Let f(x) = sin^4 x + cos^4 x. Then f is an in | vedclass

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

Question

(C) Given $f(x) = \sin^4 x + \cos^4 x$.We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos

Vedclass · Accepted Answer

$\left[ \frac{\pi}{4}, \frac{\pi}{2} \right]$

Vedclass · Answer

(C) Given $f(x) = \sin^4 x + \cos^4 x$.We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2 2x$.Now,differentiate with respect to $x$:$f'(x) = \frac{d}{dx} (1 - \frac{1}{2}\sin^2 2x) = 0 - \frac{1}{2} \cdot 2\sin 2x \cdot \cos 2x \cdot 2 = -2\sin 2x \cos 2x = -\sin 4x$.For $f(x)$ to be an increasing function,we must have $f'(x) > 0$.So,$-\sin 4x > 0 \implies \sin 4x We know that $\sin 	heta Therefore,$\pi Dividing by $4$,we get $\frac{\pi}{4} Thus,$f(x)$ is an increasing function in the interval $\left[ \frac{\pi}{4}, \frac{\pi}{2} ight]$.

Let $f(x) = \sin^4 x + \cos^4 x$. Then $f$ is an increasing function in the interval:

Similar Questions

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} -\frac{4}{3}x^3 + 2x^2 + 3x, & x > 0 \\ 3xe^x, & x \leq 0 \end{cases}$. Then $f$ is an increasing function in the interval:

Find the intervals in which the function $f(x) = x^{2} + 2x - 5$ is strictly increasing or strictly decreasing.

For a real number $a$,if a real-valued function $f(x) = 4x^3 + ax^2 + 3x - 2$ is monotonic in its domain,then the range of $a$ is

The function $f(x) = x \cdot e^{x(1-x)}$ is

Given $f(x) = \int_{-2}^{x} t \cdot g'(t) \, dt$ for $x \geq -2$,where $g$ is an increasing function,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module