The number which exceeds its square by

(a) $\frac{1}{2}$

Let the required number be $x$. Then,

$f(x)=x-x^{2}$

$\Rightarrow f^{\prime}(x)=1-2 x$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow 1-2 x=0$

$\Rightarrow 2 x=1$

$\Rightarrow x=\frac{1}{2}$

Now,

$f^{\prime \prime}(x)=-2<0$

So, $x=\frac{1}{2}$ is a local maxima.

Hence, the required number is $\frac{1}{2}$.