JEE Mains · Maths · STD 11 - Trigonometrical equations
The number of elements in the set \(S=\left\{x \in R : 2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\right\}\) is\(.....\)
- A \(1\)
- B \(3\)
- C \(0\)
- D \(\infty\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\(2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\) L.H.S \(\leq 2 . \&\) R.H.S. \(\geq 2\) Hence L.H.S \(=2\) \& R.H.S \(=2\) \(2 \cos \left(\frac{x^{2}+x}{6}\right)=2 \quad 4^{x}+4^{-x}=2\) Check \(x=0\) Possible hence only one solution.
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