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Question

If limx0asinxbx+cx2+x32x2log(1+x)2x3+x4=I and I is finite, then

A
a=6
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B
b=3
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C
c=0
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D
I=340
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Solution

The correct option is A a=6
limx0{asinxbx+cx2+x32x2log(1+x)2x3+x4}

this is 00 form so applying L-hospital rule, we get

limx0⎪ ⎪ ⎪⎪ ⎪ ⎪acosxb+2cx+3x22x21+x+4xlog(1+x)6x2+4x3⎪ ⎪ ⎪⎪ ⎪ ⎪

this is also 00 form so applying L-hospital rule, we get

limx0⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪asinx+2c+6x22x(1+x)2x2(1)(1+x)2+4x(1+x)+4log(1+x)12x+12x2⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪

this is again 00 form so applying L-hospital rule, we get

limx0⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪acosx+6ddx(22x(1+x)2x2(1+x)2+4x(1+x)+4log(1+x)12x+12x2)⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪

putting x=0, we get

=a612

now the value of a can be founded by equation the above equation equal to 0 i.e,

=a612=0

=a6=0

a=6.

Hence, option (a) is correct.

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