ધારો કે f(x) = cos^{-1}left(frac{2x}{1+x^2}ri | vedclass

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

Question

(D) આપણે જાણીએ છીએ કે $\cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y)$ અને $\sin^{-1}(y) = \frac{\pi}{2} - \cos^{-1}(y)$.આપેલ છે $f(x) = \cos^{-1}\left(\

Vedclass · Accepted Answer

$2\pi$

Vedclass · Answer

(D) આપણે જાણીએ છીએ કે $\cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y)$ અને $\sin^{-1}(y) = \frac{\pi}{2} - \cos^{-1}(y)$.આપેલ છે $f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}ight) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}ight)$.$x = 	an 	heta$ આદેશ લેતા,આપણને મળે $\frac{2x}{1+x^2} = \sin(2	heta)$ અને $\frac{1-x^2}{1+x^2} = \cos(2	heta)$.$x \ge 1$ માટે,$f(x) = \left(\frac{\pi}{2} - \sin^{-1}\left(\frac{2x}{1+x^2}ight)ight) + \left(\frac{\pi}{2} - \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)ight)$.$x \ge 1$ માટે,$\sin^{-1}\left(\frac{2x}{1+x^2}ight) = \pi - 2	an^{-1}(x)$ અને $\cos^{-1}\left(\frac{1-x^2}{1+x^2}ight) = 2	an^{-1}(x)$.આમ,$f(x) = (\pi - 2	an^{-1}(x)) + (2	an^{-1}(x)) = \pi$,જ્યાં $x \ge 1$.તેથી,$f(1) = \pi$ અને $f(2) = \pi$.આમ,$f(1) + f(2) = \pi + \pi = 2\pi$.

ધારો કે $f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,તો $f(1) + f(2)$ ની કિંમત શોધો.

Similar Questions

કિંમત શોધો: $\cos^{-1}(\cos \frac{4\pi}{3}) + \sin^{-1}(\sin \frac{4\pi}{3}) = \dots$

$\frac{d}{dx} \left( \cos^{-1} \sqrt{\frac{1 + \cos x}{2}} \right) = $

$\tan \left(2 \tan ^{-1} \frac{1}{5} + \sec ^{-1} \frac{\sqrt{5}}{2} + 2 \tan ^{-1} \frac{1}{8}\right)$ ની કિંમત શોધો.

જો ${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x$ હોય,તો $x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module