Question

If the 5th term of \( \left( 2x^2 + \frac{3}{x} \right)^5 \) is 10, then \( x = \)

• A. 6
• B. +9
• C. -6
• D. +8

Hint:

In binomial expansions, carefully identify the correct term and simplify using the powers of the variables involved.

Answer 1

The Correct Option is B

The general term in the expansion of \( (a + b)^n \) is given by: \[ T_{r+1} = \binom{n}{r} a^{n-r} b^r. \] For the expansion of \( \left( 2x^2 + \frac{3}{x} \right)^5 \), the 5th term corresponds to \( r = 4 \). Thus: \[ T_5 = \binom{5}{4} (2x^2)^{5-4} \left( \frac{3}{x} \right)^4 = 5 \times (2x^2) \times \frac{81}{x^4}. \] Simplifying: \[ T_5 = 5 \times 2x^2 \times \frac{81}{x^4} = 5 \times 2 \times 81 \times \frac{1}{x^2}. \] We are given that \( T_5 = 10 \), so: \[ 5 \times 2 \times 81 \times \frac{1}{x^2} = 10 \quad \Rightarrow \quad \frac{810}{x^2} = 10 \quad \Rightarrow \quad x^2 = 81 \quad \Rightarrow \quad x = 9. \]