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Question

Let f(x)={1|x|:|x|1ax2+b:|x|<1 be continuous and differentiable every where. Then a and b are:

A
12,32
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B
12,32
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C
12,32
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D
13,12
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Solution

The correct option is A 12,32
Let f(x)=⎪ ⎪⎪ ⎪1x:x1ax2+b:1<x<11x:x1
f(x) is continuous a+b=1.
f(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪1x2:x12ax:1<x<11x2:x1
f(x) is differentiable at x=1,11=2aa=12.
a=12,b=32

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