JEE Main 2025 Jan 29 Shift 2, Mathematics Q22: Differentiation Under the Integral Sign

Let $F(t)={\int }_0^1(3x+5{)}^{t}dx=\frac{8^{t+1}-5^{t+1}}{3(t+1)}$. As $t\to 0$, $F(t)\to 1$, so the limit is of the form $1^{\infty }$.

Write $L={\lim }_{t\to 0}F(t{)}^{1/t}$, $\ln L={\lim }_{t\to 0}\frac{\ln F(t)}{t}$. Since $\ln F(0)=0$, this is a $0/0$ limit equal to $F^{\prime }(0)/F(0)=F^{\prime }(0)$.

$F(t)=\frac{8\cdot 8^t-5\cdot 5^t}{3(t+1)}$. Differentiate using quotient rule and evaluate at $t=0$:

$F^{\prime }(0)=\frac{(8\ln 8-5\ln 5)\cdot 3-(8-5)\cdot 3}{9}=\frac{8\ln 8-5\ln 5-3}{3}$.

Therefore

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

Match against $\frac{\alpha }{5e}{\left(\frac{8}{5}\right)}^{2/3}=\frac{\alpha \cdot 8^{2/3}}{e\cdot 5^{5/3}}$:

$\alpha =\frac{8^{8/3}}{8^{2/3}}=8^{6/3}=8^2=64$.