JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(K\) be the set of all real values of \(x\) where the function \(f\left( x \right) = \sin \,\left| x \right| - \left| x \right| + 2\,\left( {x - \pi } \right)\,\cos \,\left| x \right|\) is not differentiable. Then the set \(K\) is equal to
- A \(\phi \) (en empty set)
- B \(\left\{ \pi \right\}\)
- C \(\left\{ 0 \right\}\)
- D \(\left\{ {0,\pi } \right\}\)
Answer & Solution
Correct Answer
(A) \(\phi \) (en empty set)
Step-by-step Solution
Detailed explanation
At \(x = \pi + \) or \(x = \pi - \) \(f\left( x \right) = \sin x - x + 2\left( {x - \pi } \right)\cos x\) \(f(x)\) is differentiablr For \(x=0+\) \(f\left( x \right) = \sin x - x + 2\left( {x - \pi } \right)\cos x\)…
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