The function f(x)=4 sin ^3 x-6 sin ^2 x+12 si | vedclass

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

Question

(D) Given,$f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$. First,we find the derivative $f'(x)$: $f'(x) = 12 \sin^2 x \cos x - 12 \sin x \cos x + 12 \cos

Vedclass · Accepted Answer

decreasing in $\left(\frac{\pi}{2}, \pi\right)$

Vedclass · Answer

(D) Given,$f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$. First,we find the derivative $f'(x)$: $f'(x) = 12 \sin^2 x \cos x - 12 \sin x \cos x + 12 \cos x$. Factoring out $12 \cos x$: $f'(x) = 12 \cos x (\sin^2 x - \sin x + 1)$. Consider the quadratic expression $g(t) = t^2 - t + 1$ where $t = \sin x$. The discriminant $D = (-1)^2 - 4(1)(1) = 1 - 4 = -3 Since the coefficient of $t^2$ is positive and $D Thus,the sign of $f'(x)$ depends only on $\cos x$. $f'(x) > 0$ when $\cos x > 0$,which occurs in $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. $f'(x) Therefore,$f(x)$ is strictly decreasing in the interval $(\frac{\pi}{2}, \pi)$.

The function $f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$ is strictly

Similar Questions

The function $f(x)=2 x^3-9 x^2+12 x+29$ is monotonically increasing in the interval

Show that the function given by $f(x) = 7x - 3$ is strictly increasing on $R$.

For the function $f(x) = \sin 3x$,where $x \in [0, \frac{\pi}{2}]$,which of the following is true?

If $f(x) = \int_{x^2}^{x^2 + 1} e^{-t^2} dt$,then $f(x)$ increases in

The complete set of values of $x$ for which $f(x) = 2 \log_e(x - 2) - x^2 + 4x + 1$ is increasing is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module