QJEEM16Ma9ef2: JEE Main 2016 Mathematics - Class 11 JEEM16Ma9ef2 Solution: The number of x ∈ [0,   2 π ] for which

Question

Verified Solution • 2025–26 Session moderate

The number of x

\in

[0, 2

π

] for which

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

is :

Accepted Answer

We have,

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

That is,

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

Now,

f(x)=f\left(\frac{\pi}{2}-x\right)

\Rightarrow f(x)

is symmetric about

x=\frac{\pi}{4}

.

If

f(x)

has solution in

(0, \pi / 4)

, then in

(0,2 \pi)

, there are eight solutions.

QJEEM16Ma9ef2: JEE Main 2016 Mathematics - Class 11 JEEM16Ma9ef2 Solution: The number of x ∈ [0, &nbsp; 2 π ] for which