Question

Find the value of $a$:

• A. $a=\frac{3}{4}$
• B. $a=\frac{1}{3}$
• C. $a=\frac{1}{4}$
• D. $a=\frac{1}{2}$

Answer 1

Answer: (C)

$a=\frac{1}{4}$

We have $f\left(x\right)=a{x}^2-bx+c$
and given $f\left(0\right)=2\Rightarrow c=2$ (i)
$f\left(1\right)=1\Rightarrow a-b+c=1$ (ii)
and $f^{\prime }\left(5/2\right)=0\Rightarrow 2a\left(5/2\right)-b=0\Rightarrow 5a-b=0$
So by (i), (ii) and (ii)
$\Delta =\left|\begin{array}{ccc}0& 0& 1\\ 1& -1& 1\\ 5& -1& 0\end{array}\right|=4$
${\Delta }_a=\left|\begin{array}{ccc}2& 0& 1\\ 1& -1& 1\\ 0& -1& 0\end{array}\right|=1$
${\Delta }_b=\left|\begin{array}{ccc}0& 2& 1\\ 1& 1& 1\\ 5& 0& 0\end{array}\right|=5$
${\Delta }_c=\left|\begin{array}{ccc}0& 0& 2\\ 1& -1& 1\\ 5& -1& 0\end{array}\right|=8$
Hence by cramers rule $a=\frac{{\Delta }_a}{\Delta }=\frac{1}{4},b=\frac{{\Delta }_b}{\Delta }=\frac{5}{4},c=\frac{{\Delta }_c}{\Delta }=2$
And thus $f\left(x\right)=\frac{1}{4}(x^2-5x+8)$