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Question

If the derivative of the function f(x)=bx2+ax+4;x1ax2+b;x<1, is everywhere continuous, then

A
a = 2, b = 3
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B
a = 3, b = 2
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C
a = - 2, b = - 3
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D
a = - 3, b = - 2
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Solution

The correct option is A a = 2, b = 3
We have, f(x)={ax2+b,x<1bx2+ax+4,x1f(x)={2ax,<12bx+a,x1Since, f(x) is differentiable at x=1, therefore it is continuous at x=1 and hence,limx1f(x)=limx1+f(x)a+b=ba+4a=2and also, limx1f(x)=limx1+f(x)2a=2b+a3a=2bb=3 (a=2)Hence, a=2,b=3

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