The curve f(x) = e^x sin x is defined in the | vedclass

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

Question

(B) Given: $f(x) = e^x \sin x$. The slope of the tangent is given by $m(x) = f'(x)$. $f'(x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x)$. Let $g

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(B) Given: $f(x) = e^x \sin x$. The slope of the tangent is given by $m(x) = f'(x)$. $f'(x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x)$. Let $g(x) = f'(x) = e^x (\sin x + \cos x)$. To find the maximum slope,we find $g'(x) = 0$. $g'(x) = e^x (\sin x + \cos x) + e^x (\cos x - \sin x) = 2 e^x \cos x$. Setting $g'(x) = 0$,we get $2 e^x \cos x = 0$. Since $e^x 
eq 0$ for all $x$,we have $\cos x = 0$. In the interval $[0, 2 \pi]$,$x = \frac{\pi}{2}$ or $x = \frac{3 \pi}{2}$. Now,we check the second derivative $g''(x) = 2 e^x \cos x - 2 e^x \sin x = 2 e^x (\cos x - \sin x)$. At $x = \frac{\pi}{2}$,$g''(\frac{\pi}{2}) = 2 e^{\frac{\pi}{2}} (0 - 1) = -2 e^{\frac{\pi}{2}} At $x = \frac{3 \pi}{2}$,$g''(\frac{3 \pi}{2}) = 2 e^{\frac{3 \pi}{2}} (0 - (-1)) = 2 e^{\frac{3 \pi}{2}} > 0$,which indicates a local minimum. Thus,the slope is maximum at $x = \frac{\pi}{2}$.

The curve $f(x) = e^x \sin x$ is defined in the interval $[0, 2 \pi]$. The value of $x$ for which the slope of the tangent drawn to the curve at $x$ is maximum,is

Similar Questions

If the sum of two numbers is $3$,then the maximum value of the product of the first and the square of the second is:

The shortest distance of the point $(0,0)$ from the curve $y = e^x + e^{-x}$ is-

The function $f(x) = \int_{-1}^x {t({e^t} - 1)(t - 1){(t - 2)}^3{(t - 3)}^5} dt$ has a local minimum at $x = $ ..........

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is

If the area of a right-angled triangle with hypotenuse $5$ is maximum,then its perimeter is

Vedclass Test Series

Exam Paper Generator

Online Exam Module