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Question

Let limx0sinx+aex+bex+cln(1+x)x2=1
The roots of the quadratic equation ax2+bx+c=0 are

A
both positive
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B
both negative
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C
one positive and one negative
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D
complex conjugate
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Solution

The correct option is B both positive
L=limx0sinx+aex+bex+cl(1+x)x3

=limx0(xx331)+a(1+x1!+x22!+x33!)x3

+b(1x1!+x22!x33!)+c(xx22+x33)x3
=limx0(a+b)+(1+ab+c)x+(a2+b2c2)x2x3+(13!+a3!b3!+c3)x3x3
or a+b=0,1+ab+c=0,a2+b2c2=0
and L=13!+a3!a3!+c3
Solving the first three equations, we get c=0,a=1/2,b=1/2
Then, L=1/3.
The given equation ax2+bx+c=0 reduces to x2x=0 or x=0,1.

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