The derivative of the function f(x) = cos^4 x | vedclass

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

Question

(B) Given $f(x) = \cos^4 x + \sin^4 x$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$, we have $f(x) = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x =

Vedclass · Accepted Answer

$\frac{\pi}{4} < x < \frac{\pi}{2}$

Vedclass · Answer

(B) Given $f(x) = \cos^4 x + \sin^4 x$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$, we have $f(x) = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2(2x)$. Differentiating with respect to $x$: $f'(x) = 0 - \frac{1}{2} \cdot 2\sin(2x) \cdot \cos(2x) \cdot 2 = -\sin(4x)$. We want $f'(x) > 0$, so $-\sin(4x) > 0$, which implies $\sin(4x) For $0 \le x \le 2\pi$, $0 \le 4x \le 8\pi$. $\sin(4x) For the first interval: $\pi This matches option $B$.

The derivative of the function $f(x) = \cos^4 x + \sin^4 x$ for $0 \le x \le 2\pi$ is positive for:

Similar Questions

If $f$ is a real-valued differentiable function such that $f(x) f^{\prime}(x) < 0$ for all real $x,$ then

In the interval $(-3,3)$,the function $f(x) = \frac{x}{3} + \frac{3}{x}, x \neq 0$ is :

Show that the function given by $f(x) = \sin x$ is neither increasing nor decreasing in $(0, \pi)$.

Let $f$ be a function defined on $[a, b]$ such that $f^{\prime}(x) > 0$ for all $x \in (a, b)$. Prove that $f$ is an increasing function on $(a, b)$.

The interval containing all the real values of $x$ such that the real-valued function $f(x) = \sqrt{x} + \frac{1}{\sqrt{x}}$ is strictly increasing is

Vedclass Test Series

Exam Paper Generator

Online Exam Module