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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 25 of 50
Marks: +1, -0
If a function ff defined by
f(x)={kcos⁡xπ−2xif x≠π23if x=π2f(x)=\begin{cases} \frac{k\cos x}{\pi-2x} & \text{if } x \neq \frac{\pi}{2} \\ 3 & \text{if } x = \frac{\pi}{2} \end{cases}
is continuous at x=π2x=\frac{\pi}{2}, then the value of kk is
Solution:     👈: Video Solution
Explanation: Since, f(x)f(x) is continuous at x=π2x=\frac{\pi}{2}
Therefore, lim⁡x→π2f(x)=f(π2)\lim\limits_{x\rightarrow\frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)
⇒lim⁡x→π2kcos⁡xπ−2x=3\Rightarrow \lim\limits_{x\rightarrow\frac{\pi}{2}} \frac{k\cos x}{\pi-2x} = 3
⇒klim⁡x→π2sin⁡(π2−x)2(π2−x)=3\Rightarrow k \lim\limits_{x\rightarrow\frac{\pi}{2}} \frac{\sin\left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)} = 3
⇒k2lim⁡x→π2sin⁡(π2−x)π2−x=3\Rightarrow \frac{k}{2} \lim\limits_{x\rightarrow\frac{\pi}{2}} \frac{\sin\left(\frac{\pi}{2}-x\right)}{\frac{\pi}{2}-x} = 3
⇒k2×1=3⇒k=6\Rightarrow \frac{k}{2} \times 1 = 3 \Rightarrow k = 6
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