 2x\(\sqrt{1-x^2}\)
∴ \(f\left(\sin \frac{x}{2}\right)=2 \sin \frac{x}{2} \sqrt{1-\sin ^2 \frac{x}{2}}=2 \sin \frac{x}{2} \cos \frac{x}{2}=\sin x\)

Question 7.
If f(x) = cos (log x), then prove that f\(\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]\) = 0
Solution:

Question 8.
If y = f (x) = \(\frac{5 x+3}{4 x-5}\), then show that f (y) = x.
Solution:
Given y = f (x) = \(\frac{5 x+3}{4 x-5}\)
⇒ y (4x – 5) = 5x + 3 ⇒ 4xy – 5y = 5x + 3 ⇒ 4xy – 5x = 5y + 3
⇒ x (4y – 5) = 5y + 3 ⇒ x = \(\frac{5 y+3}{4 y-5}\) = f (y) [using (1)]

Question 9.
If f (x) = x² + kx + 1, for all x and if it is an even function, find k.
Solution:
Given f (x) = x² + kx + 1
since f (x) be an even function
∴ f (- x) = f (x)
⇒ (-x)² – kx + 1 = x² + kx + 1
⇒ x² – kx + 1 = x² + kx + 1
⇒ 2kx = 0
⇒ k = 0

Question 10.
If f(x) = x³ – (k – 2) x² + 2x, for all x and if it is an odd function, find k.
Solution:
Given f (x) = x³ – (k – 2) x² + 2x
∴ f(-x) = (-x)³ – (k – 2) (-x)² – 2x = -x³ – (k – 2) x² – 2x
Since f be an odd function
∴ f(- x) = – f(x)
⇒ – x³ – (k – 2) x² – 2x = – [x³ – (k – 2) x² + 2x]
⇒ – x³ – (k – 2) x² – 2x = – x³ + (k – 2) x² – 2x
⇒ 2(k – 2) x² = 0 ⇒ k – 2 = 0 ⇒ k = 2 ∀x

Question 11.
Is there a function f which is both even and odd ?
Solution:
Given f is an even function
∴ f (- x) = f (x) ∀ x … (1)
Also f is given to be an odd function.
∴ f(- x) = – f(x) … (2)
From eqn. (1) and eqn. (2) ; we have
f(x) = – f(x) ⇒ 2 f (x) = 0 ⇒ f (x) = 0 ∀ x
Thus f be a zero function.

Question 12.
The function f (x) = log (x + \(\sqrt{x^2+1}\)), is (a) an even function (A) an odd function (c) a periodic function (d) Neither an even nor an odd function.
Solution:

Question 13.
Prove that f (x) = (1/x)log\(\sqrt{x+\sqrt{x^2+1}}\) is an even function.
Solution: