Let tan ^{-1}(x) inleft(-frac{pi}{2}, frac{pi | vedclass

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

Question

(D) Given equation: $\sqrt{1+\cos (2 x)}=\sqrt{2} 	an ^{-1}(	an x)$Using the identity $1+\cos(2x) = 2\cos^2 x$,we get $\sqrt{2\cos^2 x} = \sqrt{2} \

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given equation: $\sqrt{1+\cos (2 x)}=\sqrt{2} 	an ^{-1}(	an x)$Using the identity $1+\cos(2x) = 2\cos^2 x$,we get $\sqrt{2\cos^2 x} = \sqrt{2} 	an^{-1}(	an x)$.This simplifies to $\sqrt{2}|\cos x| = \sqrt{2} 	an^{-1}(	an x)$,or $|\cos x| = 	an^{-1}(	an x)$.We need to find the number of intersection points of $y = |\cos x|$ and $y = 	an^{-1}(	an x)$ in the domain $D = \left(-\frac{3 \pi}{2},-\frac{\pi}{2}ight) \cup\left(-\frac{\pi}{2}, \frac{\pi}{2}ight) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}ight)$.$1$. In $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$,$	an^{-1}(	an x) = x$. The equation is $|\cos x| = x$. Since $|\cos x| > 0$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}ight) \setminus \{0\}$ and $x=0$ gives $|\cos 0| = 1 
eq 0$,there is one solution for $x > 0$ and no solution for $x \leq 0$.$2$. In $\left(\frac{\pi}{2}, \frac{3 \pi}{2}ight)$,$	an^{-1}(	an x) = x - \pi$. The equation is $|\cos x| = x - \pi$. There is one intersection point in this interval.$3$. In $\left(-\frac{3 \pi}{2}, -\frac{\pi}{2}ight)$,$	an^{-1}(	an x) = x + \pi$. The equation is $|\cos x| = x + \pi$. There is one intersection point in this interval.Total number of solutions is $1 + 1 + 1 = 3$.

Similar Questions

For $x \in (-1, 1]$,the number of solutions of the equation $\sin^{-1} x = 2 \tan^{-1} x$ is equal to

The sum $\sum\limits_{n = 1}^\infty {{\cot }^{ - 1}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to

$\cos ^{-1} \frac{3}{5} + \sin ^{-1} \frac{5}{13} + \tan ^{-1} \frac{16}{63} = $

If $\alpha$ and $\beta$ are the roots of the quadratic equation $3x^2 - 16x + 5 = 0$,then $\tan^{-1} \alpha + \tan^{-1} \beta - \tan^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}\right) = $

The number of solutions of the equation $2\tan^{-1}(\cos^2 x) = \tan^{-1}(2\csc^2 x)$ in the interval $[0, 5\pi]$ is $m$. Then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module