Hint to The Answer Is Not

Observe that the denominator is always positive. Thus, the LHS is minimized when the denominator is maximized. Since this is a quadratic, the maximum of $- a^2 + a$ occurs when $a = - \frac{ 1 } { 2 \times (-1) } = \frac{1}{2}$ , which is within range.

Thus, the minimum is $\sqrt{ \frac{ 1}{ \frac{1}{2} ( 1 - \frac{1}{2} ) } } = 2$ .

I did a calculus solution, but we can also use the GM-HM inequality since our domain of $a(1-a)$ is positive.

Let $x_{1} = \frac{1}{a}$ and $x_{2} = \frac{1}{1-a}$ . Since $\sqrt{x_{1}x_{2}} \geq \frac{2}{\frac{1}{x_{1}}+\frac{1}{x_{2}}}$ , substituting the above values will give $\frac{2}{a+1-a} = 2$ .

We can also diffrentiate the funtion' k' w.r.t. ' a' and equate it to 0 . we get 'a'= 0.5 thus 'k' =2