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7.Binomial Theorem
medium
If the term independent of $x$ in the expansion of $\left(\sqrt{\mathrm{ax}}{ }^2+\frac{1}{2 \mathrm{x}^3}\right)^{10}$ is 105 , then $\mathrm{a}^2$ is equal to :
A
$4$
B
$9$
C
$6$
D
$2$
(JEE MAIN-2024)
Solution
$ \left(\sqrt{\mathrm{a}} \mathrm{x}^2+\frac{1}{2 \mathrm{x}^3}\right)^{10} $
$ \text { General term }={ }^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{ax}})^{10-\mathrm{r}}\left(\frac{1}{2 \mathrm{x}^3}\right)^{\mathrm{r}} $
$ 20-2 \mathrm{r}-3 \mathrm{r}=0 $
$ \mathrm{r}=4 $
$ { }^{10} \mathrm{C}_4 \mathrm{a}^3 \cdot \frac{1}{16}=105 $
$ \mathrm{a}^3=8 $
$ \mathrm{a}^2=4$
Standard 11
Mathematics
