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Let f:[1,1]R be defined as f(x)=ax2+bx+c for all x[1,1], where a,b,cR such that f(1)=2,f(1)=1 and for x(1,1) the maximum value of f′′(x) is 12. If f(x)α,x[1,1], then the least value of α is equal to

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Solution

f(x)=ax2+bx+cf(x)=2ax+bf′′(x)=2a
We know
f(1)=2ab+c=2(1)f(1)=1b2a=1(2)f′′(x)12a14(3)
From equations (1) and (2), we get
b=1+2a, c=3+af(x)=ax2+(1+2a)x+(3+a)f(x)=a(x+1)2+(x+3)
To get the maximum value of f(x), a should be maximum, so
a=14f(x)=(x+1)24+(x+3)f(x)[2,5], x[1,1]
As f(x)α,x[1,1]
α=5

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