Question

Suppose, f(x) is a function satisfying the following conditions
(a) f(0) = 2, f(1) = 1
(b) f has a minimum value at $x=\frac{5}{2}$, and
(c) 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 constants. Determine the constants a, b and the function f(x).

• A. $f\left(x\right)=\frac{1}{4}{x}^2-\frac{5}{4}x+2$
• B. $f\left(x\right)=-x^3+\frac{75}{16}{x}^2-\frac{75}{16}x+2$
• C. $f\left(x\right)=x$
• D. $f\left(x\right)=-\frac{1}{4}{x}^2-\frac{5}{4}x+2$

Answer 1

The correct option is A $f\left(x\right)=\frac{1}{4}{x}^2-\frac{5}{4}x+2$

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