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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits3 Marks
Question
Let $\text{f(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, & \text{where x} \ne\frac\pi2\\3, & \text{where x} \ne\frac\pi2\end{cases}$ and if $\lim\limits_{\text{x}\rightarrow\frac\pi2}{\text{f(x)}}=\text{f}\Big(\frac\pi2\Big),$ find the value of k.
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Answer

We have,
$\text{f(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, & \text{where x} \ne\frac\pi2\\3, & \text{where x} \ne\frac\pi2\end{cases}$
It is given that,
$\lim\limits_{\text{x}\rightarrow\frac\pi2}{\text{f(x)}}={{\text{f}\big(\frac{\pi}{2}\big)}}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow\frac\pi2}\ \frac{{\text{k}\cos{\text{x}}}}{\pi-2\text{x}}=3$
$\Rightarrow\frac{{\text{k}}}{2}\times\lim\limits_{\text{x}-\frac\pi2\rightarrow0}\ \frac{\sin\big(\frac{\pi}{2}-{\text{x}}\big)}{\frac\pi2-{\text{x}}}=3$
$\Rightarrow\frac{\text{k}}{2}\times\lim\limits_{\text{h}\rightarrow0}\ \frac{\sin(-\text{h)}}{-\text{h}}=3$ $\big(​​\text{Put }\text{x}-\frac\pi2=\text{h}\big)$
$\Rightarrow\frac{\text{k}}{2}\times\lim\limits_{\text{h}\rightarrow0}\ \frac{\sin\text{h}}{\text{h}}=3$ $[\sin(-\theta)=-\sin\theta]$
$\Rightarrow\frac{\text{k}}{2}=3$ $\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big)$
$\Rightarrow\text{k}=6$
Hence, the value of k is 6.

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