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7.Binomial Theorem
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यदि $\left(\sqrt{ x }-\frac{ k }{ x ^{2}}\right)^{10}$ के द्विपद प्रसार में अचर में पद $405$ , है तो $| k |$ बराबर है
A
$2$
B
$1$
C
$3$
D
$9$
(JEE MAIN-2020)
Solution
$\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$
$T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r}\left(\frac{-k}{x^{2}}\right)^{r}$
$T_{r+1}={ }^{10} C_{r} \cdot x^{\frac{10-r}{2}} \cdot(-k)^{r} \cdot x^{-2 r}$
$T_{r+1}={ }^{10} C_{r} x^{\frac{10-5 r}{2}}(-k)^{r}$
Constant term $: \frac{10-5 r}{2}=0 \Rightarrow r=2$
$T_{3}={ }^{10} C_{2} \cdot(-k)^{2}=405$
$k^{2}=\frac{405}{45}=9$
$k=\pm 3 \Rightarrow|k|=3$
Standard 11
Mathematics
