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7.Binomial Theorem
hard
$\left(\frac{ x +1}{ x ^{2 / 3}- x ^{1 / 3}+1}-\frac{ x -1}{ x - x ^{1 / 2}}\right)^{10}, x \neq 0,1$ के प्रसार में ' $x$ ' से स्वतंत्र पद बराबर है
A
$110$
B
$210$
C
$300$
D
$400$
(JEE MAIN-2021)
Solution
$\left(\left(x^{1 / 3}+1\right)-\left(\frac{x^{1 / 2}+1}{x^{1 / 2}}\right)\right)^{10}$
$=\left(x^{1 / 3} \frac{1}{x^{1 / 2}}\right)^{10}$
Now General Term
$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{r}\left(\mathrm{x}^{1 / 3}\right)^{10-\mathrm{r}} \cdot\left(-\frac{1}{\mathrm{x}^{1 / 2}}\right)^{\mathrm{r}}$
For independent term
$\frac{10-r}{3}-\frac{r}{2}=0 \Rightarrow r=4$
$\Rightarrow T_{5}={ }^{10} C_{4}=210$
Standard 11
Mathematics
