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Question

The least value of αR for which 4αx2+1x1, for all x>0, is:

A
164
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B
132
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C
127
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D
125
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Solution

The correct option is C 127
αx14x3
αmax(x14x3)

Let f(x)=x14x3

f(x)=14[2x3+3x4]=0x=32
f′′(x)=14[6x412x5]
f′′(32)<0
f(x) is maximum at x=32
f(32)=127

α127

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