Extremal value problems: Very Difficult Problems with Solutions

Solution:
[tex]f(x)[/tex] is a parabola, which is opened up (since its leading coefficient is [tex]a^2+1>0)[/tex], so it has only one extremum and it is a (global) minimum. [tex]f'(x)=0 <=> 2(a^2+1)x-2a=0[/tex], or [tex]x=\frac{a}{a^2+1}[/tex]. Luckily for us, [tex]\frac{a}{a^2+1}=\frac{1}{2} \times \frac{2a}{a^2+1} \le \frac{1}{2}[/tex] (since [tex]0 \le \frac{2a}{a^2+1} \le 1[/tex] for any positive [tex]a)[/tex]. So the minimum in the interval is reached for [tex]x=\frac{a}{a^2+1}[/tex]. We substitute into [tex]f(x)[/tex] to reach
[tex]f_{min}(x)=f(\frac{a}{a^2+1})=(a^2+1) \times (\frac{a}{a^2+1})^2-2a \times \frac{a}{a^2+1}+10=\frac{a^2}{a^2+1}-\frac{2a^2}{a^2+1}+10=10-\frac{a^2}{a^2+1}=\frac{9a^2+10}{a^2+1}[/tex]. We want this value to be equal to [tex]\frac{451}{50}[/tex]:
[tex]\frac{9a^2+10}{a^2+1}=\frac{451}{50}[/tex], we cross multiply:
[tex]450a^2+500=451a^2+451[/tex], or [tex]a^2=49[/tex]. Which means that [tex]a=7[/tex], since [tex]a[/tex] is positive.