Question

Suppose $f\left(x\right)$ is a function satisfying the following conditions:
For all $x,f^{\prime }\left(x\right)=\left|\begin{array}{ccc}2ax& 2ax-1& 2ax+b+1\\ b& b+1& -1\\ 2(ax+b)& 2ax+2b+1& 2ax+b\end{array}\right|$
where $a,b$ are some constant. Determine the constant $a,b$, and the function $f\left(x\right)$.

Answer 1

Applying $R_3\to R_3-R_1-2R_2$
we get $f^{\prime }\left(x\right)=\left|\begin{array}{ccc}2ax& 2ax-a& 2ax+b+1\\ b& b+1& -1\\ 0& 0& 1\end{array}\right|$
$=\left|\begin{array}{cc}2ax& 2ax-1\\ b& b+1\end{array}\right|=\left|\begin{array}{cc}2ax& -1\\ b& 1\end{array}\right|$ [Using $C_2\to C_2-C_1$]
$\Rightarrow f^{\prime }\left(x\right)=2ax+b$
Integrating, we get $f\left(x\right)=a{x}^2+bx+c$ where $c$ is an arbitrary constant.
Since, $f$ has a maximum at $x=5/2$
$f^{\prime }\left(5/2\right)=0\Rightarrow 5a+b=0$
Also $f\left(0\right)=2\Rightarrow c=2$
Ans $f\left(1\right)=1\Rightarrow a+b+c=1$
$\therefore a+b=1$
Solving $\left(1\right)$ and $\left(2\right)$ for $a,b$ we get, $a=1/4,b=-5/4$
Thus $f\left(x\right)=\frac{1}{4}{x}^2-\frac{5}{4}x+2$