The maximum value of sin x(1 + cos x) occurs | vedclass

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

Question

(C) Let $f(x) = \sin x(1 + \cos x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$.To find the maximum,we find the derivative $f'(x)$ and set

Vedclass · Accepted Answer

$x = \frac{\pi}{3}$

Vedclass · Answer

(C) Let $f(x) = \sin x(1 + \cos x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$.To find the maximum,we find the derivative $f'(x)$ and set it to $0$:$f'(x) = \cos x + \cos 2x = 0$.Using the identity $\cos 2x = 2 \cos^2 x - 1$,we get:$2 \cos^2 x + \cos x - 1 = 0$.Let $u = \cos x$,then $2u^2 + u - 1 = 0$,which factors as $(2u - 1)(u + 1) = 0$.So,$\cos x = \frac{1}{2}$ or $\cos x = -1$.For $\cos x = \frac{1}{2}$,$x = \frac{\pi}{3}$ (in the first quadrant).For $\cos x = -1$,$x = \pi$,but at $x = \pi$,$f(x) = 0$,which is the minimum.Checking the second derivative $f''(x) = -\sin x - 2\sin 2x$:At $x = \frac{\pi}{3}$,$f''(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) - 2\sin(\frac{2\pi}{3}) = -\frac{\sqrt{3}}{2} - 2(\frac{\sqrt{3}}{2}) = -\frac{3\sqrt{3}}{2} Since the second derivative is negative,the function has a maximum at $x = \frac{\pi}{3}$.

The maximum value of $\sin x(1 + \cos x)$ occurs at

Similar Questions

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

The maximum value of the function $f(x)=3x^3-18x^2+27x-40$ on the set $S=\{x \in R : x^2+30 \leq 11x\}$ is

The maximum value of ${\left( {\frac{1}{x}} \right)^x}$ is

$A$ wire of length $2$ units is cut into two parts,which are bent respectively to form a square of side $x$ units and a circle of radius $r$ units. If the sum of the areas of the square and the circle so formed is minimum,then:

The maximum value of $f(x) = \frac{x}{4 + x + x^2}$ on the interval $[-1, 1]$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module