If f(x)=max(y2−2xy+1) for 0≤y≤2, then the minimum value of f(x)∀x∈R is
A
1
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B
2
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C
0
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D
Can not be determined.
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Solution
The correct option is A1 Let g(y)=y2−2xy+1{0≤y≤2} g′(y)=2y−2x g′(y)=0⇒y=x. g(0)=1,g(2)=5−4x g(x)=1−x2 f(x)=max{1,5−4x,1−x2} f(x)={1,x≥15−4x,x<1 ∴ Minimum value of f(x) = 1.