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Question

# If f(x)=min{x2,āx+1,sgn|āx|} then f(x) is (where sgn(x) denotes signum function of x)

A
continuous at x=0
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B
discontinuous at one point
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C
non-differentiable at 3 points
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D
differentiable at x=12
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Solution

## The correct option is D differentiable at x=12By the figure given above we can say that f(x) graph will be The curves x2 and −x+1 intersect at ⇒x2=−x+1⇒ x2+x−1=0 ⇒x=1±√52 f(x) is continuous for all real numbers but not differentiable at x=−1,1+√52 So, f(x) is continous at x=0 and differentiable at x=12.

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