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 integral_from_log_e(alpha-1)_to_log_e(alpha) dx / (e^(2x) - e^(-2x)) is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q18 – JEE-MAIN 2026

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Home Maths Sequences and SeriesQuestion #399

Let

(2^{1 - a} + 2^{1 + a})

f (a)

(3^{a} + 3^{- 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:

Clue 1 of 3: Arithmetic Progression Property

Recall the property of terms in an Arithmetic Progression (A.P.) to express $f (a)$ in terms of the other two given expressions.

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Step 1: Determine

f (a)

and its minimum value

α

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The terms $(2^{1 - a} + 2^{1 + a})$ , $f (a)$ , and $(3^{a} + 3^{- a})$ are in A.P. Using the property $2 B = A + C$ , we have:

2 f (a) = (2^{1 - a} + 2^{1 + a}) + (3^{a} + 3^{- a})

2 f (a) = (2 \cdot 2^{- a} + 2 \cdot 2^{a}) + (3^{a} + 3^{- a})

2 f (a) = 2 (2^{a} + 2^{- a}) + (3^{a} + 3^{- a})

f (a) = (2^{a} + 2^{- a}) + \frac{1}{2} (3^{a} + 3^{- a})

By AM-GM inequality, for $x > 0$ , $x + \frac{1}{x} \geq 2$ . Thus, $2^{a} + 2^{- a} \geq 2$ and $3^{a} + 3^{- a} \geq 2$ . The minimum value of $f (a)$ is $α = 2 + \frac{1}{2} (2) = 2 + 1 = 3$ .

Step 2: Set up the definite integral with the correct limits○ Expand

The integral is $\int_{\log_{e} (α - 1)}^{\log_{e} (α)} \frac{d x}{(e^{2 x} - e^{- 2 x})}$ . Substitute $α = 3$ :

Lower limit: $\log_{e} (3 - 1) = \log_{e} (2)$ .

Upper limit: $\log_{e} (3)$ .

The integral becomes $\int_{\log_{e} (2)}^{\log_{e} (3)} \frac{d x}{(e^{2 x} - e^{- 2 x})}$ . Rewrite the integrand:

\frac{1}{e^{2 x} - e^{- 2 x}} = \frac{1}{e^{2 x} - \frac{1}{e^{2 x}}} = \frac{e^{2 x}}{e^{4 x} - 1}

Step 3: Evaluate the integral using substitution○ Expand

Let $u = e^{2 x}$ . Then $d u = 2 e^{2 x} d x$ , so $e^{2 x} d x = \frac{1}{2} d u$ . Adjust the limits of integration:

When $x = \log_{e} (2)$ , $u = e^{2 \log_{e} (2)} = e^{\log_{e} (2^{2})} = 4$ .

When $x = \log_{e} (3)$ , $u = e^{2 \log_{e} (3)} = e^{\log_{e} (3^{2})} = 9$ .

The integral transforms to:

\int_{4}^{9} \frac{1}{u^{2} - 1} \cdot \frac{1}{2} d u = \frac{1}{2} \int_{4}^{9} \frac{1}{u^{2} - 1} d u

Using the standard integral $\int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \log_{e} | \frac{x - a}{x + a} |$ , with $a = 1$ :

\frac{1}{2} {[\frac{1}{2} \log_{e} | \frac{u - 1}{u + 1} |]}_{4}^{9} = \frac{1}{4} {[\log_{e} (\frac{u - 1}{u + 1})]}_{4}^{9}

Evaluate the limits:

\frac{1}{4} [\log_{e} (\frac{9 - 1}{9 + 1}) - \log_{e} (\frac{4 - 1}{4 + 1})]

= \frac{1}{4} [\log_{e} (\frac{8}{10}) - \log_{e} (\frac{3}{5})] = \frac{1}{4} [\log_{e} (\frac{4}{5}) - \log_{e} (\frac{3}{5})]

Using $\log_{e} A - \log_{e} B = \log_{e} (\frac{A}{B})$ :

= \frac{1}{4} \log_{e} (\frac{4 / 5}{3 / 5}) = \frac{1}{4} \log_{e} (\frac{4}{3})

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Maths Question 18 – JEE-MAIN 2026

Clue 1 of 3: Arithmetic Progression Property

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