ધારો કે \((2^{1-a} + 2^{1+a})\), \(f(a)\), \((3^a + 3^{-a})\) સમાંતર શ્રેણીમાં છે અને \(\alpha\) એ \(f(a)\) નું ન્યૂનતમ મૂલ્ય છે. તો સંકલિત \(\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \dfrac{dx}{(e^{2x} - e^{-2x})}\) નું મૂલ્ય શોધો :

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 \(\dfrac{1}{4} \log_e\left(\dfrac{8}{5}\right)\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\dfrac{1}{4} \log_e\left(\dfrac{4}{3}\right)\)

Step-by-step Solution

Detailed explanation

જેহেতু \((2^{1-a} + 2^{1+a})\), \(f(a)\) અને \((3^a + 3^{-a})\) સમાંતર શ્રેણીમાં છે, તેથી આપણી પાસે છે: \(2f(a) = 2^{1-a} + 2^{1+a} + 3^a + 3^{-a}\) \(f(a) = (2^{-a} + 2^a) + \dfrac{1}{2}(3^a + 3^{-a})\) AM-GM અસમાનતાનો ઉપયોગ કરીને, \(x + \dfrac{1}{x} \ge 2\) જ્યાં \(x > 0\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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