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If the function f(x)={1−cos⁡kxsin⁡2x,x≠012,x=0f(x)=\begin{cases}\frac{1-\cos kx}{\sin^2 x}, & x\ne0\\\frac{1}{2}, & x=0\end{cases} is continuous at x=0x=0, then the value of kk is

Solution

✅ Correct Option: 3

Continuity needs lim⁡x→01−cos⁡kxsin⁡2x=12\lim_{x\to0}\frac{1-\cos kx}{\sin^2x}=\frac{1}{2}.

Using 1−cos⁡kx≈k2x221-\cos kx\approx\frac{k^2x^2}{2} and sin⁡2x≈x2\sin^2x\approx x^2, the limit is k22\frac{k^2}{2}, so k2=1k^2=1 and k=±1k=\pm1.

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