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7.Binomial Theorem
hard
The term independent of $x$ in the expansion of $\left(\frac{(x+1)}{\left(x^{2 / 3}+1-x^{1 / 3}\right)}-\frac{(x+1)}{\left(x-x^{1 / 2}\right)}\right)^{10}, x>1$ is:
A$210$
B$150$
C$240$
D$120$
(JEE MAIN-2025)
Solution
$\left(\frac{(x+1)}{\left(x^{\frac{2}{3}}+1-x^{\frac{1}{3}}\right)}-\frac{(x-1)}{\left(x-x^{\frac{1}{2}}\right)}\right)^{10}$
$=\left(\left(x^{\frac{1}{3}}+1\right)-\left(\frac{\sqrt{ x }+1}{\sqrt{ x }}\right)\right)^{10}$
$=\left(x^{\frac{1}{3}}-\frac{1}{\sqrt{x}}\right)^{10}$
$T_{ r +1}={ }^{10} C _{ r }( x )^{\frac{10- r }{3}}(-1)^{ r }( x )^{-\frac{ r }{2}}$
$\frac{10- r }{3}-\frac{ r }{2}=0$
$(20-2 r )-3 r =0$
$r =4$
$\Rightarrow{ }^{10} C _4(-1)^4=210$
$=\left(\left(x^{\frac{1}{3}}+1\right)-\left(\frac{\sqrt{ x }+1}{\sqrt{ x }}\right)\right)^{10}$
$=\left(x^{\frac{1}{3}}-\frac{1}{\sqrt{x}}\right)^{10}$
$T_{ r +1}={ }^{10} C _{ r }( x )^{\frac{10- r }{3}}(-1)^{ r }( x )^{-\frac{ r }{2}}$
$\frac{10- r }{3}-\frac{ r }{2}=0$
$(20-2 r )-3 r =0$
$r =4$
$\Rightarrow{ }^{10} C _4(-1)^4=210$
Standard 11
Mathematics
