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TS EAMCET · Maths · Limits

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

  1. A does not exist at all points in \(\left[0, \frac{\pi}{2}\right]\)
  2. B \(=1\) when \(a=\frac{\pi}{3}\)
  3. C \(=-1\), when \(a=\frac{\pi}{3}\)
  4. D \(=1\), when \(0 \leq a \lt \frac{\pi}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(=1\), when \(0 \leq a \lt \frac{\pi}{3}\)

Step-by-step Solution

Detailed explanation

\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…