Q71 / 150 · MHT CET 2024 (16 May Shift 1) · Mathematics
Let $\mathrm{f}(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $\mathrm{f}\left(\frac{\pi}{4}\right)$ is
- Correct answerA$-\frac{1}{2}$
- B$\frac{1}{2}$
- C1
- D-1
Solution
Since $\mathrm{f}(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$.
$\therefore \quad$ it is continuous at $x=\frac{\pi}{4}$.
$\therefore \quad \mathrm{f}\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} \mathrm{f}(x)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}$
Applying L'Hospital rule on R.H.S., we get $\begin{aligned} & \mathrm{f}\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^2 x}{4} \\ & \Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{-2}{4}=\frac{-1}{2} \end{aligned}$
Applying L'Hospital rule on R.H.S., we get $\begin{aligned} & \mathrm{f}\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^2 x}{4} \\ & \Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{-2}{4}=\frac{-1}{2} \end{aligned}$
Hence, the correct answer is option: 'A'.