If the term independent of x in the expansion of ( a x 2 + 1 2 x 3 ) 10 is 105, then a 2 is equal to : [2024]

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Solution

Q.
If the term independent of $x$ in the expansion of ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$ is 105, then $a^{2}$ is equal to : [2024]

1 2

2 6

3 4

4 9

Ans.
(3)

We have, ${(\sqrt{a} x^{2} + \frac{1}{2 x^{3}})}^{10}$

$T_{r + 1} = {}^{10}C_{r} {(\sqrt{a} x^{2})}^{10 - r} {(\frac{1}{2 x^{3}})}^{r}$

$= {}^{10}C_{r} {(\sqrt{a})}^{10 - r} {(\frac{1}{2})}^{r} x^{20 - 2 r - 3 r}$

For the term to be independent of $x,$ we have $20 - 2 r - 3 r = 0$

$\Rightarrow r = 4$

$∴$ Required term $= T_{5} = {}^{10}C_{4} {(\sqrt{a})}^{6} {(\frac{1}{2})}^{4} = 105 (given)$

$\Rightarrow \frac{210}{16} a^{3} = 105 \Rightarrow a^{3} = 8 \Rightarrow a = 2$

Hence, $a^{2} = 4$

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