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The number of elements in the set S = ${x \in R : 2 c o s (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} i s :$

Option 1 -

Option 2 -

Option 3 -

Option 4 -

Infinite

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

2cos $(\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$

$- 2 \leq L H S \leq 2 L H S = 2 & R H S = 2 \Rightarrow x = 0 o n l y \to t h e n L H S = 2 a l s o$

RHS $\geq$ 2

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The number of elements in the set S = ${x \in R : 2 c o s (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} i s :$

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The number of elements in the set S = { x ∈ R : 2 c o s ( x 2 + x 6 ) = 4 x + 4 − x } i s :

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The number of elements in the set S = ${x \in R : 2 c o s (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} i s :$