- Home
- Standard 11
- Mathematics
7.Binomial Theorem
hard
$\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}$ के प्रसार में $x$ से स्वतंत्र पद है
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