Limits – Integration Based – JEE Main 29 Jan 2025 Shift 2

Solution:

Let $L = \lim_{t \to 0} \left( \int_{0}^{1} (3x+5)^{t} dx \right)^{\frac{1}{t}}$.
First, evaluate the integral $I(t) = \int_{0}^{1} (3x+5)^{t} dx$.
$$ I(t) = \left[ \frac{(3x+5)^{t+1}}{3(t+1)} \right]_{0}^{1} = \frac{8^{t+1} – 5^{t+1}}{3(t+1)} $$

Now, find the limit of the form $1^{\infty}$.
$$ L = \lim_{t \to 0} \left( \frac{8^{t+1} – 5^{t+1}}{3(t+1)} \right)^{\frac{1}{t}} $$
$$ L = e^{\lim_{t \to 0} \frac{1}{t} \left( \frac{8^{t+1} – 5^{t+1}}{3(t+1)} – 1 \right)} $$

Simplify the exponent:
$$ E = \lim_{t \to 0} \frac{8^{t+1} – 5^{t+1} – 3(t+1)}{3t(t+1)} $$
Using L’Hopital’s Rule (differentiating with respect to $t$):
$$ \text{Numerator}’ = 8^{t+1} \ln 8 – 5^{t+1} \ln 5 – 3 $$
$$ \text{Denominator}’ = 3(t+1) + 3t = 6t + 3 $$

Substituting $t=0$:
$$ E = \frac{8 \ln 8 – 5 \ln 5 – 3}{3} = \frac{\ln(8^8) – \ln(5^5) – 3}{3} $$
$$ E = \frac{8}{3} \ln 8 – \frac{5}{3} \ln 5 – 1 $$

So, $L = e^{E} = e^{\frac{8}{3} \ln 8} \cdot e^{-\frac{5}{3} \ln 5} \cdot e^{-1}$.
$$ L = \frac{8^{8/3}}{5^{5/3} \cdot e} $$

We manipulate this to match the required form $\frac{\alpha}{5e} \left( \frac{8}{5} \right)^{\frac{2}{3}}$.
$$ L = \frac{1}{e} \cdot \frac{8^{2} \cdot 8^{2/3}}{5^{1} \cdot 5^{2/3}} = \frac{64}{5e} \left( \frac{8^{2/3}}{5^{2/3}} \right) = \frac{64}{5e} \left( \frac{8}{5} \right)^{\frac{2}{3}} $$

Comparing with the given expression, $\alpha = 64$.

Ans. 64