f(x)=x3+ax2+bx+cf(−2)=−8+4a−2b+c=−10⇒4a−2b+c+2=0 ⋯(1)
f′(x)=3x2+2ax+bf′(23)=0∴3×49+2a×23+b=0⇒43+4a3+b=0⇒4+4a+3b=0 ⋯(2)
f(23)=5027
This gives 827+4a9+2b3+c=5027
⇒8+12a+18b+27c=50⇒12a+18b+27c=42⇒4a+6b+9c=14 ⋯(3)
Solving (1),(2) and (3), we get
a=−1,b=0 and c=2