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7.Binomial Theorem
hard
$\left(x \sin \alpha+a \frac{\cos \alpha}{x}\right)^{10}$ ના વિસ્તરણમાં જો અચળ પદ $\frac{10 !}{(5 !)^{2}}$ હોય તો $' a^{\prime}$ ની કિમંત મેળવો.
A
$2$
B
$-1$
C
$1$
D
$-2$
(JEE MAIN-2021)
Solution
$T_{r+1}={ }^{10} C_{r}(x \sin \alpha)^{10-t}\left(\frac{a \cos \alpha}{x}\right)^{r}$
$r=0,1,2, \ldots, 10$
$T_{r+1}$ will be independent of $x$
When $10-2 r=0 \Rightarrow r=5$
$T_{6}={ }^{10} C_{5}(x \sin \alpha)^{5} \times\left(\frac{a \cos \alpha}{x}\right)^{5}$
$=^{10} C_{5} \times a^{5} \times \frac{1}{2^{5}}(\sin 2 \alpha)^{5}$
will be greatest when $\sin 2 \alpha=1$
$\Rightarrow^{10} C_{5} \frac{a^{5}}{2^{5}}={ }^{10} C_{5} \Rightarrow a=2$
Standard 11
Mathematics