The number of elements in the set $$S=\left\{x \in \mathbb{R}: 2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\right\}$$ is :

<p>Given,</p> <p>$2\cos \left( {{{{x^2} + x} \over 6}} \right) = {4^x} + {4^{ - x}}$</p> <p>We know,</p> <p>A.M $\ge$ G.M</p> <p>$\therefore$ for 4^x and 4^$-$x</p> <p>${{{4^x} + {4^{ - x}}} \over 2} \ge \sqrt {{4^x}\,.\,{4^{ - x}}}$</p> <p>$\Rightarrow {{{4^x} + {4^{ - x}}} \over 2} \ge 1$</p> <p>$\Rightarrow {4^x} + {4^{ - x}} \ge 2$ ...... (1)</p> <p>And we know,</p> <p>$- 1 \le \cos x \le 1$</p> <p>$\therefore$ $- 1 \le \cos \left( {{{{x^2} + x} \over 6}} \right) \le 1$</p> <p>$- 2 \le 2\cos \left( {{{{x^2} + x} \over 6}} \right) \le 2$ ...... (2)</p> <p>(1) and (2) both satisfies only when both equal to 2.</p> <p>$\therefore$ $2\cos \left( {{{{x^2} + x} \over 6}} \right) = 2$</p> <p>$\Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = 1$</p> <p>$\Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = \cos 0$</p> <p>$\Rightarrow {{{x^2} + x} \over 6} = 0$</p> <p>$\Rightarrow {x^2} + x = 0$</p> <p>$\Rightarrow x(x + 1) = 0$</p> <p>$\Rightarrow x = 0, - 1$</p> <p>When $x = 0$,</p> <p>L.H.S $= 2\cos \left( {{{{x^2} + x} \over 6}} \right)$</p> <p>$= 2\cos \left( {{0 \over 6}} \right)$</p> <p>$= 2\cos 0$</p> <p>$= 2\,.\,1$</p> <p>$= 2$</p> <p>R.H.S $= {4^x} + {4^{ - x}}$</p> <p>$= {4^0} + {4^0}$</p> <p>$= 2$</p> <p>$\therefore$ $x = 0$ is accepted.</p> <p>Now, when $x = - 1$,</p> <p>L.H.S $= 2\cos \left( {{{{x^2} + x} \over 6}} \right)$</p> <p>$= 2\cos \left( {{{1 - 1} \over 6}} \right)$</p> <p>$= 2\cos 0 = 2$</p> <p>R.H.S $= {4^x} + {4^{ - x}}$</p> <p>$= {4^{ - 1}} + {4^1}$</p> <p>$= {1 \over 4} + 4$</p> <p>$= {{17} \over 4}$</p> <p>$\therefore$ L.H.S $\ne$ R.H.S</p> <p>$\therefore$ $x = - 1$ is not a solution.</p> <p>$\therefore$ Only one solution possible which is $x = 0$.</p>