Question

Suppose f(x) is a function satisfying the following conditions.
(a) f(0)=2, f(1)=1
(b) f(x) has a maximum at x=5/2, and
(c) for all $xϵR$
$f^{\prime }\left(x\right)=\left|\begin{array}{ccc}2ax& 2ax-1& 2ax+b+1\\ b& b+1& -1\\ 2(ax+b)& 2ax+2b+1& 2ax+b\end{array}\right|$
where a, b are constants, then

Answer 1

$f^{\prime }\left(x\right)=\left|\begin{array}{ccc}2ax& 2ax-1& 2ax+b+1\\ b& b+1& -1\\ 2(ax+b)& 2ax+2b+1& 2ax+b\end{array}\right|$
Applying $R_3\to R_3-2R_1-R_2$, we get
$f^{\prime }\left(x\right)=\left|\begin{array}{cc}2ax& 2ax-1\\ b& b+1\end{array}\right|=\left|\begin{array}{cc}2ax& 2ax\\ b& b\end{array}\right|+\left|\begin{array}{cc}2ax& -1\\ b& 1\end{array}\right|$
$\Rightarrow f^{\prime }\left(x\right)=2ax+b$
Integrating w.r.t. x
$f\left(x\right)=a{x}^2+bx+c$
As $f\left(x\right)$ attain maximum at $x=\frac{5}{2}$, then
$f^{\prime }\left(\frac{5}{2}\right)=0\Rightarrow 5a+b=0$ ...(1)
And $f\left(0\right)=2\Rightarrow c=2$ ...(2)
$f\left(1\right)=1\Rightarrow a+b+c=1$ ...(3)
From (2) and (3)
$\Rightarrow a+b=-1$
Now from (1)
A) $a=\frac{1}{4}$
B) $b=-\frac{5}{4}$
C) $f\left(2\right)=\frac{1}{4}.2^2-\frac{5}{4}.2+2=1-\frac{5}{2}+2=\frac{1}{2}$
D) $f^{\prime }\left(0\right)=b=-\frac{5}{4}$