Let f be a biquadratic function of x given by f(x)=Ax4+Bx3+Cx2+Dx+E, where A,B,C,D,E∈R and A≠0. If limx→0(f(−x)2x3)1/x=e−3, then
A
A+4B=0
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B
A−3B=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 Df′(1)=−30 limx→0(f(−x)2x3)1/x=e−3 ⇒limx→0(Ax4−Bx3+Cx2−Dx+E2x3)1/x=e−3
For given limit to be finite, C=D=E=0
Also, −B2=1⇒B=−2 ∴limx→0(Ax4+2x32x3)1/x=e−3 ⇒limx→0(1+Ax2)1/x=e−3 ∴limx→0Ax2⋅1x=−3 ⇒A2=−3⇒A=−6