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Question

If the function f(x)=x3+bx2+ax+5 on [1,3] satisfies the conditions of Rolle's Theorem with c=2+13, then find a + b.

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Solution

f(x)=x3+bx2+ax+5
f(1)=f(3)
1+b+a+5=27+9b+3a+5
a+4b=13
f(x)=3x22bx+a
f(c)=0
Now c=2+13
3(2+13)2+b(2+13)+a=0
13+43+2b(6+3)3+a=0
solving the above equation
x=11
y=6
x+y=5

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