[Roorkee 1990]
A) - 1
B) 0
C) $ \frac{1}{2} $
D) 1
Correct Answer: C
Let the number is x, then $ y=x-x^{2} $
therefore $ \frac{dy}{dx}=1-2x $ and $ \frac{d^{2}y}{dx^{2}}=-2(<0) $
therefore $ 1-2x=0\Rightarrow x=\frac{1}{2} $ .
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