Consider the equation 1+ 3 2 x ^4 x+4 ^2 x + ^4 x+4 ^2 x = 2 ^ -1 ( 2 x) x ∈(-π, π)

Correct answer: D. If x ∈[0, n π] n ∈ N , number of roots will never be odd aligned & ^4 x+4 ^2 x + ^4 x+4 ^2 x & = ^4 x-4 ^2 x+4 + ^4 x+4-4 ^2 x & =2- ^2 x+2- ^2 x=3 aligned equation reduces to = 1+ 2 x = 2 ^ -1 ( 2 x) aligned & | x|= ^ -1 ( 2 x) & f(π / 8)=| π / 8|- ^ -1 π / 4 aligned = π / 8-π / 4= (1+ (π)/(4))/(2) -π / 4 0f(π / 6)=| (π)/(6)|- ^ -1 (π)/(3)= 3 2 -(π)/(3) 0 so one root exists between (π / 8, π / 6) .

JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Trigonometry

JEE Advanced · Mathematics · Trigonometry

Consider the equation 1+ 3 2 x ^4 x+4 ^2 x + ^4 x+4 ^2 x = 2 ^ -1 (...

Consider the equation

\sqrt{1+\frac{3 \cos 2 x}{\sqrt{\sin ^4 x+4 \cos ^2 x}+\sqrt{\cos ^4 x+4 \sin ^2 x}}}=\sqrt{2} \sin \sin ^{-1}(\sin 2 x) x \in(-\pi, \pi)

Options

ANumber of roots of this equation is 2
BNumber of roots of this equation is 4
COne root lies in the interval \left(\frac{\pi}{8}, \frac{\pi}{6}\right)
DIf $x \in[0, n \pi] n \in N$ , number of roots will never be odd

Correct answer

D. If $x \in[0, n \pi] n \in N$ , number of roots will never be odd

Step-by-step solution

\begin{aligned} & \sqrt{\sin ^4 x+4 \cos ^2 x}+\sqrt{\cos ^4 x+4 \sin ^2 x} \\ & =\sqrt{\sin ^4 x-4 \sin ^2 x+4}+\sqrt{\cos ^4 x+4-4 \cos ^2 x} \\ & =2-\sin ^2 x+2-\cos ^2 x=3\end{aligned}

equation reduces to

=\sqrt{1+\cos 2 x}=\sqrt{2} \sin ^{-1}(\sin 2 x)

\begin{aligned} & |\cos x|=\sin ^{-1}(\sin 2 x) \\ & f(\pi / 8)=|\cos \pi / 8|-\sin ^{-1} \sin \pi / 4\end{aligned}

=\cos \pi / 8-\pi / 4=\sqrt{\frac{1+\cos \frac{\pi}{4}}{2}}-\pi / 4\gt0f(\pi / 6)=\left|\cos \frac{\pi}{6}\right|-\sin ^{-1} \frac{\pi}{3}=\frac{\sqrt{3}}{2}-\frac{\pi}{3} \lt 0

so one root exists between

(\pi / 8, \pi / 6)

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