JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f(\mathrm{x})=\left|2 \mathrm{x}^2+5\right| \mathrm{x}|-3|, \mathrm{x} \in \mathrm{R}\). If \(\mathrm{m}\) and \(\mathrm{n}\) denote the number of points where \(f\) is not continuous and not differentiable respectively, then \(m+n\) is equal to :
- A \(5\)
- B \(2\)
- C \(0\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\( f(\mathrm{x})=\left|2 \mathrm{x}^2+5\right| \mathrm{x}|-3| \) \( \text {Graph of } \mathrm{y}=\left|2 \mathrm{x}^2+5 \mathrm{x}-3\right|\) Number of points of discontinuity \(=0=\mathrm{m}\) Number of points of non-differentiability \(=3=n\)
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