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Question

Let f(x)=⎪ ⎪⎪ ⎪[x];2x122x21;12<x2 and g(x)=f(|x|)+|f(x)|, where [.] represents the greatest integer function. The number of points where g(x) is non-differentiable is

A
1
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B
2
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C
3
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D
4
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Solution

The correct option is C 3
f(x)=⎪ ⎪⎪ ⎪[x];2x122x21;12<x2

f(x)=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪2;2x11;1x122x21;12<x2

|f(x)|=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪2;2x11;1x12|2x21|;12<x2

|f(x)|=⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪2;2x11;1x1212x2;12<x122x21;12<x2

f(|x|)=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪2;2|x|11;1|x|122|x|21;12<x2
Hence,
f(|x|)=2x21; 2x2
Now, g(x)=|f(x)|+f(|x|)
g(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪2x2+1;2x<12x2;1x120;12<x<124x22;12x2

g(1)=limx1(2x2+1)=3 and g(1+)=limx12x2=2

g(12)=limx12(2x2)=12 and g(12+)=limx120=0

g(12)=limx120=0 and g(12+)=limx12(4x22)=0

Hence, g(x) is discontinuous at x=1, 12
g(x) is non-differentiable at x=1, 12
Now, g(12)=0, g(12+)=82
Hence, g(x) is non-differentiable at x=12

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