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Question

If f(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪x2+2,x<02ex,0x<2,2e2(x3),x2 then

A
|f(x)| is discontinuous at 0 points
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B
|f(x)| is discontinuous at 2 points
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C
|f(x)| is not differentiable at 2 points
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D
|f(x)| is not differentiable at 3 points
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Solution

The correct option is C |f(x)| is not differentiable at 2 points
Graph of f(x)



Graph of |f(x)|


The graph is discontinuous at 0 points
Before x=0 the function is differentiable at all points because our function is a quadratic function.
At x=2 and x=0, the function changes definition.
So we need to check differentiability.

At x=3, a linear function changes direction (sharp turn). So the function is not differentiable.

At x=0
L.H.D. =0
R.H.D =2exx=0=2
So, function is not differentiable.

At x=2
L.H.D. =2exx=2=2e2
R.H.D. =2e2

Therefore, |f(x)| is not differentiable at x=0 and x=3
There are only 2 points where function is not differentiable.

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