The solution set of the inequality (tan^{-1} | vedclass

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

Question

(C) Let $u = 	an^{-1} x$. Then $\cot^{-1} x = \frac{\pi}{2} - u$.Given inequality: $u(\frac{\pi}{2} - u) - u(1 + \frac{\pi}{2}) - 2(\frac{\pi}{2} - u

Vedclass · Accepted Answer

$(-	an 1, 	an 2)$

Vedclass · Answer

(C) Let $u = an^{-1} x$. Then $\cot^{-1} x = \frac{\pi}{2} - u$.Given inequality: $u(\frac{\pi}{2} - u) - u(1 + \frac{\pi}{2}) - 2(\frac{\pi}{2} - u) + 2(1 + \frac{\pi}{2}) > \lim_{x o \infty} [\sec^{-1} x - \frac{\pi}{2}]$.As $x o \infty$,$\sec^{-1} x o \frac{\pi}{2}$,so the limit is $[\frac{\pi}{2} - \frac{\pi}{2}] = [0] = 0$.Simplifying the left side: $\frac{\pi}{2}u - u^2 - u - \frac{\pi}{2}u - \pi + 2u + 2 + \pi > 0$.$-u^2 + u + 2 > 0 \Rightarrow u^2 - u - 2 $(u - 2)(u + 1) Since $u = an^{-1} x$,we have $-1 Taking tangent on all sides: $ an(-1) Thus,$- an 1 < x < an 2$.

The solution set of the inequality $(\tan^{-1} x)(\cot^{-1} x) - (\tan^{-1} x)(1 + \frac{\pi}{2}) - 2\cot^{-1} x + 2(1 + \frac{\pi}{2}) > \lim_{x \to \infty} [\sec^{-1} x - \frac{\pi}{2}]$ is (where $[.]$ denotes the greatest integer function):

Similar Questions

$\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} \, - \,\sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to

If $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$,where $-1 \le x \le 1$,$-2 \le y \le 2$,and $x \le \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to

The sum of the maximum and the minimum values of $2(\cos ^{-1} x)^2-\pi \cos ^{-1} x+\frac{\pi^2}{4}$ is

If $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$,then the value of $\frac{dy}{dx}$ at $x = \sqrt{3}$ is

If $p, q, r$ are in $G.P.$ and $\tan^{-1} p, \tan^{-1} q, \tan^{-1} r$ are in $A.P.$,then $p, q, r$ satisfy the relation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module