QJEEM22Mae0e4: JEE Main 2022 Mathematics - Class 11 JEEM22Mae0e4 Solution: The number of elements in the set is :

Given,

$2\cos \left( {{{{x^2} + x} \over 6}} \right) = {4^x} + {4^{ - x}}$

We know,

A.M $\ge$ G.M

$\therefore$ for 4^x and 4 ^{$-$ x}

${{{4^x} + {4^{ - x}}} \over 2} \ge \sqrt {{4^x}\,.\,{4^{ - x}}}$

$\Rightarrow {{{4^x} + {4^{ - x}}} \over 2} \ge 1$

$\Rightarrow {4^x} + {4^{ - x}} \ge 2$ ...... (1)

And we know,

$- 1 \le \cos x \le 1$

$\therefore$ $- 1 \le \cos \left( {{{{x^2} + x} \over 6}} \right) \le 1$

$- 2 \le 2\cos \left( {{{{x^2} + x} \over 6}} \right) \le 2$ ...... (2)

(1) and (2) both satisfies only when both equal to 2.

$\therefore$ $2\cos \left( {{{{x^2} + x} \over 6}} \right) = 2$

$\Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = 1$

$\Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = \cos 0$

$\Rightarrow {{{x^2} + x} \over 6} = 0$

$\Rightarrow {x^2} + x = 0$

$\Rightarrow x(x + 1) = 0$

$\Rightarrow x = 0, - 1$

When $x = 0$ ,

L.H.S $= 2\cos \left( {{{{x^2} + x} \over 6}} \right)$

$= 2\cos \left( {{0 \over 6}} \right)$

$= 2\cos 0$

$= 2\,.\,1$

$= 2$

R.H.S $= {4^x} + {4^{ - x}}$

$= {4^0} + {4^0}$

$= 2$

$\therefore$ $x = 0$ is accepted.

Now, when $x = - 1$ ,

L.H.S $= 2\cos \left( {{{{x^2} + x} \over 6}} \right)$

$= 2\cos \left( {{{1 - 1} \over 6}} \right)$

$= 2\cos 0 = 2$

R.H.S $= {4^x} + {4^{ - x}}$

$= {4^{ - 1}} + {4^1}$

$= {1 \over 4} + 4$

$= {{17} \over 4}$

$\therefore$ L.H.S $\ne$ R.H.S

$\therefore$ $x = - 1$ is not a solution.

$\therefore$ Only one solution possible which is $x = 0$ .