Q28 / 75 · JEE Main 2025 (2 Apr Shift 1) · Mathematics

The term independent of $x$ in the expansion of $\left(\frac{(x+1)}{\left(x^{2 / 3}+1-x^{1 / 3}\right)}-\frac{(x+1)}{\left(x-x^{1 / 2}\right)}\right)^{10}, x\gt1$ is:

  • Correct answer
    A
    $210$
  • B
    $150$
  • C
    $240$
  • D
    $120$

Solution

$\begin{aligned} & \left(\frac{(x+1)}{\left(x^{\frac{2}{3}}+1-x^{\frac{1}{3}}\right)}-\frac{(x-1)}{\left(x-x^{\frac{1}{2}}\right)}\right)^{10} \\ & =\left(\left(x^{\frac{1}{3}}+1\right)-\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)\right)^{10} \\ & =\left(x^{\frac{1}{3}}-\frac{1}{\sqrt{x}}\right)^{10} \\ & T_{r+1}={ }^{10} C_r(x)^{\frac{10-r}{3}}(-1)^r(x)^{-\frac{r}{2}} \\ & \frac{10-r}{3}-\frac{r}{2}=0 \\ & (20-2 r)-3 r=0 \\ & r^{10}=4 \\ & \Rightarrow{ }^{10} C_4(-1)^4=210\end{aligned}$

Hence, the correct answer is option: 'A'.

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