Let f(x) = cot left( sin^{-1} sqrt{frac{2}{3 | vedclass

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

Question

(C) Given $f(x) = \cot \left( \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}} \right)$.We know that $\cos 2x = 2\cos^2 x - 1$,so $3 + \cos 2x = 3 + 2\cos^2 x -

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(C) Given $f(x) = \cot \left( \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}} ight)$.We know that $\cos 2x = 2\cos^2 x - 1$,so $3 + \cos 2x = 3 + 2\cos^2 x - 1 = 2 + 2\cos^2 x = 2(1 + \cos^2 x)$.Substituting this into the expression,we get $\sqrt{\frac{2}{2(1 + \cos^2 x)}} = \frac{1}{\sqrt{1 + \cos^2 x}}$.However,let us simplify the argument of $\cot$ differently. Let $	heta = \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}}$. Then $\sin 	heta = \sqrt{\frac{2}{3 + \cos 2x}}$.$\sin^2 	heta = \frac{2}{3 + \cos 2x} \implies \csc^2 	heta = \frac{3 + \cos 2x}{2} = \frac{3 + 2\cos^2 x - 1}{2} = 1 + \cos^2 x$.Since $\cot^2 	heta = \csc^2 	heta - 1$,we have $\cot^2 	heta = (1 + \cos^2 x) - 1 = \cos^2 x$.Thus,$f(x) = \cot 	heta = \sqrt{\cos^2 x} = |\cos x|$.At $x = \frac{2\pi}{3}$,$\cos x = -\frac{1}{2}$,so $f(x) = |\cos x| = -\cos x$ in the neighborhood of $\frac{2\pi}{3}$.$f'(x) = \frac{d}{dx}(-\cos x) = \sin x$.$f'\left( \frac{2\pi}{3} ight) = \sin \left( \frac{2\pi}{3} ight) = \frac{\sqrt{3}}{2}$.

Let $f(x) = \cot \left( \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}} \right)$. Then,the value of $f'\left( \frac{2\pi}{3} \right)$ is:

Similar Questions

If $0 \leq A \leq \frac{\pi}{4},$ then $\tan ^{-1}\left(\frac{1}{2} \tan 2 A\right)+\tan ^{-1}(\cot A)+\tan ^{-1}(\cot ^{3} A)$ is equal to

Let $f : R \to R$,$f(x) = \max\{|\tan^{-1}x|, \cot^{-1}x\}$. Consider the following statements:
$I.$ The function is continuous and derivable $\forall x \in R$.
$II.$ The range of the function is $\left[ \frac{\pi}{4}, \pi \right]$.
$III.$ $f(x)$ is many-one into.
Identify the correct option.

The numerical value of $\tan \left(2 \tan ^{-1}\left(\frac{1}{5}\right)+\frac{\pi}{4}\right)$ is:

The derivative of ${\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ with respect to ${\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ is

The value of $\cos^{-1}(\cos 12) - \sin^{-1}(\sin 14)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module