Q101 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

If $0 \leq x \leq \pi / 2$, then $\lim _{x \rightarrow a} \frac{|2 \cos x-1|}{2 \cos x-1}$

  • A
    does not exist at all points in $\left[0, \frac{\pi}{2}\right]$
  • B
    $=1$ when $a=\frac{\pi}{3}$
  • C
    $=-1$, when $a=\frac{\pi}{3}$
  • Correct answer
    D
    $=1$, when $0 \leq a \lt \frac{\pi}{3}$

Solution

$\begin{aligned} & f(x)=\frac{|2 \cos x-1|}{2 \cos x-1}, 0 \leq x \leq \frac{\pi}{2} \\ &=\left\{\begin{array}{ll}\frac{2 \cos x-1}{2 \cos x-1}, & 0 \leq x \lt \frac{\pi}{3} \\ \frac{1-2 \cos x}{2 \cos x-1}, & \frac{\pi}{3} \lt x \leq \frac{\pi}{2}\end{array}= \begin{cases}1, & 0 \leq x \lt \frac{\pi}{3} \\ -1, & \frac{\pi}{3} \lt x \leq \frac{\pi}{2}\end{cases} \right.\end{aligned}$ $\therefore \lim _{x \rightarrow a} \frac{|2 \cos x-1|}{2 \cos x-1}=1$ when $0 \leq a \lt \frac{\pi}{3}$ and $\lim _{x \rightarrow \frac{\pi}{3}} \frac{|2 \cos x-1|}{2 \cos x-1}$ does not exist

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

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