Question

Match the entries of Column - I and Column - II.
Number of real solutions of

	Column - I		Column - II
a	tan $(\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}x)$ + tan $(\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}x)$ = 1	p	0
b	$ta{n}^{-1}$ $\frac{1}{2x+1}$ + $ta{n}^{-1}$ $\frac{1}{4x+1}$ = $ta{n}^{-1}$ $\frac{2}{x^2}$	q	2
c	$ta{n}^{-1}$ $(x+\frac{2}{x})$ - $ta{n}^{-1}$ $(x-\frac{2}{x})$ - $ta{n}^{-1}$ $\frac{4}{x}$ = 0	r	3
d	$ta{n}^{-1}$ (1 - x) + $ta{n}^{-1}$ (1 + x) = $ta{n}^{-1}$ 2x	s	1

• A. (a) $\to$ (p),(b) $\to$ (s), (c) $\to$ (q), (d) $\to$ (r)
• B. (a) $\to$ (s),(b) $\to$ (r), (c) $\to$ (q), (d) $\to$ (p)
• C. (a) $\to$ (c),(b) $\to$ (p), (c) $\to$ (q), (d) $\to$ (s)
• D. (a) $\to$ (p),(b) $\to$ (r), (c) $\to$ (q), (d) $\to$ (s)

Answer 1

The correct option is D (a) $\to$ (p),(b) $\to$ (r), (c) $\to$ (q), (d) $\to$ (s)

(a) $\to$ (p),(b) $\to$ (r), (c) $\to$ (q), (d) $\to$ (s)
Let $\frac{1}{2}co{s}^{-1}$ x = $\theta$ $\therefore$ x = cos 2$\theta$
(a) $\frac{1+tan\theta }{1-tan\theta }$ + $\frac{1-tan\theta }{1+tan\theta }$ = $\frac{2(1+ta{n}^2\theta )}{(1-ta{n}^2\theta )}$
L.H.S = $\frac{2}{cos2\theta }$ = $\frac{2}{x}=1$ $\therefore$ x = 2 = cos 2 $\theta$
Since cos 2 $\theta$ $\le$ $\therefore$ there does not exist any solution.
(b) $ta{n}^{-1}$ $\frac{\frac{1}{2x+1}+\frac{1}{4x+1}}{1-\frac{1}{(2x+1)(4x+1)}}$
= $ta{n}^{-1}$ $\frac{6x+2}{8x^2+6x}$ = $ta{n}^{-1}$ $\frac{2}{x^2}$
$\therefore$ $x^2$ (3x + 1) = $8x^2+6x$ or $x(3^2-7x+2)$ = 0 or x(x - 2) (3x - 1)
$\therefore$ x = 0, $\frac{1}{3}$, 2 $\therefore$ (b) $\to$ (r)
(c) $\frac{(x+\frac{2}{x})-(x-\frac{2}{x})}{1+(x^2-\frac{4}{x^2})}$ = $\frac{4}{x}$
$\frac{4}{x}$ = $\frac{4}{x}$ $(1+x^2-\frac{4}{x^2})$
1 = 1 + $x^2$ - $\frac{4}{x^2}$ $\therefore$ $x^4$ = 4 or $x^2$ = 2 or x = $\pm$ $\sqrt{2}$
$\therefore$ (c) $\to$ (q)
(d) $\frac{(1-x)+(1+x)}{1-(1-x^2)}$ = 2x or $\frac{2}{x^2}$ = 2x or $x^3=1$
$\therefore$ x = 1 only $\therefore$ (d) $\to$ (s)