Question

If $x={\int }_0^{t2}{e}^{\sqrt{z}}\left\{\frac{2\tan \sqrt{z}+1-{\tan }^2\sqrt{z}}{2\sqrt{z}{\sec }^2\sqrt{z}}\right\}dz$ &

y=${\int }_0^{t^2}{e}^{\sqrt{z}}\left\{\frac{1-{\tan }^2\sqrt{z}-2\tan \sqrt{z}}{2\sqrt{z}{\sec }^2\sqrt{z}}\right\}dz.$ Then the inclination of the tangent to the curve at t $=\frac{\pi }{4}$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{3\pi }{4}$

Answer 1

Answer: (D)

$\frac{3\pi }{4}$

$x={\int }_0^{t^2}{e}^{\sqrt{2}}\left\{\frac{2\tan \sqrt{z}+1-{\tan }^2\sqrt{z}}{2\sqrt{z}{\sec }^2\sqrt{z}}\right\}dz$
$\Rightarrow \frac{dx}{dt}=2t{e}^t\left\{\frac{2\tan t+1-{\tan }^{2}t}{2t{\sec }^{2}t}\right\}$
$y={\int }_0^{t^2}{e}^{\sqrt{2}}\left\{\frac{1-{\tan }^2\sqrt{z}-2\tan \sqrt{z}}{2\sqrt{z}{\sec }^2\sqrt{z}}\right\}$
$\Rightarrow \frac{dy}{dt}=2t{e}^t\left\{\frac{1-{\tan }^{2}t-2\tan t}{2t{\sec }^{2}t}\right\}$
Therefore $\frac{dy}{dx}=\frac{1-{\tan }^{2}t-2\tan t}{1-{\tan }^{2}t+2\tan t}$
At $x=\frac{\pi }{4}$ $\frac{dy}{dx}=\frac{1-1-2}{1-1+2}=-1$
Therefore angle $=\tan (-1)=\frac{3\pi }{4}$