Question

Let $y=e^{x^2}$ and $y=e^{x^2}\sin x$ be two given curves. Then, angle between the tangents to the curves at any point their intersection is

• A. $0$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (A)

$0$

For intersecting points, $e^{x^2}=e^{x^2}\sin x$
$\Rightarrow e^{x^2}(\sin x-1)=0$
$\Rightarrow e^{x^2}=0$ or $\sin x=1$

But $e^{x^2}\ne 0\Rightarrow \sin x=1$
$\Rightarrow x=\frac{\pi }{2}$

Now,

$y=e^{x^2}$

$\therefore \frac{dy}{dx}=e^{x^2}.2x=2x{e}^{x^2}$ $[\because \sin x=1,\cos x=0]$

Since, both the curves has equal slope.
Hence, angle between the tangents at intersecting point is $0$.