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JEE Mathematics 2025 Question with Solution

If lim⁡t→∞(∫01(3x+5)tdx)=α5e(85)32,\lim_{t \to \infty} \left( \int_0^{1} \left( 3x + 5 \right)^t dx \right) = \frac{\alpha}{5e} \left( \frac{8}{5} \right)^{\frac{3}{2}}, then α\alpha is equal to _____ :

Answer

Correct answer:64

Step-by-step solution

Standard Method

Given:

lim⁡t→∞(∫01(3x+5)t dx)=α5e(85)32\lim_{t \to \infty} \left( \int_0^{1} (3x + 5)^t \, dx \right) = \frac{\alpha}{5e} \left( \frac{8}{5} \right)^{\frac{3}{2}}

Find: α\alpha

From the solution, the integral is first written as

I(t)=∫01(3x+5)t dxI(t) = \int_0^{1} (3x + 5)^t \, dx

and integrated directly as

I(t)=(3x+5)t+13(t+1)∣01=13(t+1)[8t+1−5t+1]I(t) = \frac{(3x + 5)^{t+1}}{3(t+1)} \Big|_0^1 = \frac{1}{3(t+1)} \left[ 8^{t+1} - 5^{t+1} \right]

Taking 5t+15^{t+1} common,

I(t)=5t+13(t+1)[(85)t+1−1]I(t) = \frac{5^{t+1}}{3(t+1)} \left[ \left( \frac{8}{5} \right)^{t+1} - 1 \right]

The provided solution then states that, by matching the asymptotic form with

α5e(85)32,\frac{\alpha}{5e} \left( \frac{8}{5} \right)^{\frac{3}{2}},

we obtain

α=64\alpha = 64

Therefore, the required value of α\alpha is 6464.

Note: The working shown in the solution is internally inconsistent because the displayed integral expression itself does not yield a finite constant limit in the stated form. However, both solution approaches explicitly conclude that the final answer is 6464, and that is the extracted answer.

Common mistakes

  • Treating ∫01(3x+5)t dx\int_0^1 (3x+5)^t \, dx as a standard finite limit without checking growth as t→∞t \to \infty is incorrect, because the integrand increases exponentially with tt. First examine whether any normalization is present before applying limit formulas.

  • After integrating, dropping the term −1-1 in (85)t+1−1\left(\frac{8}{5}\right)^{t+1}-1 without tracking the overall magnitude can lead to a wrong conclusion about convergence. Simplifying dominant terms is valid only after confirming the scaled expression being analyzed.

  • Using Laplace’s method mechanically is a mistake here if the original problem statement has no explicit normalizing factor. Laplace asymptotics describe growth, but they do not by themselves convert a divergent expression into a finite constant.

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