Question

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

• A. 4
• B. 9
• C. 6
• D. 2

Answer 1

Answer: (A)

4

Answer 2

Consider the given expression:

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

The general term in the binomial expansion of $(x + y)^n$ is given by:

$T_{r+1} = \binom{n}{r} x^{n-r} y^r.$

For the given expression, the general term becomes:

$T_{r+1} = \binom{10}{r} \left( \sqrt{ax^2} \right)^{10-r} \left( \frac{1}{2x^3} \right)^r.$

Simplify the powers of $x$:

$T_{r+1} = \binom{10}{r} \left( \sqrt{a} \right)^{10-r} x^{(20-2r)} \times \frac{1}{(2x^3)^r}$

Combine the powers of $x$:

$T_{r+1} = \binom{10}{r} \left( \sqrt{a} \right)^{10-r} \times \frac{1}{2^r} \times x^{20-5r}$

To find the term independent of $x$, set the power of $x$ to zero:

$20 - 5r = 0$

Solve for $r$:

$r = 4$

Substitute $r = 4$ into the general term:

$T_5 = \binom{10}{4} \left( \sqrt{a} \right)^{10-4} \times \frac{1}{2^4}$

Simplify:

$T_5 = \binom{10}{4} \left( \sqrt{a} \right)^6 \times \frac{1}{16}$

Substitute $\binom{10}{4} = 210$:

$T_5 = 210 \times \left( \sqrt{a} \right)^6 \times \frac{1}{16}$

$T_5 = 210 \times \frac{a^3}{16}$

The value of $T_5$ is given as 105:

$210 \times \frac{a^3}{16} = 105$

Solve for $a^3$:

$a^3 = \frac{105 \times 16}{210}$

$a^3 = 8$

Take the cube root of both sides:

$a = \sqrt[3]{8}$

$a = 2$

Finally, $a^2 = 4$.

Answer: $(1) \ 4$