Calculus
General
Grade 12
Question:
<p>Suppose $f(x)$ is a function satisfying the following conditions:\n(i) $f(0) = 2, f(1) = 1$,\n(ii) $f$ has a minimum value at $x = 5/2$\n(iii) For all $x, f'(x) = \begin{vmatrix} 2ax & 2ax - 1 & 2ax + b + 1 \\ b & b + 1 & -1 \\ 2(ax + b) & 2ax + 2b + 1 & 2ax + b \end{vmatrix}$\n$f(x) = 0$ has</p>
<p>Both roots positive </p>
<p>Both roots negative </p>
<p>Roots of opposite sign </p>
<p>imaginary roots </p>
Step-by-Step Solution
Key Concept: General
From the function $f(x) = \frac{1}{4}x^2 - \frac{5}{4}x + 2$, the discriminant is $D = b^2 - 4ac = (-\frac{5}{4})^2 - 4(\frac{1}{4})(2) = \frac{25}{16} - 2 = -\frac{7}{16}$. Since $D < 0$, the roots are imaginary.
Correct Answer: D