JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \((2^{1-a} + 2^{1+a})\), \(f(a)\), \((3^a + 3^{-a})\) be in A.P. and \(\alpha\) be the minimum value of \(f(a)\). Then the value of the integral \(\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \dfrac{dx}{(e^{2x} - e^{-2x})}\) is :
- A \(\dfrac{1}{2} \log_e\left(\dfrac{4}{3}\right)\)
- B \(\dfrac{1}{4} \log_e\left(\dfrac{4}{3}\right)\)
- C \(\dfrac{1}{2} \log_e\left(\dfrac{8}{5}\right)\)
- D \(\dfrac{1}{4} \log_e\left(\dfrac{8}{5}\right)\)
Answer & Solution
Correct Answer
(B) \(\dfrac{1}{4} \log_e\left(\dfrac{4}{3}\right)\)
Step-by-step Solution
Detailed explanation
Since \((2^{1-a} + 2^{1+a})\), \(f(a)\), and \((3^a + 3^{-a})\) are in A.P., we have: \(2f(a) = 2^{1-a} + 2^{1+a} + 3^a + 3^{-a}\) \(f(a) = (2^{-a} + 2^a) + \dfrac{1}{2}(3^a + 3^{-a})\) Using the AM-GM inequality, \(x + \dfrac{1}{x} \ge 2\) for \(x > 0\). \(2^a + 2^{-a} \ge 2\)…
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