The correct option is B 14
Given : limx→−1f(x)(x+1)3=1
By observation, we get
f(x)=(x+1)3[ax+b]
Now,
limx→−1f(x)(x+1)3=1⇒b−a=1⇒f(x)=(x+1)3[ax+1+a]⇒f(x)=a(x+1)4+(x+1)3
Also,
f′′′(0)=−12⇒24a+6=−12⇒a=−34∴f(x)=(x+1)3−34(x+1)4
Now, for maxima and minima
f′(x)=0⇒3(x+1)2−3(x+1)3=0⇒(x+1)2[1−x−1]=0⇒x=−1,0
f(−1)=0, f(0)=14
Hence, the maximum value of the function is 14