The correct option is A 12
Let f(x)=ax4+bx3+cx2+dx+e
Since, limx→0(1+f(x)x2)=3
⇒limx→0(1+ax4+bx3+cx2+dx+ex2)=3
In order to limit exists, d,e must be 0
⇒limx→0(1+ax2+bx+c)=3
⇒1+c=3
⇒c=2
∴f(x)=ax4+bx3+2x2
Now f′(1)=0
⇒f′(1)=4a+3b+4=0 ⋯(1)
f′(2)=0⇒32a+12b+8=0 ⋯(2)
From equation (1) and (2), we have
a=0.5, b=−2
⇒f(x)=0.5x4−2x3+2x2
For minimum/maximum value
f′(x)=0⇒2x3−6x2+4x=0⇒x(x2−3x+2)=0⇒x(x−1)(x−2)=0⇒x=0,1,2
Now, f′′(x)=6x2−12x+4
⇒f′′(0)=f′′(2)=4>0
and f′′(1)=−2<0
Hence, the maximum value of f(x) in the given interval is 12.