Question

The smallest positive root of the equation $\sqrt{sin(1-x)}=\sqrt{cosx}$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The given equation is possible if sin$(1-x)>0$and cos$\left(x\right)>0$

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

$\Rightarrow$ 1 - x = $n\pi$ + $(-1{)}^n$ $(\frac{\pi }{2}-x)$ where n $\in$ l

But for n = 2m (m $\in$ l) we get no value of x

1 - x = (2m + 1) $\pi$ - $(\frac{\pi }{2}-x)$

$\Rightarrow$ x = $\frac{1}{2}$ - $\left(\frac{4m+1}{4}\right)\pi (m\in l)$

If $m\ge 0$, x < 0

for, m = -1, x = $\frac{1}{2}+\frac{3\pi }{4}$,

1 - x = $\frac{1}{2}-\frac{3\pi }{4}$

So that sin(1-x) = sin $(\frac{1}{2}+\frac{\pi }{4}-\pi )$

= -sin $(\frac{\pi }{4}+\frac{1}{2})<0$

for m = -2, x = $\frac{1}{2}+\frac{7\pi }{4}$, 1 - x = $\frac{1}{2}-\frac{7\pi }{4}$

so that, sin(1-x) = sin $(\frac{1}{2}+\frac{\pi }{4}-2\pi )>0$

and cosx = cos $(2\pi -\frac{\pi }{4}+\frac{1}{2})>0$

Hence, x = $\frac{1}{2}+\frac{7\pi }{4}$ is the smallest positive root of the given equation.