The number of elements in the set S = {x in R | vedclass

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(A) Given equation: $2 \cos \left(\frac{x^{2}+x}{6}\right) = 4^{x} + 4^{-x}$We know that the range of the cosine function is $[-1, 1]$,so $2 \cos \lef

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$1$

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(A) Given equation: $2 \cos \left(\frac{x^{2}+x}{6}\right) = 4^{x} + 4^{-x}$We know that the range of the cosine function is $[-1, 1]$,so $2 \cos \left(\frac{x^{2}+x}{6}\right) \leq 2$.For the $R$.$H$.$S$.,by the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$,we have $\frac{4^{x} + 4^{-x}}{2} \geq \sqrt{4^{x} \cdot 4^{-x}} = 1$,which implies $4^{x} + 4^{-x} \geq 2$.For the equation to hold,both sides must be equal to $2$.$2 \cos \left(\frac{x^{2}+x}{6}\right) = 2 \implies \cos \left(\frac{x^{2}+x}{6}\right) = 1$$4^{x} + 4^{-x} = 2 \implies (2^{x} - 2^{-x})^{2} + 2 = 2 \implies 2^{x} = 2^{-x} \implies x = 0$.Substituting $x = 0$ into the cosine part: $\cos \left(\frac{0^{2}+0}{6}\right) = \cos(0) = 1$. This satisfies the equation.Thus,there is only $1$ element in the set $S$.

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

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