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Question

Let f′′(x) be continous at x=0.

If limx02f(x)3af(2x)+bf(8x)sin2x exists and f(0)0,f(0)0, then

A
a=79
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B
b=13
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C
a=79
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D
b=13
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Solution

The correct options are
B b=13
C a=79
We have,
limx02f(x)3af(2x)+bf(8x)sin2x

=limx02f(x)3af(2x)+bf(8x)x2

For the limit to exist, we have

2f(0)3af(0)+bf(0)=0 i.e., 3ab=2 [f(0)0, given] ...(1)

=limx02f(x)6af(2x)+8bf(8x)2x

For the limit to exist, we have

2f(0)6af(0)=8bf(0)=0

i.e., 3a4b=1 [f(0)0, given] ...(2)

Solving equations (1) and (2), we have

a=79 and b=13

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