TS EAMCET · Maths · Binomial Theorem
If the coefficient of \(x^3\) in the binomial expansion of \(x^3\left(2 \sqrt{3} x^2+\frac{1}{k x}\right)^{12}\) is 880 , then \(k\) is equal to
- A \(2 \sqrt{2}\)
- B \(4 \sqrt{3}\)
- C \(2 \sqrt{3}\)
- D \(\sqrt{3}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & x^3\left[{ }^{12} C_4\left(2 \sqrt{3} x^2\right)^4 \cdot\left(\frac{1}{k x}\right)^8\right]=880 x^3 \\ & \Rightarrow \quad \frac{12 !}{4 ! 8 !} \times(16 \times 9) \times \frac{1}{k^8}=880\end{aligned}\)…
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