If tan^{-1} (x^2 + 3|x|-4) = tan^{-1} (4pi + | vedclass

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

Question

(D) Given the equation: $	an^{-1} (x^2 + 3|x|-4) = 	an^{-1} (4\pi + \sin^{-1}(\sin 14))$.Since $	an^{-1}$ is a one-to-one function,we have $x^2 + 3

Vedclass · Accepted Answer

$9 - 2\pi$

Vedclass · Answer

(D) Given the equation: $	an^{-1} (x^2 + 3|x|-4) = 	an^{-1} (4\pi + \sin^{-1}(\sin 14))$.Since $	an^{-1}$ is a one-to-one function,we have $x^2 + 3|x| - 4 = 4\pi + \sin^{-1}(\sin 14)$.We know that $\sin^{-1}(\sin 14) = 14 - 4\pi$ because $4\pi Substituting this into the equation: $x^2 + 3|x| - 4 = 4\pi + (14 - 4\pi) = 14$.So,$x^2 + 3|x| - 18 = 0$.Let $t = |x|$,where $t \ge 0$. Then $t^2 + 3t - 18 = 0$.$(t+6)(t-3) = 0$. Since $t \ge 0$,we have $t = 3$,so $|x| = 3$.We need to find $\cos^{-1}(\cos 3|x|) = \cos^{-1}(\cos 9)$.Since $2\pi Thus,$\cos^{-1}(\cos 9) = 9 - 2\pi$.

If $\tan^{-1} (x^2 + 3|x|-4) = \tan^{-1} (4\pi + \sin^{-1}(\sin 14))$,then the value of $\cos^{-1}(\cos 3|x|)$ is equal to

Similar Questions

If $f(x) = 2 \sin^{-1} \sqrt{1-x} + \sin^{-1} (2 \sqrt{x(1-x)})$ where $x \in (0, 1/2)$,then $f'(x)$ has the value equal to

$\cos ^{-1} \frac{3}{5} + \sin ^{-1} \frac{5}{13} + \tan ^{-1} \frac{16}{63} = $

The number of positive integral solutions of $\tan ^{-1} x+\cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$ are

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = $

If $x=\sin \left(2 \tan ^{-1} 2\right)$ and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module