JEE Main 2025 — 29 January, Evening Shift — Question 63

Given

$$\lim_{t\to 0} \left( \int_0^1 (3x+5)^t dx \right)^{1/t} = \frac{\alpha}{5e} \left(\frac{8}{5}\right)^{2/3}$$

Step 1: Use standard limit result

We use the identity:

$$\lim_{t\to 0} \left(\int_a^b (f(x))^t dx \right)^{1/t} = \exp\left( \int_a^b \ln f(x)\,dx \right)$$

Here,

$$f(x)=3x+5$$

So limit becomes

$$L= \exp\left( \int_0^1 \ln(3x+5)\,dx \right)$$

Step 2: Evaluate the integral

Let

$$I=\int_0^1 \ln(3x+5)\,dx$$

Substitute:

$$u=3x+5, \quad du=3dx, \quad dx=\frac{du}{3}$$

When $x=0$, $u=5$ When $x=1$, $u=8$

$$I= \frac{1}{3} \int_5^8 \ln u \,du$$

We know:

$$\int \ln u\,du = u\ln u - u$$

Thus,

$$I= \frac{1}{3} \left[ u\ln u - u \right]_5^8$$ $$= \frac{1}{3} \left( 8\ln 8 - 8 - 5\ln 5 + 5 \right)$$ $$= \frac{1}{3} \left( 8\ln 8 - 5\ln 5 - 3 \right)$$

Step 3: Compute the limit

$$L= \exp\left( \frac{8\ln 8 - 5\ln 5 - 3}{3} \right)$$ $$= \exp\left(\frac{8\ln 8 - 5\ln 5}{3}\right) \cdot e^{-1}$$ $$= e^{-1} \left( \frac{8^8}{5^5} \right)^{1/3}$$ $$= \frac{1}{e} \cdot \frac{8^{8/3}}{5^{5/3}}$$

Now write:

$$8^{8/3} = 8^2 \cdot 8^{2/3} = 64\cdot 8^{2/3}$$

Thus,

$$L= \frac{64}{5e} \left(\frac{8}{5}\right)^{2/3}$$

Step 4: Compare with given expression

Given:

$$L= \frac{\alpha}{5e} \left(\frac{8}{5}\right)^{2/3}$$

Hence,

$$\boxed{\alpha=64}$$