Question

Fill in the blanks.
(i) The number of solutions of the equation
$2sin^{-1}\left(\sqrt{x^{2}-x+1}\right)+cos^{-1}\left(\sqrt{x^{2}-x}\right) = \frac{3\pi}{2}$ is P.
(ii) The number of real roots of
$tan^{-1}\left(x+\frac{2}{x}\right)-tan^{-1} \frac{4}{x}-tan^{-1}\left(x-\frac{2}{x}\right) = 0$ is Q.
(iii) If $cos\left(tan^{-1}\,x+cot^{-1}\,\sqrt{3}\right) = 0$, then value of $x$ is R.
(iv) The number of solutions of
$sin[2 \,cos^{-1} \{cot(2\, tan^{-1} \,x\}] = 0$ is S.

	P	Q	R	S
(a)	3	2	$3\sqrt{2}$	6
(b)	2	2	$\sqrt{3}$	6
(c)	2	1	$2\sqrt{3}$	4
(d)	3	1	$\sqrt{2}$	4

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

(i) $sin^{-1} \sqrt{x}$ and $cos^{-1} \sqrt{x}$ defined for $x \le 1$ and $x \ge 0$
$\therefore \sqrt{x^{2}-x+1} \le 1$ and $\sqrt{x^{2}-x} \ge 0$
$\Rightarrow x^{2}-x \le 0$ and $x^{2}-x \ge 0$
$\Rightarrow x^{2}-x = 0$
$\Rightarrow x = 1, 0$
$\therefore $ There are two solutions, both satisfies the equation,
(ii) The equation can be written as
$tan^{-1} \left(x+\frac{2}{x}\right) - tan^{-1} \left(x-\frac{2}{x}\right) = tan^{-1} \frac{4}{x}$
$\Rightarrow tan^{-1} \left[\frac{\left(x+\frac{2}{x}\right)-\left(x--\frac{2}{x}\right)}{1+\left(x+\frac{2}{x}\right)\left(x-\frac{2}{x}\right)}\right] = tan^{-1} \frac{4}{x}$
$\Rightarrow tan^{-1}\left[\frac{\frac{4}{x}}{1+\left(x^{2}-\frac{4}{x^{2}}\right)}\right] = tan^{-1} \frac{4}{x}$
$\Rightarrow \frac{\frac{4}{x}}{1+x^{2}-\frac{4}{x^{2}}} = \frac{4}{x}$
$\Rightarrow 1+x^{2}-\frac{4}{x^{2}} = 1$
$\Rightarrow x^{4} - 4 = 0$
$\Rightarrow x^{2} =2$
$\Rightarrow x = \pm \sqrt{2}$
Thus there are two real roots.
(iii) Since, $cos\left(tan^{-1}\, x + cot^{-1}\sqrt{3}\right) = 0$
$\Rightarrow tan^{-1}\, x + cot^{-1}\sqrt{3} = cos^{-1}\,0$
$\Rightarrow tan^{-1}\, x + cot^{-1}\sqrt{3} = \frac{\pi}{2}$
$\Rightarrow tan^{-1}\, x = \frac{\pi}{2}- cot^{-1}\sqrt{3}$
$\Rightarrow tan^{-1}\, x = tan^{-1}\,\sqrt{3}$
$\Rightarrow x = \sqrt{3}$
(iv) We have, $sin[2cos^{-1} \{cot(2tan^{-1}\, x)\}] = 0$
$\Rightarrow sin\left[2\,cos^{-1}\left\{cot\left(tan^{-1}\left(\frac{2x}{1-x^{2}}\right)\right)\right\}\right] = 0$
$\Rightarrow sin\left[2\,cos^{-1}\left\{cot\left(cot^{-1}\left(\frac{1-x^{2}}{2x}\right)\right)\right\}\right] = 0$
$\Rightarrow sin \left[2\,cos^{-1}\left\{\frac{1-x^{2}}{2x}\right\}\right] = 0$
$\Rightarrow sin\left[sin^{-1}\left[2\left\{\frac{1-x^{2}}{2x}\right\}\sqrt{1-\left\{\frac{1-x^{2}}{2x}\right\}^{2}}\right]\right] = 0$
$\Rightarrow \left[\frac{1-x^{2}}{2x}\right] \sqrt{1-\left\{\frac{1-x^{2}}{2x}\right\}^{2}} = 0$
$\Rightarrow \frac{1-x^{2}}{2x} = 0$ or $\left(1-x^{2}\right)^{2} = 4x^{2}$
$\Rightarrow 1 - x^{2} = 0$ or $\left(1- x^{2}\right)^{2} - \left(2x\right)^{2} = 0$
$\Rightarrow x = \pm1$ or $\left[\left(1 - x^{2}\right) - \left(2x\right)\right] \left[\left(1 - x^{2}\right) + \left(2x\right)\right] = 0$
$\Rightarrow x = \pm1$ or $\left[\left(1 - x^{2}\right) - \left(2x\right)\right] = 0$ or $\left[\left(1 - x^{2}\right) + \left(2x\right)\right] = 0$
$\Rightarrow x = \pm1$ or $x^{2} + 2x - 1 = 0$ or $x^{2} - 2x - 1 = 0$
$\Rightarrow x = \pm 1$ or $x = -1\pm \sqrt{2}, x = 1\pm \sqrt{2}$