Question

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

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

Answer 1

The correct option is C $n=0,2,4$

$x=\frac{n\pi }{2},$ satisfied $\sin \frac{x}{2}-\cos \frac{x}{2}=1-\sin x$

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

at $x=0,|\frac{0}{2}-\frac{\pi }{2}|=\frac{\pi }{2}<\frac{3\pi }{4}$

$x=1,|\frac{\pi }{2}-\frac{\pi }{2}|=0$

$\Rightarrow \sin 0-\cos 0=1-\sin 0$

$\Rightarrow -1\ne 1$

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

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

$\Rightarrow \sin \frac{3\pi }{4}-\cos \frac{3\pi }{4}=1-\sin \frac{3\pi }{2}$

$\Rightarrow \sqrt{2}-\sqrt{2}=1-1$

$\Rightarrow 0=0$

Hence, the answer is $x=0,2,4.$