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16 Sep 2014, 01:21
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E
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Difficulty:
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Question Stats:
44% (01:48) correct
56%
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wrong
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If \(ac \ne 0\), do graphs \(y=ax^2+b\) and \(y=cx^2+d\) intersect?
(1) \(a = -c\)
(2) \(b \gt d\)
(1) \(a = -c\)
(2) \(b \gt d\)
16 Sep 2014, 01:21
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Official Solution:
If \(ac \ne 0\), do graphs \(y=ax^2+b\) and \(y=cx^2+d\) intersect?
Notice that graphs of \(y=ax^2+b\) and \(y=cx^2+d\) are parabolas.
Algebraic approach:
(1) \(a = -c\).
Given: \(y_1= ax^2 + b\) and \(y_2=-ax^2 + d\). Now, if these two parabolas cross, then for some \(x\), \(ax^2+b=-ax^2 + d\) should be true, which means that equation \(2ax^2+(b-d)=0\) must have a solution, some real root(s), or in other words, the discriminant of this quadratic equation must be \(\ge 0\). So, the question becomes: is \(discriminant=-8a(b-d) \ge 0\)? Since we know nothing about \(a\), \(b\), and \(d\), the answer is no. Not sufficient..
(2) \(b \gt d\).
The same steps apply: if \(y_1= ax^2 + b\) and \(y_2= cx^2 + d\) cross, then for some \(x\), \(ax^2 +b=cx^2+d\) should be true, which means that equation \((a-c)x^2+(b-d)=0\) must have a solution, or in other words, the discriminant of this quadratic equation must be \(\ge 0\). So, the question becomes: is \(discriminant=-4(a-c)(b-d) \ge 0\)? Now, can we determine whether this is true? We know that \(b-d \gt 0\) but what about \(a-c\)? Hence no. Not sufficient.
(1)+(2) We have that \(a=-c\) and \(b > d\), so \(y_1= ax^2 + b\) and \(y_2=-ax^2 + d\). The same steps as above: \(2ax^2+(b-d)=0\) and the same question remains: is \(discriminant=-8a(b-d) \ge 0\) true? \(b-d > 0\), but what about \(a\)? Not sufficient.
Else consider two cases.
First case: \(y=-x^2+1\) and \(y=x^2+0\) (upward and downward parabolas). Notice that these parabolas satisfy both statements and they cross each other;
Second case: \(y=x^2+1\) and \(y=-x^2+0\) (also upward and downward parabolas). Notice that these parabolas satisfy both statements and they do not cross each other.
Answer: E
If \(ac \ne 0\), do graphs \(y=ax^2+b\) and \(y=cx^2+d\) intersect?
Notice that graphs of \(y=ax^2+b\) and \(y=cx^2+d\) are parabolas.
Algebraic approach:
(1) \(a = -c\).
Given: \(y_1= ax^2 + b\) and \(y_2=-ax^2 + d\). Now, if these two parabolas cross, then for some \(x\), \(ax^2+b=-ax^2 + d\) should be true, which means that equation \(2ax^2+(b-d)=0\) must have a solution, some real root(s), or in other words, the discriminant of this quadratic equation must be \(\ge 0\). So, the question becomes: is \(discriminant=-8a(b-d) \ge 0\)? Since we know nothing about \(a\), \(b\), and \(d\), the answer is no. Not sufficient..
(2) \(b \gt d\).
The same steps apply: if \(y_1= ax^2 + b\) and \(y_2= cx^2 + d\) cross, then for some \(x\), \(ax^2 +b=cx^2+d\) should be true, which means that equation \((a-c)x^2+(b-d)=0\) must have a solution, or in other words, the discriminant of this quadratic equation must be \(\ge 0\). So, the question becomes: is \(discriminant=-4(a-c)(b-d) \ge 0\)? Now, can we determine whether this is true? We know that \(b-d \gt 0\) but what about \(a-c\)? Hence no. Not sufficient.
(1)+(2) We have that \(a=-c\) and \(b > d\), so \(y_1= ax^2 + b\) and \(y_2=-ax^2 + d\). The same steps as above: \(2ax^2+(b-d)=0\) and the same question remains: is \(discriminant=-8a(b-d) \ge 0\) true? \(b-d > 0\), but what about \(a\)? Not sufficient.
Else consider two cases.
First case: \(y=-x^2+1\) and \(y=x^2+0\) (upward and downward parabolas). Notice that these parabolas satisfy both statements and they cross each other;
Second case: \(y=x^2+1\) and \(y=-x^2+0\) (also upward and downward parabolas). Notice that these parabolas satisfy both statements and they do not cross each other.
Answer: E
General Discussion
