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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2019 · Mathematics · Binomial Theorem

If the constant term in the expansion of ( x -(k)/(x^2))^ 10 is 405...

If the constant term in the expansion of (x−kx2)10\left(\sqrt{x}-\frac{k}{x^2}\right)^{10} is 405 , then what can be the values of kk ?

Options

  1. A±2\pm 2
  2. B±3\pm 3
  3. C±5\pm 5
  4. D±9\pm 9

Correct answer

B. ±3\pm 3

Step-by-step solution

(x−kx2)10\left(\sqrt{x}-\frac{k}{x^2}\right)^{10} according to question. Term independent of xx can be calculated as 10Cr(x)10−r⋅(−kx2)r⇒x5−r/2−2r=x0⇒10−4r−r=0⇒r=2∴10C2⋅(x)8⋅(−kx2)2=405⇒10C2⋅k2=405⇒10!2!⋅8!⋅k2=405⇒k2=9⇒k=±3\begin{aligned} & { }^{10} C_r(\sqrt{x})^{10-r} \cdot\left(-\frac{k}{x^2}\right)^r \\ & \Rightarrow x^{5-r / 2-2 r}=x^0 \\ & \Rightarrow 10-4 r-r=0 \\ & \Rightarrow r=2 \\ & \therefore \quad{ }^{10} C_2 \cdot(\sqrt{x})^8 \cdot\left(\frac{-k}{x^2}\right)^2=405 \\ & \Rightarrow \quad{ }^{10} C_2 \cdot k^2=405 \\ & \Rightarrow \frac{10!}{2!\cdot 8!} \cdot k^2=405 \\ & \Rightarrow k^2=9 \\ & \Rightarrow k= \pm 3 \end{aligned}
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