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12 Sep 2022, 08:08
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
52% (01:54) correct
48%
(02:04)
wrong
based on 131
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If f(x) = ax^2 + 3x + 1, where a ≠ 0, is a < 0 ?
(1) f(1) = -1.
(2) In the xy‐plane, the graph of f(x) = ax^2 + 3x + 1 does not intersect with the graph of g(x) = 2x + 3.
(1) f(1) = -1.
(2) In the xy‐plane, the graph of f(x) = ax^2 + 3x + 1 does not intersect with the graph of g(x) = 2x + 3.
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13 Sep 2022, 02:43
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According to me, option D is correct.
Statement 1: f(1) = -1
f(1) = -1 on substitution gives you a+4 = -1 => a = -5. Hence a<0. Therefore, this statement alone is sufficient.
Statement 2: does not intersect with the graph of 2x+ 3.
This means that the equation f(x) = g(x) should have no real solution. Equating both sides, we get ax^2 + 3x + 1 = 2x + 3. This gives us the equation ax^2 + x -2 =0. Using the quadratic formula, we get x = (-1 + \(\sqrt{1+8a}\))/(2a) or x = (-1 - \(\sqrt{1+8a}\))/(2a). In order to get no real solution, 1+8a should be less than 0 [to get a negative expression under the square root]. Hence, a < 0.
therefore this statement is sufficient alone as well.
another way to look at this part:
f(x) = ax^2 + 3x +1 is a parabola, with a y intercept at (0,1). For any parabola, a>0 indicates a parabola that opens upwards and a<0 indicates one that opens downwards. If you were to plot g(x) = 2x+3, you would realize that any upward pointing parabola (a>0) which passes through (0,1) will definitely intersect the given line at some point. Hence, only a downward parabola will not intersect the given line g(x). Hence, a<0.
Statement 1: f(1) = -1
f(1) = -1 on substitution gives you a+4 = -1 => a = -5. Hence a<0. Therefore, this statement alone is sufficient.
Statement 2: does not intersect with the graph of 2x+ 3.
This means that the equation f(x) = g(x) should have no real solution. Equating both sides, we get ax^2 + 3x + 1 = 2x + 3. This gives us the equation ax^2 + x -2 =0. Using the quadratic formula, we get x = (-1 + \(\sqrt{1+8a}\))/(2a) or x = (-1 - \(\sqrt{1+8a}\))/(2a). In order to get no real solution, 1+8a should be less than 0 [to get a negative expression under the square root]. Hence, a < 0.
therefore this statement is sufficient alone as well.
another way to look at this part:
f(x) = ax^2 + 3x +1 is a parabola, with a y intercept at (0,1). For any parabola, a>0 indicates a parabola that opens upwards and a<0 indicates one that opens downwards. If you were to plot g(x) = 2x+3, you would realize that any upward pointing parabola (a>0) which passes through (0,1) will definitely intersect the given line at some point. Hence, only a downward parabola will not intersect the given line g(x). Hence, a<0.
13 Sep 2022, 03:44
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\(ax^2 + 3x + 1\) is a formula for a parabola. The \(a\) will dictate how the direction of the parabola. If \(a>0\), then the parabola will open upward, while if \(a<0\) then it will open downward.
(1) f(1) = -1.
\( -1 = a(1)^2 + 3(1) + 1\)
\(-1 = a + 4\)
\(a = -5\)
SUFFICIENT
(2) In the xy‐plane, the graph of f(x) = ax^2 + 3x + 1 does not intersect with the graph of g(x) = 2x + 3.
The straight line graph \(y = 2x + 3\) has a y-intercept of 3, while the parabola has a y-intercept of 1. As the straight line's y-intercept is above the parabola's y-intercept, and the two never intersect then the direction of the parabola has to be downward. Therefore, \(a<0\).
Attached below is the line-graph plotted to visually depict that the only possible direction of the parabola is opening downward.
SUFFICIENT
Answer D
(1) f(1) = -1.
\( -1 = a(1)^2 + 3(1) + 1\)
\(-1 = a + 4\)
\(a = -5\)
SUFFICIENT
(2) In the xy‐plane, the graph of f(x) = ax^2 + 3x + 1 does not intersect with the graph of g(x) = 2x + 3.
The straight line graph \(y = 2x + 3\) has a y-intercept of 3, while the parabola has a y-intercept of 1. As the straight line's y-intercept is above the parabola's y-intercept, and the two never intersect then the direction of the parabola has to be downward. Therefore, \(a<0\).
Attached below is the line-graph plotted to visually depict that the only possible direction of the parabola is opening downward.
SUFFICIENT
Answer D
Attachments
Parabola and Straight Line.jpg [ 57.51 KiB | Viewed 2554 times ]
