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Given,$3^{49}(x+iy)=(\large\frac{3}{2}+\frac{\sqrt 3}{2})^{100}$$\quad y\in N$ and $x=ky$,then value of k is

$\begin{array}{1 1}(A)\;-\large\frac{1}{3}&(B)\;-\large\frac{1}{\sqrt 3}\\(C)\;2\sqrt 2&(D)\;\large\frac{1}{4}\end{array} $

1 Answer

We have
$3^{49}(x+iy)=(\large\frac{3}{2}+\frac{\sqrt 3}{2})^{100}$
$\Rightarrow 3^{49}\sqrt{x^2+y^2}=(\large\frac{9}{4}+\frac{ 3}{4})^{50}$
$\Rightarrow \sqrt{x^2+y^2}=3$
$\Rightarrow y\sqrt{1+k^2}=3$
$\Rightarrow \sqrt{1+k^2}=3,\large\frac{3}{2},$$1$ as $y \in N$
$\Rightarrow k=\pm 2\sqrt 2,\pm \large\frac{\sqrt 5}{2}$$,0$
Out of given options,only $2\sqrt 2$ satisfies
Hence (C) is the correct answer.
answered Apr 9, 2014 by sreemathi.v

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