For how many distinct values of x,the equatio | vedclass

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

Question

(C) Given equation is $\sin [2 \cos^{-1} \cot (2 	an^{-1} x)] = 0$. Let $	heta = 2 	an^{-1} x$. Then $\cot 	heta = \cot (2 	an^{-1} x) = \frac{1

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Given equation is $\sin [2 \cos^{-1} \cot (2 	an^{-1} x)] = 0$. Let $	heta = 2 	an^{-1} x$. Then $\cot 	heta = \cot (2 	an^{-1} x) = \frac{1 - 	an^2(	an^{-1} x)}{2 	an(	an^{-1} x)} = \frac{1 - x^2}{2x}$. The equation becomes $\sin [2 \cos^{-1} (\frac{1 - x^2}{2x})] = 0$. This implies $2 \cos^{-1} (\frac{1 - x^2}{2x}) = n\pi$ for some integer $n$. So,$\cos^{-1} (\frac{1 - x^2}{2x}) = \frac{n\pi}{2}$. Taking cosine on both sides,$\frac{1 - x^2}{2x} = \cos(\frac{n\pi}{2})$. For the expression to be defined,the argument of $\cos^{-1}$ must be in $[-1, 1]$,so $|\frac{1 - x^2}{2x}| \le 1$. Possible values for $\cos(\frac{n\pi}{2})$ are $0, 1, -1$. Case $1$: $\frac{1 - x^2}{2x} = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$. Case $2$: $\frac{1 - x^2}{2x} = 1 \Rightarrow x^2 + 2x - 1 = 0 \Rightarrow x = -1 \pm \sqrt{2}$. Case $3$: $\frac{1 - x^2}{2x} = -1 \Rightarrow x^2 - 2x - 1 = 0 \Rightarrow x = 1 \pm \sqrt{2}$. All these $6$ values satisfy the condition $|\frac{1 - x^2}{2x}| \le 1$. Thus,there are $6$ distinct values of $x$.

For how many distinct values of $x$,the equation $\sin [2 \cos^{-1} \cot (2 \tan^{-1} x)] = 0$ holds?

Similar Questions

The number of solutions of the equation $\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}$ for $x \in[-1,1],$ where $[x]$ denotes the greatest integer function,is:

If $f(x) = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $g(x) = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$,then the derivative of $f(x)$ with respect to $g(x)$ is

If $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$,then $x=$ . . . . . . .

If $f(x) = 2 \sin^{-1} \sqrt{1-x} + \sin^{-1} (2 \sqrt{x(1-x)})$ where $x \in (0, 1/2)$,then $f'(x)$ has the value equal to

If $y = \tan^{-1}\left(\frac{1}{1+x+x^2}\right) + \tan^{-1}\left(\frac{1}{x^2+3x+3}\right) + \tan^{-1}\left(\frac{1}{x^2+5x+7}\right)$,then the value of $y'(0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module