Mathematics | VITEEE 2017 | Discussion Page

The set of points of discontinuity of the function

$f(x)=\lim_{n \rightarrow \alpha}\frac{(2\sin x)^{2n}}{3^{n}-(2\cos x)^{2n}}$ is given by

A) R

B) ${n\pi\pm\frac{\pi}{3},n\epsilon I}$

C) ${n\pi\pm\frac{\pi}{6},n\epsilon I}$

D) None of these

Option C

We have, $f(x)=\lim_{n \rightarrow \alpha}\frac{(2\sin x)^{2n}}{3^{n}-(2\cos x)^{2n}}$

$=\lim_{n \rightarrow \alpha}\frac{(2\sin x)^{2n}}{(\sqrt{3})^{2n}-(2\cos x)^{2n}}$

f(x) is discountinuous when

${(\sqrt{3})^{2n}-(2\cos x)^{2n}=0}$

i.e. $\cos x = \pm\frac{\sqrt{3}}{2}\Rightarrow x =n\pi\pm\frac{\pi}{6}$

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