The maximum possible value of y = min (1/2 – 3x2/4, 5x2/4) for the range 0 < x < 1 is
Explanation:
So maximum possible value will be at the point of intersection of the two graphs.
∴ 12-3x24=5x24⇒x2=14
Hence, required maximum value = 5x24=54×14=516.
Hence, option (d).
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