Binomial Theorem – Term Independent of x – JEE Main 2 Apr 2025 Shift 1

Solution:

Simplify the terms inside the bracket.

First term: Let $x^{1/3} = t \Rightarrow x = t^3$.
$$ \frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{t^3+1}{t^2-t+1} = \frac{(t+1)(t^2-t+1)}{t^2-t+1} = t+1 = x^{1/3} + 1 $$

Second term: Let $\sqrt{x} = u \Rightarrow x = u^2$.
$$ \frac{x-1}{x-\sqrt{x}} = \frac{u^2-1}{u^2-u} = \frac{(u-1)(u+1)}{u(u-1)} = \frac{u+1}{u} = 1 + \frac{1}{u} = 1 + x^{-1/2} $$

Now substitute these back into the expression:
$$ \left( (x^{1/3} + 1) – (1 + x^{-1/2}) \right)^{10} = \left( x^{1/3} – x^{-1/2} \right)^{10} $$

The general term $T_{r+1}$ in the expansion is:
$$ T_{r+1} = \binom{10}{r} (x^{1/3})^{10-r} (-x^{-1/2})^r $$

$$ T_{r+1} = \binom{10}{r} (-1)^r x^{\frac{10-r}{3} – \frac{r}{2}} $$

For the term independent of $x$, the exponent of $x$ must be 0:
$$ \frac{10-r}{3} – \frac{r}{2} = 0 \Rightarrow 2(10-r) – 3r = 0 $$
$$ 20 – 5r = 0 \Rightarrow r = 4 $$

Calculate the coefficient:
$$ T_5 = \binom{10}{4} (-1)^4 = \frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1} = 210 $$

Ans. (1)