If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$,then $f'\left( \frac{\pi}{4} \right)$ is

  • A
    $\sqrt{2}$
  • B
    $\frac{1}{\sqrt{2}}$
  • C
    $1$
  • D
    $\frac{\sqrt{3}}{2}$

Explore More

Similar Questions

Let $f:\left[-\frac{1}{2}, 2\right] \rightarrow R$ and $g:\left[-\frac{1}{2}, 2\right] \rightarrow R$ be functions defined by $f(x)=\left[x^2-3\right]$ and $g(x)=|x| f(x)+|4 x-7| f(x)$,where $[y]$ denotes the greatest integer less than or equal to $y$ for $y \in R$. Then
$(A)$ $f$ is discontinuous exactly at three points in $\left[-\frac{1}{2}, 2\right]$
$(B)$ $f$ is discontinuous exactly at four points in $\left[-\frac{1}{2}, 2\right]$
$(C)$ $g$ is $NOT$ differentiable exactly at four points in $\left(-\frac{1}{2}, 2\right)$
$(D)$ $g$ is $NOT$ differentiable exactly at five points in $\left(-\frac{1}{2}, 2\right)$

If $F(x) = \left(f\left(\frac{x}{2}\right)\right)^2 + \left(g\left(\frac{x}{2}\right)\right)^2$,where $f^{\prime \prime}(x) = -f(x)$ and $g(x) = f^{\prime}(x)$,and given $F(5) = 5$,then $F(10)$ is equal to

Let $f:R \to R$ be a continuous function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$.
Statement-$1$: $f(c) = \frac{1}{3}$ for some $c \in R$.
Statement-$2$: $0 < f(x) < \frac{1}{2\sqrt{2}}$ for all $x \in R$.

Let $f(x) = (\sin(\tan^{-1} x) + \sin(\cot^{-1} x))^2 - 1$ for $|x| > 1$. If $\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx}(\sin^{-1}(f(x)))$ and $y(\sqrt{3}) = \frac{\pi}{6}$,then $y(-\sqrt{3})$ is equal to

If $\alpha$ is the angle of intersection between the curves $y = a^x$ and $y = b^x$,then what is $\tan \alpha$ equal to?

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo