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JUNE 2024 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsJUNE 20242 Marks
Question
Find the values of $k$ so that the function $f$ is continuous at $x=\frac{\pi}{2}$
$f(x)=\left\{\begin{array}{cc}
\frac{k \cos x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\
3, & x=\frac{\pi}{2}
\end{array}\right.$
✓

Answer

$f$ is continuous at $x=\frac{\pi}{2}$
$\therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)$
Now, $\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{k \cos x}{\pi-2 x}\right)=3$
$\therefore \lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{2\left(\frac{\pi}{2}-x\right)}=3$
$\begin{array}{l}\therefore \lim _{\frac{\pi}{2}-x \rightarrow 0} \frac{k}{2} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)}=3\binom{\because x \rightarrow \frac{\pi}{2}}{\Rightarrow \frac{\pi}{2}-x \rightarrow 0} \\ \therefore \frac{k(1)}{2}=3 \\ \therefore k=6\end{array}$

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