JEE Main 2023 (Online) 15th April Morning ShiftQuestion 1
If $\int_{0}^{1}\frac{1}{\left(5+2x-2x^{2}\right)\left(1+e^{(2-4x)}\right)}dx=\frac{1}{\alpha}\log_e\left(\frac{\alpha+1}{\beta}\right)$, $\alpha,\beta>0$, then $\alpha^{4}-\beta^{4}$ is equal to
Definite Integration
Solution
Concept. The quadratic is unchanged by $x\to1-x$ while $e^{2-4x}\to e^{4x-2}$; adding removes the exponential. The rest is a standard $\frac1{a^2-t^2}$ integral. Formulas used. (1) $\dfrac1{1+e^{g}}+\dfrac1{1+e^{-g}}=1$(2) $\displaystyle\int\frac{dt}{a^2-t^2}=\frac1{2a}\ln\frac{a+t}{a-t}$Step 1 - $x\to1-x$. $5+2(1-x)-2(1-x)^2=5+2x-2x^2$ and $2-4(1-x)=-(2-4x)$. By (1), $$J=\frac12\int_0^1\frac{dx}{5+2x-2x^2}.$$ Step 2 - Complete the square. With $t=x-\frac12$: $5+2x-2x^2=\frac{11}2-2t^2=2\big(\frac{11}4-t^2\big)$, $t:-\frac12\to\frac12$. Step 3 - Use (2). $a=\frac{\sqrt{11}}2$: $$J=\frac14\cdot\frac1{\sqrt{11}}\Big[\ln\frac{a+t}{a-t}\Big]_{-1/2}^{1/2}=\frac1{2\sqrt{11}}\ln\frac{\sqrt{11}+1}{\sqrt{11}-1}.$$ Step 4 - Simplify. $\frac{\sqrt{11}+1}{\sqrt{11}-1}=\frac{(\sqrt{11}+1)^2}{10}$, so $$J=\frac1{\sqrt{11}}\ln\frac{\sqrt{11}+1}{\sqrt{10}}.$$ So $\alpha=\sqrt{11}$, $\beta=\sqrt{10}$, and $\alpha^4-\beta^4=121-100=21$. Answer: (D)