Question

If $x=\frac{n\pi }{2}$, satisfies the equation $\sin \frac{x}{2}-\cos \frac{x}{2}=1-\sin x$ & the inequality $\mid \frac{x}{2}-\frac{\pi }{2}\mid \le \frac{3\pi }{4}$, then:

• A. $n=-1,0,3,5$
• B. $n=1,2,4,5$
• C. $n=0,2,4$
• D. $m=-1,1,3,5$

Answer 1

Answer: (B)

$n=1,2,4,5$

Given, $\sin \frac{x}{2}-\cos \frac{x}{2}=1-\cos (\frac{\pi }{2}-x)$

$\sqrt{2}\left[\sin \right(\frac{x}{2}-\frac{\pi }{4}\left)\right]=2{\sin }^2(\frac{x}{2}-\frac{\pi }{4})$

$\sqrt{2}\left[\sin \right(\frac{x}{2}-\frac{\pi }{4}\left)\right][1-\sqrt{2}[\sin (\frac{x}{2}-\frac{\pi }{4})]=0$

$\sqrt{2}\left[\sin \right(\frac{x}{2}-\frac{\pi }{4}\left)\right]=0$ $⟹$ $x=\frac{\pi }{2}$ and

$1-\sqrt{2}\left[\sin \right(\frac{x}{2}-\frac{\pi }{4}\left)\right]=0$

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

$\frac{x}{2}-\frac{\pi }{4}=\frac{\pi }{4},\frac{3\pi }{4}$

$x=\pi ,2\pi$.

$|\frac{x}{2}-\frac{\pi }{2}|\le \frac{3\pi }{4}$.

Hence

$-\frac{3\pi }{4}\le \frac{x}{2}-\frac{\pi }{2}\le \frac{3\pi }{4}$

$-\frac{\pi }{2}\le x\le \frac{5\pi }{2}$

Hence $n=1,2,3,4,5$.