Question

Fill in the blanks.
(i) The number of solutions of the equation
$2si{n}^{-1}\left(\sqrt{x^2-x+1}\right)+co{s}^{-1}\left(\sqrt{x^2-x}\right)=\frac{3\pi }{2}$ is P.
(ii) The number of real roots of
$ta{n}^{-1}\left(x+\frac{2}{x}\right)-ta{n}^{-1}\frac{4}{x}-ta{n}^{-1}\left(x-\frac{2}{x}\right)=0$ is Q.
(iii) If $cos\left(ta{n}^{-1}x+co{t}^{-1}\sqrt{3}\right)=0$, then value of $x$ is R.
(iv) The number of solutions of
$sin[2co{s}^{-1}\{cot(2ta{n}^{-1}x\}]=0$ 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) $si{n}^{-1}\sqrt{x}$ and $co{s}^{-1}\sqrt{x}$ 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
$ta{n}^{-1}\left(x+\frac{2}{x}\right)-ta{n}^{-1}\left(x-\frac{2}{x}\right)=ta{n}^{-1}\frac{4}{x}$
$\Rightarrow ta{n}^{-1}\left[\frac{\left(x+\frac{2}{x}\right)-\left(x--\frac{2}{x}\right)}{1+\left(x+\frac{2}{x}\right)\left(x-\frac{2}{x}\right)}\right]=ta{n}^{-1}\frac{4}{x}$
$\Rightarrow ta{n}^{-1}\left[\frac{\frac{4}{x}}{1+\left(x^2-\frac{4}{x^2}\right)}\right]=ta{n}^{-1}\frac{4}{x}$
$\Rightarrow \frac{\frac{4}{x}}{1+x^2-\frac{4}{x^2}}=\frac{4}{x}$
$\Rightarrow 1+x^2-\frac{4}{x^2}=1$
$\Rightarrow x^4-4=0$
$\Rightarrow x^2=2$
$\Rightarrow x=\pm \sqrt{2}$
Thus there are two real roots.
(iii) Since, $cos\left(ta{n}^{-1}x+co{t}^{-1}\sqrt{3}\right)=0$
$\Rightarrow ta{n}^{-1}x+co{t}^{-1}\sqrt{3}=co{s}^{-1}0$
$\Rightarrow ta{n}^{-1}x+co{t}^{-1}\sqrt{3}=\frac{\pi }{2}$
$\Rightarrow ta{n}^{-1}x=\frac{\pi }{2}-co{t}^{-1}\sqrt{3}$
$\Rightarrow ta{n}^{-1}x=ta{n}^{-1}\sqrt{3}$
$\Rightarrow x=\sqrt{3}$
(iv) We have, $sin[2co{s}^{-1}\{cot(2ta{n}^{-1}x)\}]=0$
$\Rightarrow sin\left[2co{s}^{-1}\left\{cot\left(ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)\right)\right\}\right]=0$
$\Rightarrow sin\left[2co{s}^{-1}\left\{cot\left(co{t}^{-1}\left(\frac{1-x^2}{2x}\right)\right)\right\}\right]=0$
$\Rightarrow sin\left[2co{s}^{-1}\left\{\frac{1-x^2}{2x}\right\}\right]=0$
$\Rightarrow sin\left[si{n}^{-1}\left[2\left\{\frac{1-x^2}{2x}\right\}\sqrt{1-{\left\{\frac{1-x^2}{2x}\right\}}^2}\right]\right]=0$
$\Rightarrow \left[\frac{1-x^2}{2x}\right]\sqrt{1-{\left\{\frac{1-x^2}{2x}\right\}}^2}=0$
$\Rightarrow \frac{1-x^2}{2x}=0$ or ${\left(1-x^2\right)}^2=4x^2$
$\Rightarrow 1-x^2=0$ or ${\left(1-x^2\right)}^2-{\left(2x\right)}^2=0$
$\Rightarrow x=\pm 1$ or $\left[\left(1-x^2\right)-\left(2x\right)\right]\left[\left(1-x^2\right)+\left(2x\right)\right]=0$
$\Rightarrow x=\pm 1$ or $\left[\left(1-x^2\right)-\left(2x\right)\right]=0$ or $\left[\left(1-x^2\right)+\left(2x\right)\right]=0$
$\Rightarrow x=\pm 1$ or $x^2+2x-1=0$ or $x^2-2x-1=0$
$\Rightarrow x=\pm 1$ or $x=-1\pm \sqrt{2},x=1\pm \sqrt{2}$