Question

If the coefficient of \(x^m\) in the expansion of \(\left(\sqrt{2x} + \sqrt[3]{\frac{3}{x^2}}\right)^9\) is equal to \(k\), then \(k\) is:

• A. \(1008\)
• B. \(2016\)
  % option (C) \(3024\)
• C. \(1016\)
• D. "\(1816\) "

Hint:

Always include constants \(2^{(\cdot)}\) and \(3^{(\cdot)}\) while finding coefficient.

Answer 1

The Correct Option is B

Concept: General term: \[ T_r = \binom{9}{r} (\sqrt{2x})^{9-r} \left(\sqrt[3]{\frac{3}{x^2}}\right)^r \]

Step 1:Simplify terms \[ (\sqrt{2x})^{9-r} = (2x)^{\frac{9-r}{2}} = 2^{\frac{9-r}{2}} x^{\frac{9-r}{2}} \] \[ \left(\sqrt[3]{\frac{3}{x^2}}\right)^r = \left(\frac{3}{x^2}\right)^{r/3} = 3^{r/3} x^{-2r/3} \]

Step 2:Power of \(x\) \[ x^{\frac{9-r}{2} - \frac{2r}{3}} = x^{\frac{27 - 7r}{6}} \]

Step 3:For integral power \[ \frac{27 - 7r}{6} \in \mathbb{Z} \Rightarrow 27 - 7r \equiv 0 \ (\text{mod }6) \] \[ 27 \equiv 3 \Rightarrow 7r \equiv 3 \ (\text{mod }6) \Rightarrow r \equiv 3 \ (\text{mod }6) \]

Step 4:Possible values \[ r = 3,\ 9 \]

Step 5:Coefficient calculation For \(r = 3\): \[ T_4 = \binom{9}{3} \cdot 2^{3} \cdot 3^{1} = 84 \cdot 8 \cdot 3 = 2016 \] For \(r = 9\): \[ T_{10} = \binom{9}{9} \cdot 2^{0} \cdot 3^{3} = 1 \cdot 1 \cdot 27 = 27 \]

Step 6:Required coefficient Coefficient of \(x^m\) (integral power term with highest contribution): \[ k = 2016 \] Conclusion : 2016