Question

The smallest positive number p for which the equation cos (p sin x) = sin (p cos x) has a solution in $[0,2\pi ]$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4\sqrt{2}}$
• D. $\frac{\pi }{2\sqrt{2}}$

Answer 1

The correct option is D $\frac{\pi }{2\sqrt{2}}$

$\begin{array}{cc}& \begin{array}{cc}& -\cos \left(p\sin x\right)=\sin \left(p\cos x\right)\\ \Rightarrow & \cos \left(p\sin x\right)=\cos (\frac{\pi }{2}-p\cos x)\end{array}\\ & 2sinx\pm (\frac{\pi }{2}-p\cos x)(n⩾0)\\ & \Rightarrow p\sin x+p\cos x=2n\pi +\frac{\pi }{2}\\ & 4\\ & p\sin x-p\cos x=2n\pi -\frac{\pi }{2}\\ & \Rightarrow p\sqrt{2}\left(\sin (x+\frac{\pi }{4})\right)=\frac{(4n+1)}{2}\pi \end{array}$

or $\frac{\sqrt{2}\left(\sin (x-\frac{\pi }{4})\right)}{=}\frac{(4n-1)n}{2}$

As $-1\le \sin (x+\frac{\pi }{4})⩽1$

$\Rightarrow -p\sqrt{2}\le p\sqrt{2}\sin (x+\frac{\pi }{4})⩽p\sqrt{2}$

$\Rightarrow -p\sqrt{2}\le \left(\frac{4n+1}{2}\right)\pi \le p\sqrt{2}$... (i)

And similarly,

$\Rightarrow -p\sqrt{2}\le {\left(\frac{4n-1}{2}\right)}^n\le p\sqrt{2}$

(ii) is subset of (i),

$\therefore$ We can consider only (i).

If $\begin{array}{cc}n⩾0,& \left(\frac{4n+1}{2}\right)\pi \Rightarrow \left(\frac{4n+1}{2}\right)n⩽\sqrt{2}p\end{array}$

$\Rightarrow \sqrt{2}p⩾\frac{\pi }{2}$

$\Rightarrow p⩾\frac{\pi }{2\sqrt{2}}\Rightarrow p⩾\frac{\pi \sqrt{2}}{4}$

$\therefore$ option $D$ is correct.