If the function f(x) = sin x - cos^2 x is def | vedclass

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

Question

(D) Given $f(x) = \sin x - \cos^2 x = \sin x - (1 - \sin^2 x) = \sin^2 x + \sin x - 1$. Let $t = \sin x$. Since $x \in [-\pi, \pi]$,$t \in [-1, 1]$. T

Vedclass · Accepted Answer

$(-\frac{5\pi}{6}, -\frac{\pi}{2}) \cup (-\frac{\pi}{6}, \frac{\pi}{2})$

Vedclass · Answer

(D) Given $f(x) = \sin x - \cos^2 x = \sin x - (1 - \sin^2 x) = \sin^2 x + \sin x - 1$. Let $t = \sin x$. Since $x \in [-\pi, \pi]$,$t \in [-1, 1]$. Then $g(t) = t^2 + t - 1$. For $f(x)$ to be strictly increasing,we need $f'(x) > 0$. $f'(x) = \cos x + 2 \cos x \sin x = \cos x(1 + 2 \sin x)$. Setting $f'(x) > 0$: Case $1$: $\cos x > 0$ and $1 + 2 \sin x > 0$. $\cos x > 0 \implies x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. $1 + 2 \sin x > 0 \implies \sin x > -\frac{1}{2} \implies x \in (-\frac{\pi}{6}, \frac{7\pi}{6})$. Intersection: $x \in (-\frac{\pi}{6}, \frac{\pi}{2})$. Case $2$: $\cos x $\cos x $1 + 2 \sin x Intersection: $x \in (-\frac{5\pi}{6}, -\frac{\pi}{2})$. Combining both cases,$f(x)$ is strictly increasing in $(-\frac{5\pi}{6}, -\frac{\pi}{2}) \cup (-\frac{\pi}{6}, \frac{\pi}{2})$. Thus,the correct option is $D$.

If the function $f(x) = \sin x - \cos^2 x$ is defined on the interval $[-\pi, \pi]$,then $f$ is strictly increasing in the interval

Similar Questions

If $f(x) = \log(1 + x) - \frac{x}{1 + x}$, then the values of $x$ for which $f(x)$ is monotonically increasing and monotonically decreasing are respectively.....

If $2f(x) + 3f\left(\frac{1}{x}\right) = x^2 + 1, x \neq 0$ and $y = 5x^2 f(x)$,then $y$ is strictly increasing in

The function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$ is increasing,if

The interval in which $y=x^{2} e^{-x}$ is increasing is

For what value of $a$ is the function $f(x) = (a + 2)x^3 - 3ax^2 + 9ax - 1$ a decreasing function for all $x \in R$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module