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Question

Suppose f(x)=x3+ax2+bx+c satisfies f(2)=10 and takes the extreme value 5027 at x=23. Then the value of a+b+c is

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Solution

f(x)=x3+ax2+bx+cf(2)=8+4a2b+c=104a2b+c+2=0 (1)

f(x)=3x2+2ax+bf(23)=03×49+2a×23+b=043+4a3+b=04+4a+3b=0 (2)

f(23)=5027
This gives 827+4a9+2b3+c=5027
8+12a+18b+27c=5012a+18b+27c=424a+6b+9c=14 (3)

Solving (1),(2) and (3), we get
a=1,b=0 and c=2

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