Q53 / 150 · MHT CET 2024 (03 May Shift 1) · Mathematics

Let $\mathrm{f}(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{1}{2}\right], \quad \mathrm{f}(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f\left(\frac{\pi}{4}\right)$ is

  • Correct answer
    A
    $-\frac{1}{2}$
  • B
    $\frac{1}{2}$
  • C
    1
  • D
    -1

Solution

$\begin{aligned} & f(x)=\frac{1-\tan x}{4 x-\pi} \\ & \text { Let } x=\frac{\pi}{4}+\mathrm{t} \\ & \therefore \quad \text { when } x \rightarrow \frac{\pi}{4}, \mathrm{t} \rightarrow 0 \\ & \therefore f(x)=\frac{1-\tan \left(\frac{\pi}{4}+t\right)}{4\left(\frac{\pi}{4}+t\right)-\pi} \end{aligned}$ $\begin{aligned} \therefore \quad f(t) & =\frac{1-\tan \left(\frac{\pi}{4}+t\right)}{4 t} \\ & =\frac{1-\left(\frac{1+\tan t}{1-\tan t}\right)}{4 t} \\ & =\frac{-2 \tan t}{(1-\tan t) 4 t}=\frac{-1}{2} \times \frac{\tan t}{t} \times \frac{1}{1-\tan t} \\ \therefore \quad f\left(\frac{\pi}{4}\right) & =\lim _{t \rightarrow 0} f(t)=\frac{-1}{2}\end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions