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Question

Let f be a biquadratic function of x given by f(x)=Ax4+Bx3+Cx2+Dx+E, where A,B,C,D,ER and A0. If limx0(f(x)2x3)1/x=e3, then

A
A+4B=0
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B
A3B=0
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C
f(1)=8
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D
f(1)=30
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Solution

The correct option is D f(1)=30
limx0(f(x)2x3)1/x=e3
limx0(Ax4Bx3+Cx2Dx+E2x3)1/x=e3
For given limit to be finite,
C=D=E=0
Also, B2=1B=2
limx0(Ax4+2x32x3)1/x=e3
limx0(1+Ax2)1/x=e3
limx0Ax21x=3
A2=3A=6

So, f(x)=6x42x3


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