Question

If ${\int }_{\pi /2}^x\sqrt{3-2{\sin }^{2}t}dt+{\int }_0^y\cos tdt=0$, then $\left(\frac{dy}{dx}\right)$ at $x=\pi$ and $y=\pi$ is

• A. $\sqrt{3}$
• B. $-\sqrt{2}$
• C. $-\sqrt{3}$
• D. $Noneofthese$

Answer 1

Answer: (A)

$\sqrt{3}$

Here, ${\int }_{\pi /2}^{\pi }\sqrt{(3-2si{n}^{2}t)dt}+{\int }_y^{0}costdt$=0
Differentiating both the sides, we get
$(\sqrt{3-2si{n}^{2}x)}.1+(cosy)\left(\frac{dy}{dx}\right)=0$
$\Rightarrow \frac{dy}{dx}=\frac{-\sqrt{3-2si{n}^{2}x}}{cosy}$
$\therefore {\left(\frac{dy}{dx}\right)}_{(\pi ,\pi )}=\frac{-\sqrt{3}}{-1}=\sqrt{3}$