QJEEM23Maaa5a: JEE Main 2023 Mathematics - Class 12 JEEM23Maaa5a Solution: If , then is equal to :

1. The given integral is:

$I=\int_0^1 \frac{dx}{\left(5+2x-2x^2\right)\left(1+e^{2-4x}\right)}$ .............(i)

1. Perform a substitution, $x \rightarrow 1-x$ . This gives:

$I=\int_0^1 \frac{dx}{\left(5+2(1-x)-2(1-x)^2\right)\left(1+e^{2-4(1-x)}\right)}$

1. Simplify the expression inside the integral:

$I=\int_0^1 \frac{dx}{\left(5+2-2x-2(1-2x+x^2)\right)\left(1+e^{4x -2}\right)}$ ............(ii)

1. Add the original integral (i) and the integral after substitution (ii):

$2I=\int_0^1 \frac{dx}{5+2x-2x^2}$

1. Factor the quadratic expression in the denominator:

$2I=\int_0^1 \frac{dx}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)}$

1. To solve this integral, we can perform a change of variables using the substitution
   
   $x-\frac{1}{2}= \frac{\sqrt{11}}{2}\tan{u}$ , then $dx=\frac{\sqrt{11}}{2}\sec^2{u} du$ :

$I=\frac{1}{\sqrt{11}}\int \frac{\sec^2{u}}{1+\tan^2{u}}du$

1. Using the identity $\sec^2{u}=1+\tan^2{u}$ :

$I=\frac{1}{\sqrt{11}}\int du = \frac{1}{\sqrt{11}}(u+C)$

1. Now we need to find the limits of the integral after the substitution. If $x=0$ , then $u=\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)$ . If $x=1$ , then $u=\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)$ . So, the integral becomes:

$I=\frac{1}{\sqrt{11}}\left[\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)-\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)\right]$

1. Using the properties of the arctangent function, we can rewrite the integral as:

$I=\frac{1}{\sqrt{11}} \ln\left(\frac{\sqrt{11}+1}{\sqrt{10}}\right)$

1. From this result, we have $\alpha=\sqrt{11}$ and $\beta=\sqrt{10}$ . Now, we can find $\alpha^4 - \beta^4$ :

$\alpha^4 - \beta^4 = (11)^2 - (10)^2 = 121 - 100 = 21$

Thus, $\alpha^4 - \beta^4 = 21$ .