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(A) The general term $T_{r+1}$ in the expansion of $(\sqrt{a}x^2 + \frac{1}{2x^3})^{10}$ is given by:$T_{r+1} = {}^{10}C_r (\sqrt{a}x^2)^{10-r} (\frac

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$4$

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(A) The general term $T_{r+1}$ in the expansion of $(\sqrt{a}x^2 + \frac{1}{2x^3})^{10}$ is given by:$T_{r+1} = {}^{10}C_r (\sqrt{a}x^2)^{10-r} (\frac{1}{2x^3})^r$$T_{r+1} = {}^{10}C_r (\sqrt{a})^{10-r} x^{20-2r} \cdot (\frac{1}{2})^r x^{-3r}$$T_{r+1} = {}^{10}C_r (\sqrt{a})^{10-r} (\frac{1}{2})^r x^{20-5r}$For the term to be independent of $x$,the exponent of $x$ must be zero:$20 - 5r = 0 \implies r = 4$Substituting $r = 4$ into the expression for the term:$T_{4+1} = {}^{10}C_4 (\sqrt{a})^{10-4} (\frac{1}{2})^4 = 105$${}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$$210 \cdot a^3 \cdot \frac{1}{16} = 105$$a^3 \cdot \frac{210}{16} = 105$$a^3 \cdot \frac{105}{8} = 105$$a^3 = 8 \implies a = 2$Therefore,$a^2 = 2^2 = 4$.

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

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