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If f(x)={kcos⁡xπ−2x,x≠π23,x=π2f(x) = \begin{cases} \frac{k\cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases} is continuous at x=π2x = \frac{\pi}{2}, then k is :

Solution

✅ Correct Option: 1

Let h=π2−xh = \frac{\pi}{2} - x. Then cos⁡x=sin⁡h\cos x = \sin h and π−2x=2h\pi - 2x = 2h. So lim⁡x→π/2kcos⁡xπ−2x=lim⁡h→0ksin⁡h2h=k2\lim_{x \to \pi/2} \frac{k\cos x}{\pi - 2x} = \lim_{h \to 0} \frac{k \sin h}{2h} = \frac{k}{2}. For continuity: k2=3\frac{k}{2} = 3, so k=6k = 6.

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