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 $x, f'(x) = \begin{vmatrix} 2ax & 2ax - 1 & 2ax + b + 1 \\ b & b + 1 & -1 \\ 2(ax + b) & 2ax + 2b + 1 & 2ax + b \end{vmatrix}$\nRange of $f(x)$ is</p>
7/16, \infty
-\infty, 15/16]
3/4, \infty
none of these
Step-by-Step Solution
Key Concept: General
The function $f(x) = \frac{1}{4}x^2 - \frac{5}{4}x + 2$ has a minimum value at $x = 5/2$. The minimum value is $f(5/2) = \frac{1}{4}(\frac{25}{4}) - \frac{5}{4}(\frac{5}{2}) + 2 = \frac{7}{16}$. Since the parabola opens upwards ($a > 0$), the range is $[7/16, \infty)$.
Correct Answer: A