Let, f(x)=x3+ax2+bx+c
⇒f′(x)=3x2+2ax+b
⇒f′′(x)=6x+2a
⇒f′′′(x)=6
According to the question,
a=f′(1)=3+2a+b⇒a+b=−3...(i)
b=f′′(2)=12+2a⇒2a−b=−12...(ii)
c=f′′′(3)⇒c=6
Solving equations (i)&(ii), we get, a= -5&b=2
⇒f(x)=x3−5x2+2x+6
∴f(2)=8−20+4+6=−2.
JEE Main 2019 — Mathematics Calculus
Let, f:R→R be a function such that f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3),∀x∈R. Then f(2) equals
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