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

The Correct Option is A

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: \((1) \ 4\)