Let f(x) = cos^{-1}left(frac{2x}{1+x^2}right) | vedclass

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

Question

(D) We know that $\cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y)$ and $\sin^{-1}(y) = \frac{\pi}{2} - \cos^{-1}(y)$.Given $f(x) = \cos^{-1}\left(\frac{2x

Vedclass · Accepted Answer

$2\pi$

Vedclass · Answer

(D) We know that $\cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y)$ and $\sin^{-1}(y) = \frac{\pi}{2} - \cos^{-1}(y)$.Given $f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}ight) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}ight)$.Using the substitution $x = 	an 	heta$,we have $\frac{2x}{1+x^2} = \sin(2	heta)$ and $\frac{1-x^2}{1+x^2} = \cos(2	heta)$.For $x \ge 1$,$f(x) = \left(\frac{\pi}{2} - \sin^{-1}\left(\frac{2x}{1+x^2}ight)ight) + \left(\frac{\pi}{2} - \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)ight)$.For $x \ge 1$,$\sin^{-1}\left(\frac{2x}{1+x^2}ight) = 2	an^{-1}(x) - \pi$ is not correct; rather,for $x > 1$,$\sin^{-1}\left(\frac{2x}{1+x^2}ight) = \pi - 2	an^{-1}(x)$ and $\cos^{-1}\left(\frac{1-x^2}{1+x^2}ight) = 2	an^{-1}(x)$.Thus,$f(x) = (\pi - 2	an^{-1}(x)) + (2	an^{-1}(x)) = \pi$ for $x \ge 1$.Therefore,$f(1) = \pi$ and $f(2) = \pi$.Hence,$f(1) + f(2) = \pi + \pi = 2\pi$.

Let $f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,then the value of $f(1) + f(2)$ is -

Similar Questions

The minimum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ for $x \in [-1, 1]$ is:

$(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

The number of solutions of $\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$ where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ is equal to

If $\tan^{-1} \frac{1}{1+1(2)} + \tan^{-1} \frac{1}{1+2(3)} + \tan^{-1} \frac{1}{1+3(4)} + \dots + \tan^{-1} \frac{1}{1+n(n+1)} = \tan^{-1} \theta$,then $\theta$ =

$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}$ has the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module