QJEEM24Mae8a7: JEE Main 2024 Mathematics - Class 12 JEEM24Mae8a7 Solution: Let β ( m , n ) = ∫ 0 1 x m − 1 ( 1 − x ) n − 1   d x ,…

First, let's rewrite the given integral using the given form of the Beta function. The given integral is:

$$\underset{0}{\overset{1}{\int }}{(1-x^{10})}^{20}\mathrm{d}x$$

To use the Beta function, let us make a substitution. Let $x^{10} = t$ . Then, $dx = \frac{1}{10}t^{-\frac{9}{10}} dt$ or $dx = \frac{1}{10} t^{-\frac{9}{10}} dt$ . The limits of integration change as follows: when $x = 0$ , $t = 0$ , and when $x = 1$ , $t = 1$ .

Substituting these into the integral, we have:

$$\underset{0}{\overset{1}{\int }}(1-t{)}^{20}\cdot \frac{1}{10}{t}^{-\frac{9}{10}}dt$$

which simplifies to:

$$\frac{1}{10}\underset{0}{\overset{1}{\int }}(1-t{)}^{20}{t}^{-\frac{9}{10}}dt$$

We recognize this integral as a Beta function $\beta(m, n)$ where $m = 1 - \frac{9}{10} = \frac{1}{10}$ and $n = 20 + 1 = 21$ .

Therefore, we can write this as:

$\frac{1}{10} \beta \left( \frac{1}{10}, 21 \right)$

Comparing this to $a \times \beta(b, c)$ , we have $a = \frac{1}{10}$ , $b = \frac{1}{10}$ , and $c = 21$ .

Now we calculate $100(a + b + c)$ :

$100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{5} + 21 \right) = 100 \left( \frac{1}{5} + \frac{105}{5} \right) = 100 \left( \frac{106}{5} \right) = 100 \times 21.2 = 2120$

So, the answer is Option D, 2120.