The set of points of discontinuity of the function ⨍ (x)=Ltn→∞(2sin x)2n3n-(2cos x)2n is

f(x)=limn→∞2 sin x2n3n-2 cos x2n is discontinuous if 3n-2 cos x2n=0 ⇒4 cos2xn=3n ⇒cos2x=34=cos2π6 ⇒x=nπ±π6; n∈z

The set of points of discontinuity of the function ⨍ (x)=Ltn→∞(2sin x)2n3n-(2cos x)2n is

f(x)=limn→∞2 sin x2n3n-2 cos x2n is discontinuous if 3n-2 cos x2n=0 ⇒4 cos2xn=3n ⇒cos2x=34=cos2π6 ⇒x=nπ±π6; n∈z

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The set of points of discontinuity of the function $⨍ (x)= \underset{n \to \infty}{Lt} \frac{{(2 \sin x)}^{2 n}}{3^{n} - {(2 \cos x)}^{2 n}}$ is

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$\{nπ / n \in Z\}$

$\{nπ \pm π / 6 / n \in Z\}$

$\{nπ \pm π / 2 / n \in Z\}$

$\{nπ \pm π / 3 / n \in Z\}$

answer is B.

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Detailed Solution

$f (x) = \lim_{n \to \infty} \frac{{(2 \sin x)}^{2 n}}{3^{n} - {(2 \cos x)}^{2 n}} is discontinuous if 3^{n} - {(2 \cos x)}^{2 n} = 0 \Rightarrow {(4 \cos^{2} x)}^{n} = 3^{n} \Rightarrow \cos^{2} x = \frac{3}{4} = \cos^{2} \frac{π}{6} \Rightarrow x = nπ \pm \frac{π}{6}; n \in z$