Let , f(a), be in A.P. and be the minimum value of f(a). Then the value of the integral is: [2026]

Question

Let $(1^{1 - a} + 2^{1 + a}), f (a), (3^{a} + 3^{- a})$ , f(a), be in A.P. and $α$ be the minimum value of f(a). Then the value of the integral $\int_{\log_{e} (α - 1)}^{\log_{e} (α)} \frac{d x}{(e^{2 x} - e^{- 2 x})}$ is: [2026]

Smart Achievers · Accepted Answer

(2)

Given, $(1^{1 - a} + 2^{1 + a}), f (a), (3^{a} + 3^{- a})$ are in A.P., then

$α$

$f {(a)}_{\min} = 3 at a = 0$

So, $α = 3$

Let $I = \int_{\ln 2}^{\ln 3} \frac{d x}{e^{2 x} - e^{- 2 x}}$

Put $e^{2 x} = t$

$\Rightarrow 2 e^{2 x} d x = d t$

$I = \frac{1}{2} \int_{4}^{9} \frac{d t}{t^{2} - 1} = \frac{1}{4} (\ln \frac{4}{5} - \ln \frac{3}{5}) = \frac{1}{4} (\ln \frac{4}{3})$

Solution

Q.
Let $(1^{1 - a} + 2^{1 + a}), f (a), (3^{a} + 3^{- a})$ , f(a), be in A.P. and $α$ be the minimum value of f(a). Then the value of the integral $\int_{\log_{e} (α - 1)}^{\log_{e} (α)} \frac{d x}{(e^{2 x} - e^{- 2 x})}$ is: [2026]

1 $\frac{1}{2} \log_{e} (\frac{4}{3})$

2 $\frac{1}{4} \log_{e} (\frac{4}{3})$

3 $\frac{1}{2} \log_{e} (\frac{8}{5})$

4 $\frac{1}{4} \log_{e} (\frac{8}{5})$

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Solution

Q. Let (11–a+21+a),f(a),(3a+3–a), f(a), be in A.P. and α be the minimum value of f(a). Then the value of the integral ∫loge(α–1)loge(α)dx(e2x–e–2x) is: [2026]

1 12loge(43)

2 14loge(43)

3 12loge(85)

4 14loge(85)