The point in the interval [0, 2pi], where f(x | vedclass

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

Question

(B) To find the point where the slope is maximum, we need to maximize $f'(x)$. Let $g(x) = f'(x) = e^x(\sin x + \cos x)$.For $g(x)$ to have a maximum,

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(B) To find the point where the slope is maximum, we need to maximize $f'(x)$. Let $g(x) = f'(x) = e^x(\sin x + \cos x)$.For $g(x)$ to have a maximum, we set $g'(x) = f''(x) = 0$.$f'(x) = e^x(\sin x + \cos x)$$f''(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.Setting $f''(x) = 0$ gives $2e^x \cos x = 0$. Since $e^x 
eq 0$, we have $\cos x = 0$.In the interval $[0, 2\pi]$, $\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.We check the second derivative of $g(x)$, which is $g'(x) = f''(x) = 2e^x \cos x$.$g''(x) = f'''(x) = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x)$.At $x = \frac{\pi}{2}$, $g''(\frac{\pi}{2}) = 2e^{\pi/2}(0 - 1) = -2e^{\pi/2} At $x = \frac{3\pi}{2}$, $g''(\frac{3\pi}{2}) = 2e^{3\pi/2}(0 - (-1)) = 2e^{3\pi/2} > 0$ (Local minimum).Thus, the slope is maximum at $x = \frac{\pi}{2}$.

The point in the interval $[0, 2\pi]$, where $f(x) = e^x \sin x$ has maximum slope, is

Similar Questions

The function $S(x) = \int\limits_0^x {\sin \left( {\frac{{\pi {t^2}}}{2}} \right)\,dt} $ has two critical points in the interval $[1, 2.4]$. One of the critical points is a local minimum and the other is a local maximum. The local minimum occurs at $x =$

What is the maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$?

The height of a right circular cone of maximum volume inscribed in a sphere of diameter $a$ is-

The perimeter of a sector is a constant. If its area is to be maximum,then the sectorial angle is

Let $f(x) = x^2 e^{-2x}, x > 0$. The maximum value of $f(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module