1955 AHSME Problems/Problem 24

Revision as of 16:25, 1 August 2020 by Angrybird029 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 24

The function $4x^2-12x-1$:

$\textbf{(A)}\ \text{always increases as }x\text{ increases}\\ \textbf{(B)}\ \text{always decreases as }x\text{ decreases to 1}\\ \textbf{(C)}\ \text{cannot equal 0}\\ \textbf{(D)}\ \text{has a maximum value when }x\text{ is negative}\\ \textbf{(E)}\ \text{has a minimum value of-10}$

Solution

We can use the process of elimination to narrow down the field substantially:

$\textbf{(A)}\ \text{always increases as } x\text{ increases}$ is wrong due to the quadratic nature of the function.

$\textbf{(B)}\ \text{always decreases as } x\text{ decreases to 1}$ is wrong due to the vertex being on the line $x = 1.5$. $f(x)$ would decrease all the way to $x = 1.5$, but increase from there.

$\textbf{(C)}\ \text{cannot equal 0}$ is wrong due to the discriminant being greater than zero.

$\textbf{(D)}\ \text{has a maximum value when}\ x \text{ is negative}$ is wrong due to the function not having a maximum value (the function opens up, so the vertex is the minimum value).

Therefore, $\textbf{(E)}$ is the correct answer (this can be verified by plugging in 1.5 (the x-coordinate of the vertex) in).

See Also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png