Trigonometric Equations — Page 5 | JEE Mathematics

JEE Mathematics · 41 questions · Page 5 of 5 · Click an option or "Show Solution" to reveal answer

Q41

The number of x

\in

[0, 2

\pi

] for which

\left| {\sqrt {2{{\sin }^4}x + 18{{\cos }^2}x} - \sqrt {2{{\cos }^4}x + 18{{\sin }^2}x} } \right| = 1

is :

A 2

B 4

C 6

D 8

Correct Answer

Option D

Solution

We have,

\left|\sqrt{2 \sin ^4 x+18 \cos ^2 x}-\sqrt{2 \cos ^4 x+18 \sin ^2 x}\right|=1

That is,

f(x)=\left|\sqrt{2 \sin ^4 x+8 \cos ^2 x}-\sqrt{2 \cos ^4 x+18 \sin ^2 x}\right|

Now, $f(x)=f\left(\dfrac{\pi}{2}-x\right)$ $\Rightarrow f(x)$ is symmetric about $x=\dfrac{\pi}{4}$ .

If $f(x)$ has solution in $(0, \pi / 4)$ , then in $(0,2 \pi)$ , there are eight solutions.