Question

If ${\int }_{\pi /2}^x\sqrt{3-2{\sin }^{2}zdz}+{\int }_0^y\cos tdt=0$

then ${\left|\frac{dy}{dx}\right|}_{(\pi /2,\pi )}$ is?

• A. $1$
• B. $2$
• C. $\sqrt{3}$
• D. $0$

Answer 1

The correct option is A $1$

${\int }_{\frac{\pi }{2}}^x\sqrt{3-2{\sin }^{2}zdx}+{\int }_0^y\cot tdt=0$
Differentiating, we get
$\sqrt{3-2{\sin }^{2}x}dx+\cos ydy=0$
$\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{3-2{\sin }^{2}x}}{\cos y}$
At $(\frac{\pi }{2},\pi )$
$\frac{dy}{dx}=-\frac{\sqrt{3-2{\sin }^2\left(\frac{\pi }{2}\right)}}{\cos \left(\pi \right)}=1$