The term independent of $x$ in the expansion of $\left(\frac{(x+1)}{(x^{2/3}+1-x^{1/3})}-\frac{(x-1)}{(x-x^{1/2})}\right)^{10}$, $x>1$ is:

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Binomial Theorem

The term independent of

x

in the expansion of

(\frac{( x + 1 )}{( x ^{2/3} + 1 - x ^{1/3} )} - \frac{( x - 1 )}{( x - x ^{1/2} )})^{10}

x > 1

is:

Answer: (A)

Since

x + 1 = (x^{1/3})^{3} + 1 = (x^{1/3} + 1) (x^{2/3} - x^{1/3} + 1)

, the first fraction reduces to

\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} = x^{1/3} + 1.

For the second,

x - 1 = (x^{1/2} - 1) (x^{1/2} + 1)

and

x - x^{1/2} = x^{1/2} (x^{1/2} - 1)

, so

\frac{x - 1}{x - x ^{1/2}} = \frac{x ^{1/2} + 1}{x ^{1/2}} = 1 + x^{- 1/2} .

The bracket therefore equals

(x^{1/3} + 1) - (1 + x^{- 1/2}) = x^{1/3} - x^{- 1/2}

, and the expression is

(x^{1/3} - x^{- 1/2})^{10}

. Its general term is

T_{r + 1} = (r 10) (x^{1/3})^{10 - r} (- x^{- 1/2})^{r} = (r 10) (- 1)^{r} x^{\frac{10 - r}{3} - \frac{r}{2}} .

The power of

x

vanishes when

\frac{10 - r}{3} - \frac{r}{2} = 0

, that is

20 - 5 r = 0

, giving

r = 4

. The term independent of

x

(4 10) (- 1)^{4} = 210

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option