જો S = {x in R : sin^{-1}left(frac{x+1}{sqrt{ | vedclass

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

Question

(C) ધારો કે $f(x) = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}$.ધારો કે $\alpha

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) ધારો કે $f(x) = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}$.ધારો કે $\alpha = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right)$ અને $\beta = \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)$.તેથી $\sin \alpha = \frac{x+1}{\sqrt{(x+1)^2+1}}$ અને $\sin \beta = \frac{x}{\sqrt{x^2+1}}$.કારણ કે $\cos \alpha = \frac{1}{\sqrt{(x+1)^2+1}}$ અને $\cos \beta = \frac{1}{\sqrt{x^2+1}}$,આપણને મળે $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = \frac{1}{\sqrt{(x+1)^2+1}\sqrt{x^2+1}}$.આપેલ છે કે $\alpha - \beta = \frac{\pi}{4}$,તેથી $\sin(\alpha - \beta) = \frac{1}{\sqrt{2}}$.આમ,$\frac{1}{\sqrt{(x+1)^2+1}\sqrt{x^2+1}} = \frac{1}{\sqrt{2}} \implies ((x+1)^2+1)(x^2+1) = 2$.$(x^2+2x+2)(x^2+1) = 2 \implies x^4 + 2x^3 + 3x^2 + 2x + 2 = 2 \implies x(x+1)(x^2+x+2) = 0$.વાસ્તવિક ઉકેલો $x=0$ અને $x=-1$ મળે છે.$S = \{-1, 0\}$.$x=0$ માટે,$\sin(5\pi/2) - \cos(5\pi) = 1 - (-1) = 2$.$x=-1$ માટે,$\sin(5\pi/2) - \cos(5\pi) = 1 - (-1) = 2$.સરવાળો $= 2 + 2 = 4$.

જો $S = \{x \in R : \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}\}$,હોય તો $\sum_{x \in S} \left(\sin\left((x^2+x+5)\frac{\pi}{2}\right) - \cos((x^2+x+5)\pi)\right)$ ની કિંમત $........$ થાય.

Similar Questions

$2{\tan ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{\theta }{2}} \right] = $

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{a - x}{1 + ax} \right) \right] = $

જો $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$ હોય, તો $1-x y-y z-z x$ ની કિંમત શોધો.

$\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$ ના ઉકેલોની સંખ્યા શોધો, જ્યાં $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ છે.

જો $0 < x < 1$ હોય,તો $\cot ^{-1}\left( \frac{2x^2 - 1}{2x\sqrt{1 - x^2}} \right)$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module