Question

$\begin{array}{cccc}& \text{List - I}& & \text{List - II}\\ \text{(I)}& \text{Number of solutions of the equation}& \text{(P)}& 0\\ & e^x+e^{-x}=\tan x\forall x\in [0,\frac{\pi }{2})& & \\ \text{(II)}& \text{Number of solutions of the equations}& \text{(Q)}& 1\\ & x+y=\frac{2\pi }{3}\text{and}\cos x+\cos y=\frac{3}{2}\text{is}& & \\ \text{(III)}& \text{Number of solutions of the equation}& \text{(R)}& 2\\ & \cos x+2\sin x=1,x\in [0,2\pi )\text{is}& & \\ \text{(IV)}& \text{Number of solutions of the equation}& \text{(S)}& \text{Infinite}\\ & (\sqrt{3}\sin x+\cos x{)}^{\sqrt{\sqrt{3}\sin 2x-\cos 2x+2}}=4\text{is}& & \end{array}$

Which of the following is only INCORRECT combination?

• A. $\left(I\right)\to \left(Q\right)$
• B. $\left(II\right)\to \left(P\right)$
• C. $\left(III\right)\to \left(S\right)$
• D. $\left(IV\right)\to \left(S\right)$

Answer 1

The correct option is C $\left(III\right)\to \left(S\right)$

$e^x+e^{-x}=\tan x$

only one solution

$x+y=\frac{2\pi }{3}$
$\cos x+\cos y=\frac{3}{2}\Rightarrow \cos x+\cos (\frac{2\pi }{3}-x)=\frac{3}{2}$
$\Rightarrow \cos x+\sqrt{3}\sin x=3$
$\sin (x+\frac{\pi }{6})=6$, which is not possible.

$\cos x+2\sin x=1$
$\Rightarrow 2{\sin }^2\frac{x}{2}=2\sin x\Rightarrow \sin \frac{x}{2}(\sin \frac{x}{2}-2\cos \frac{x}{2})=0$
$\sin \left(\frac{x}{2}\right)=0,\tan \left(\frac{x}{2}\right)=2$
So two solutions

$\sqrt{3}\sin 2x-\cos 2x+2=2-2\cos (\frac{\pi }{3}+2x)=2.2{\sin }^2(\frac{\pi }{6}+2x)$
$\sqrt{\sqrt{3}\sin 2x-\cos 2x+2}=2\left|\sin (\frac{\pi }{6}+2x)\right|$
$\sqrt{3}\sin x+\cos x=2\sin (x+\frac{\pi }{6})$
So $\sin (x+\frac{\pi }{6})=y,(2y{)}^{2y}=2^2\Rightarrow y=1$
$\sin (\frac{\pi }{6}+x)=1$, which has infinite number of solutions.