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Difference between revisions of "1955 AHSME Problems/Problem 24"

(Created page with "== Problem 24== The function <math>4x^2-12x-1</math>: <math> \textbf{(A)}\ \text{always increases as }x\text{ increases}\\ \textbf{(B)}\ \text{always decreases as }x\text{...")
 
 
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<math>\textbf{(D)}\ \text{has a maximum value when}\ x \text{ is negative}</math> is wrong due to the function not having a maximum value (the function opens up, so the vertex is the minimum value).
 
<math>\textbf{(D)}\ \text{has a maximum value when}\ x \text{ is negative}</math> is wrong due to the function not having a maximum value (the function opens up, so the vertex is the minimum value).
  
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Therefore, <math>\textbf{(E)}</math> is the correct answer (this can be verified by plugging in 1.5 (the x-coordinate of the vertex) in.
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Therefore, <math>\textbf{(E)}</math> is the correct answer (this can be verified by plugging in 1.5 (the x-coordinate of the vertex) in).
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==See Also==
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{{AHSME 50p box|year=1955|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 16:25, 1 August 2020

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 (Problems • Answer Key • Resources)
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


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