Question

$$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{a. For}\theta \in R\text{, if}max\{5\sin \theta +3\sin (\theta -\alpha \left)\right\}=7\text{,}& \\ \text{then the set of possible values of}\alpha \text{is}& \text{p.}2n\pi +\frac{3\pi }{4},n\in Z\\ \text{b.}x\ne \frac{n\pi }{2}\text{and}(\cos x{)}^{{\sin }^{2}x-3\sin x+2}=1& \text{q.}2n\pi \pm \frac{\pi }{3},n\in Z\\ \text{c.}\sqrt{\sin x}+2^{1/4}\cos x=0& \text{r.}2n\pi +{\cos }^{-1}\left(\frac{1}{3}\right),n\in Z\\ \text{d.}{\log }_5\tan x=\left({\log }_{5}4\right)\left({\log }_4\right(3\sin x\left)\right)& \text{s. no solution}\end{array}$$

Which of the following is the correct combination ?

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

Answer 1

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

$a.$ $5\sin \theta +3\sin (\theta -\alpha )$
$=5\sin \theta +3(\sin \theta \cos \alpha -\cos \theta \sin \alpha )$
$=(5+3\cos \alpha )\sin \theta -3\sin \alpha \cos \theta$

$\therefore \underset{\theta \in R}{max}\{5\sin \theta +3\sin (\theta -\alpha \left)\right\}$
$=\sqrt{(5+3\cos \alpha {)}^2+9{\sin }^2\alpha }$
$=\sqrt{34+30\cos \alpha }$

Therefore, $34+30\cos \alpha =7^2=49$
$\Rightarrow \cos \alpha =\frac{1}{2}$
$\Rightarrow \alpha =2n\pi \pm \frac{\pi }{3},n\in Z$

$b.$ $(\cos x{)}^{{\sin }^{2}x-3\sin x+2}=1$
$\Rightarrow {\sin }^{2}x-3\sin x+2=0$
$\Rightarrow (\sin x-1)(\sin x-2)=0$
$\Rightarrow \sin x=1$
But this does not satisfy the equation as $0^0=1$ is absurd.

$c.$ $\sqrt{\sin x}+2^{1/4}\cos x=0$
$\because \sqrt{\sin x}>0$,
$\therefore \cos x<0$
Also, $\sin x>0\Rightarrow x$ lies in the $2$nd quadrant.

$\sqrt{\sin x}+2^{1/4}\cos x=0$
$\Rightarrow 2^{1/4}\cos x=-\sqrt{\sin x}$
$\Rightarrow \sqrt{2}{\cos }^{2}x=\sin x$
$\Rightarrow \sqrt{2}{\sin }^{2}x+\sin x-\sqrt{2}=0$
$\Rightarrow (\sqrt{2}\sin x-1)(\sin x+\sqrt{2})=0$
$\Rightarrow \sin x=\frac{1}{\sqrt{2}}$
$\Rightarrow x=2n\pi +\frac{3\pi }{4},n\in Z$

$d.$ ${\log }_5\tan x=\left({\log }_{5}4\right)\left({\log }_4\right(3\sin x\left)\right)$
$\Rightarrow {\log }_5\tan x={\log }_5\left(3\sin x\right)$
For $\log$ to be defined,
$\tan x>0,\sin x>0$
$\Rightarrow x$ lies in the $1$st quadrant.

${\log }_5\tan x={\log }_5\left(3\sin x\right)$
$\Rightarrow \tan x=3\sin x$
$\Rightarrow \cos x=\frac{1}{3}$
$\Rightarrow x=2n\pi +{\cos }^{-1}\frac{1}{3},n\in Z$