JEE Main 2024 — Definite Integration Question with Solution

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

\int_\limits0^1\left(1-x^{10}\right)^{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:

\int_\limits0^1 (1 - t)^{20} \cdot \frac{1}{10} t^{-\frac{9}{10}} dt

which simplifies to:

\frac{1}{10} \int_\limits0^1 (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.