Let y=e^{x^{2}} and y=e^{x^{2}} sin x be two | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the intersection points,we set the two equations equal: $e^{x^{2}} = e^{x^{2}} \sin x$. Since $e^{x^{2}} 
eq 0$ for any real $x$,we divide by

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) For the intersection points,we set the two equations equal: $e^{x^{2}} = e^{x^{2}} \sin x$. Since $e^{x^{2}} 
eq 0$ for any real $x$,we divide by $e^{x^{2}}$ to get $\sin x = 1$. This implies $x = \frac{\pi}{2} + 2n\pi$ for any integer $n$. Let $f(x) = e^{x^{2}}$ and $g(x) = e^{x^{2}} \sin x$. The derivative of $f(x)$ is $f'(x) = 2x e^{x^{2}}$. The derivative of $g(x)$ is $g'(x) = 2x e^{x^{2}} \sin x + e^{x^{2}} \cos x$. At the intersection point where $\sin x = 1$ and $\cos x = 0$,we have: $f'(x) = 2x e^{x^{2}}$ $g'(x) = 2x e^{x^{2}}(1) + e^{x^{2}}(0) = 2x e^{x^{2}}$. Since $f'(x) = g'(x)$ at the point of intersection,the slopes of the tangents are equal. Therefore,the angle between the tangents is $0$.

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

Similar Questions

What is the equation of the normal to the curve $y = x \log x$ which is parallel to the line $2x - 2y + 3 = 0$?

The area (in sq. units) of the triangle formed by the tangent and normal drawn to the curve $(\frac{x}{3})^n+(\frac{y}{4})^n=2$ at $(3,4)$ and the $X$-axis is

The equation of the tangent to the curve $y = \cos(x + y)$ for $-2\pi \leq x \leq 2\pi$,which is parallel to the line $x + 2y = 0$,is:

The length of the normal at any point to the curve $y=c \cosh \left(\frac{x}{c}\right)$ is

The coordinates of the point$(s)$ on the graph of the function $f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4$ where the tangent drawn cuts off intercepts from the coordinate axes which are equal in magnitude but opposite in sign,are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module