If cot[f(x)] = frac{3x - x^3}{1 - 3x^2} and s | vedclass

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

Question

(B) Given $\cot[f(x)] = \frac{3x - x^3}{1 - 3x^2} = 	an(3	heta)$ where $x = 	an 	heta$. Thus, $f(x) = 	ext{arccot}(	an(3	heta)) = \frac{\pi}{2}

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Given $\cot[f(x)] = \frac{3x - x^3}{1 - 3x^2} = 	an(3	heta)$ where $x = 	an 	heta$. Thus, $f(x) = 	ext{arccot}(	an(3	heta)) = \frac{\pi}{2} - 3\arctan(x)$.Then $f'(x) = -\frac{3}{1+x^2}$.Given $\sin[g(x)] = \frac{1 - x^2}{1 + x^2} = \cos(2	heta)$ where $x = 	an 	heta$. Thus, $g(x) = \arcsin(\cos(2	heta)) = \arcsin(\sin(\frac{\pi}{2} - 2	heta)) = \frac{\pi}{2} - 2\arctan(x)$.Then $g'(x) = -\frac{2}{1+x^2}$.Using $L$'$H$ôpital's rule, $\lim_{x 	o t} \frac{f(x) - f(t)}{g(x) - g(t)} = \frac{f'(t)}{g'(t)} = \frac{-3/(1+t^2)}{-2/(1+t^2)} = \frac{3}{2}$.

If $\cot[f(x)] = \frac{3x - x^3}{1 - 3x^2}$ and $\sin[g(x)] = \frac{1 - x^2}{1 + x^2}$, then $\lim_{x \to t} \frac{f(x) - f(t)}{g(x) - g(t)} = \dots$

Similar Questions

Find the derivative: $\frac{d}{dx} \tan^{-1}(\sec x + \tan x) = $

If $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$,then $\frac{d y}{d x}=$

If $y(x)=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$ and $\left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0$, then the sum of the roots of the equation $1+a^2 x^2+g(x)=0$ is

If $y = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}\right)$ where $|x| < 1$,then find the value of $\left(\frac{dy}{dx}\right)$ at $x = \frac{1}{2}$.

If $f:R \to R$ is a differentiable function and $f(2) = 6$,then $\lim_{x \to 2} \int_{6}^{f(x)} \frac{2t \, dt}{x - 2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module