find the range....

\hspace{-16}$Here $\bf{f:\mathbb{R}\rightarrow \left(0,\frac{\pi}{2}\right]}$ and $\bf{f(x)=\cot^{-1}\left(x^2-2ax+a+1\right)}$\\\\\\ Now If the function is Surjective i.e onto-function.\\\\\\ Then Range of $\bf{f(x)}$ is Co-domain of $\bf{f(x)}$\\\\\\ So Let $\bf{y=f(x)=\cot^{-1}\left(x^2-2ax+a+1\right)}$\\\\\\ $\bf{y=\cot^{-1}\left(x^2-2ax+a+1\right)}$\\\\\\ Here $\bf{0< y \leq \frac{\pi}{2}\Leftrightarrow 0<\cot^{-1}\left(x^2-2ax+a+1\right)\leq \frac{\pi}{2}}$\\\\\\ So $\bf{\infty >x^2-2ax+a+1\geq 0}$\\\\\\ Inequality is Reverse bcz $\bf{\cot^{-1}}$ is a Strictly Decreasing function in $\bf{\left(0,\frac{\pi}{2}\right]}$\\\\\\ So $\bf{x^2-2ax+a+1\geq 0}$\\\\\\ means its Discriminant is $\bf{\leq 0}$\\\\\\ So $\bf{4a^2-4(a+1)\leq 0\Leftrightarrow a^2-a-1\leq 0}$\\\\\\ So $\boxed{\boxed{\bf{\frac{1}{2}\left(1-\sqrt{5}\right)\leq a \leq \frac{1}{2}\left(1+\sqrt{5}\right)}}}$

oops..... incorrect........

a small error...

if ax² + bx + c ≤ 0 then D = 0 and not D ≤ 0

so only two values {1 ± √5} / 2