The term independent of x in the expansion of ( (x+1) (x^ 2 / 3 +1-x^ 1 / 3 ) - (x+1) (x-x^ 1 / 2 ) )^ 10 , x 1 is:

Correct answer: A. 210 aligned & ( (x+1) (x^ (2)/(3) +1-x^ (1)/(3) ) - (x-1) (x-x^ (1)/(2) ) )^ 10 & =((x^ (1)/(3) +1)-( x +1 x ))^ 10 & =(x^ (1)/(3) - 1 x )^ 10 & T_ r+1 = ^ 10 C_r(x)^ (10-r)/(3) (-1)^r(x)^ -(r)/(2) & (10-r)/(3)-(r)/(2)=0 & (20-2 r)-3 r=0 & r^ 10 =4 & ^ 10 C_4(-1)^4=210 aligned

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Binomial Theorem

The term independent of x in the expansion of ( (x+1) (x^ 2 / 3...

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:

Options

A $210$
B $150$
C $240$
D $120$

Correct answer

A. $210$

Step-by-step 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}

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