Question

The number \(81\) is the coefficient of \( x^k \) in the binomial expansion of \( \left(x^2 + \frac{3}{x}\right)^4 \), \( x \neq 0 \). Then the value of \( k \) equals

• A. \(-2\)
• B. \(2\)
• C. \(-4\)
• D. \(4\)
• E. \(5\)

Hint:

Always match BOTH coefficient and power when identifying a term.

Answer 1

The Correct Option is C

Concept: Use binomial expansion and track both coefficient and power of \(x\).

Step 1: Write general term. \[ T_r = \binom{4}{r}(x^2)^{4-r}\left(\frac{3}{x}\right)^r \]

Step 2: Simplify powers carefully. \[ (x^2)^{4-r} = x^{8-2r} \] \[ \left(\frac{3}{x}\right)^r = 3^r x^{-r} \] Multiply: \[ T_r = \binom{4}{r} 3^r x^{8-2r-r} \] \[ = \binom{4}{r} 3^r x^{8-3r} \]

Step 3: Match coefficient 81. We need: \[ \binom{4}{r} 3^r = 81 \] Try values: \[ r=4 \Rightarrow \binom{4}{4} 3^4 = 1 \cdot 81 = 81 \] So \(r=4\)

Step 4: Find corresponding power of \(x\). \[ k = 8 - 3r = 8 - 12 = -4 \]