QJEEM23Ma04f3: JEE Main 2023 Mathematics - Class 12 JEEM23Ma04f3 Solution: Let and . If , then is equal to _________.

Given integral: $\int \sqrt{\frac{x+7}{x}} \, dx$

Let's make the substitution $x = t^2$ . Then, $dx = 2t \, dt$ .

Substituting these values, the integral becomes :

$\int 2 \sqrt{t^2 + 7} \, dt$

Now, let's evaluate this integral :

$I(t) = 2\left(\frac{t}{2} \sqrt{t^2+7} + \frac{7}{2} \ln\left|t + \sqrt{t^2+7}\right|\right) + C$

Substituting back $t = \sqrt{x}$ , we have :

$I(x) = 2\left(\frac{\sqrt{x}}{2} \sqrt{x+7} + \frac{7}{2} \ln\left|\sqrt{x} + \sqrt{x+7}\right|\right) + C$

Simplifying further :

$I(x) = \sqrt{x} \sqrt{x+7} + 7 \ln\left|\sqrt{x} + \sqrt{x+7}\right| + C$

We are given that $I(9) = 12+7 \ln 7$ .

Let's substitute $x = 9$ and solve for the constant $C$ :

$12+7 \ln 7 = \sqrt{9} \sqrt{9+7}+7 \ln |\sqrt{9}+\sqrt{9+7}|+C$

$\Rightarrow$ $12+7 \ln 7 = 3 \sqrt{16}+7 \ln |\sqrt{9}+\sqrt{16}|+C$

$\Rightarrow$ $12+7 \ln 7 = 3 \cdot 4+7 \ln (3+\sqrt{16})+C$

$\Rightarrow$ $12+7 \ln 7 = 12+7 \ln (3+4)+C$

$\Rightarrow$ $12+7 \ln 7 = 12+7 \ln 7+C$

From this equation, we can see that $C = 0$ .

Now, we need to calculate $I(1)$ :

$I(1) = \sqrt{1} \sqrt{1+7}+7 \ln |\sqrt{1}+\sqrt{1+7}|$

$\Rightarrow$ $I(1) = 1 \sqrt{8}+7 \ln (1+\sqrt{8})$

$\Rightarrow$ $I(1) = \sqrt{8}+7 \ln (1+2 \sqrt{2})$

Therefore, $\alpha = \sqrt{8}$ .

Finally, to find $\alpha^4$ :

$\alpha^4 = \left(\sqrt{8}\right)^4$

$\Rightarrow$ $\alpha^4 = 8^2$

$\Rightarrow$ $\alpha^4 = 64$

Hence, $\alpha^4$ is equal to 64.