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7.Binomial Theorem
hard
The term independent of $x$ in the expansion of $\left[\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right]^{10}, x \neq 1,$ is equal to ....... .
A
$240$
B
$225$
C
$210$
D
$196$
(JEE MAIN-2021)
Solution
$\left.\left( x ^{1 / 3}+1\right)-\left(\frac{\sqrt{ x }+1}{\sqrt{ x }}\right)\right)^{10}$
$\left( x ^{1 / 3}- x ^{-1 / 2}\right)^{10}$
$T _{ r +1}={ }^{10} C _{ r }\left( x ^{1 / 3}\right)^{10- r }\left(- x ^{-1 / 2}\right)^{ r }$
$\frac{10- r }{3}-\frac{ r }{2}=0 \Rightarrow 20-2 r -3 r =0$
$\Rightarrow r =4$
$T _{5}={ }^{10} C _{4}=\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210$
Standard 11
Mathematics
