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Question

If f(x)=max(y22xy+1) for 0y2, 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 A 1
Let g(y)=y22xy+1 {0y2}
g(y)=2y2x
g(y)=0 y=x.
g(0)=1 ,g(2)=54x
g(x)=1x2
f(x)=max{1,54x,1x2}

f(x)={1,x154x,x<1

Minimum value of f(x) = 1.


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