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Question

# Let f(x)=⎧⎨⎩−1;−2≤x<0x2−1; 1≤x≤2 and g(x)=|f(x)|+f(|x|). Then, in the interval (−2,2), g is :

A
not continuous
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B
not differentiable at one point
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C
not differentiable at two points
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D
differentiable at all points
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Solution

## The correct option is B not differentiable at one point|f(x)|=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩1; −2≤x<01−x2; 0≤x<1x2−1; 1≤x≤2 f(|x|)=x2−1 for x∈[−2,2] ∴g(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩1+x2−1; −2≤x<01−x2+x2−1; 0≤x<1x2−1+x2−1; 1≤x≤2 ⇒g(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩ x2 ; −2≤x<0 0 ; 0≤x<12(x2−1) ; 1≤x≤2 Clearly, g is continuous in (−2,2) g′(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩ 2x ; −2≤x<0 0 ; 0≤x<1 4x ; 1≤x≤2 g′(0−)=0 and g′(0+)=0 ⇒g is differentiable at x=0 g′(1−)=0 and g′(1+)=4 ⇒g is not differentiable at x=1

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