Let f:[−1,1]→R be defined as f(x)=ax2+bx+c for all x∈[−1,1], where a,b,c∈R 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)=2⇒a−b+c=2⋯(1)f′(−1)=1⇒b−2a=1⋯(2)f′′(x)≤12⇒a≤14⋯(3)
From equations (1) and (2), we get b=1+2a,c=3+a⇒f(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