Mathematics - Continuity - Differentiability Question with Solution | TestHub
MathematicsContinuity - DifferentiabilityContinuity- MiscellaneousHard2 minQB
MathematicsHardsingle choice
On , let for , and for . Match: (A) number of discontinuities; (B) number of interior points where does not exist; (C) number of nondifferentiability points; (D) if the minimum is , then . Values: (P) 2, (Q) 3, (R) 4, (S) prime, (T) neither prime nor composite.
Options:
Answer:
A
Solution:
Piecewise evaluation gives four discontinuities/nondifferentiability points: . The two-sided limit fails at the three interior points , so this count is 3, a prime. The minimum is -1 , and , neither prime nor composite.
Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Continuity- Miscellaneous
⏱ 2mℹ️ Source: QB
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