AP EAMCET · Maths · Continuity and Differentiability
The number of points in the interval \((0,2)\) at which \(f(x)=|x-0.5|+|x-1|+\tan x\) is not differentiable is
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
Given, \(f(x)=|x-0.5|+|x-1|+\tan x\) Clearly \(f(x)\) is not differentiable at \(x=0.5,1\) and \(\frac{\pi}{2}\) for \(x \in(0,2)\). \(\therefore\) Number of Non differentiable points of \(f(x)\) are 3 . \(\therefore\) Hence, option (c) is correct.
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