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

Mathematics · Binomial Theorem

JEE Main 2024 — 8 April, Shift 2 — Question 18

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

$\left(\sqrt{\mathrm{ax}}{ }^{2}+\frac{1}{2 \mathrm{x}^{3}}\right)^{10}$

General term $={ }^{10} C_{r}\left(\sqrt{a} x^{2}\right)^{10-r}\left(\frac{1}{2 x^{3}}\right)^{r}$

$20-2 r-3 r=0$

$\mathrm{r}=4$

${ }^{10} \mathrm{C}_{4} \mathrm{a}^{3} \cdot \frac{1}{16}=105$

$\mathrm{a}^{3}=8$

$\mathrm{a}^{2}=4$

Answer key and solution verified before publishing.

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