Question

The coefficient of \( x^3 \) in the binomial expansion of \( \left( \frac{1}{\sqrt{x}} - x \right)^6 \) is:

• A. 12
• B. 15
• C. 10
• D. 30
• E. 20

Hint:

In binomial expansions, express terms in powers of \( x \) and solve for \( k \) to find the required coefficient.

Answer 1

The Correct Option is B

In the expansion of \( (a - b)^n \), the general term is: \[ T_{k+1} = \binom{6}{k} a^{6-k} b^k \] For \( \left( \frac{1}{\sqrt{x}} - x \right)^6 \), we have: \[ T_k = \binom{6}{k} \left(\frac{1}{\sqrt{x}}\right)^{6-k} (-x)^k \] \[ = \binom{6}{k} x^{\frac{-(6-k)}{2}} (-1)^k x^k \] Setting exponent of \( x \) to 3: \[ k - \frac{(6-k)}{2} = 3 \] \[ 3k - 6 + k = 6 \Rightarrow 4k = 12 \Rightarrow k = 4 \] Substituting \( k = 4 \): \[ T_5 = \binom{6}{4} (-1)^4 x^3 \] \[ \binom{6}{4} = 15 \] Thus, the coefficient of \( x^3 \) is 15.