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7.Binomial Theorem
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જો દ્રીપદી વિસ્તરણ $\left(\frac{\mathrm{x}}{4}-\frac{12}{\mathrm{x}^{2}}\right)^{12}$ માં $\left(\frac{3^{6}}{4^{4}}\right) \mathrm{k}$ એ $\mathrm{x}$ થી સ્વતંત્ર છે તો $\mathrm{k}$ ની કિમંત મેળવો.
A
$22$
B
$11$
C
$55$
D
$99$
(JEE MAIN-2021)
Solution
$\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$
$T_{r+1}=(-1)^{r} \cdot{ }^{12} C_{r}\left(\frac{x}{4}\right)^{12-\mathrm{r}}\left(\frac{12}{x^{2}}\right)^{r}$
$T_{r+1}=(-1)^{r} \cdot{ }^{12} C_{r}\left(\frac{1}{4}\right)^{12-r}(12)^{r} \cdot(x)^{12-3 r}$
Term independent of $x \Rightarrow 12-3 r=0 \Rightarrow r=4$
$\mathrm{T}_{5}=(-1)^{4} \cdot{ }^{12} \mathrm{C}_{4}\left(\frac{1}{4}\right)^{8}(12)^{4}=\frac{3^{6}}{4^{4}} \cdot \mathrm{k}$
$\Rightarrow \mathrm{k}=55$
Standard 11
Mathematics
