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7.Binomial Theorem
hard
${\left( {\frac{{x + 1}}{{{x^{2/3}} - {x^{\frac{1}{3}}} + 1\;}}--\frac{{x - 1}}{{x - {x^{1/2}}}}} \right)^{10}}$ના વિસ્તરણમાં અચળ પદ મેળવો.
A
$4$
B
$120$
C
$210$
D
$310$
(JEE MAIN-2013)
Solution
$\left(\left(x^{1 / 3}+1\right)-\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)\right)^{10}$
$\left(x^{13}-x^{-1 / 2}\right)^{10}$
${T_{r + 1}}{ = ^{10}}{C_r}{({x^{1/3}})^{10 – r}}{( – {x^{ – 1/2}})^r}$
$\frac{10-r}{3}-\frac{r}{2}=0$
$\Rightarrow \quad 20-2 r-3 r=0$
$\Rightarrow \quad r=4$
$T_{5}=^{10} C_{4}=\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210$
Standard 11
Mathematics
