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11 Jun 2022, 11:45
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The graph of x = -y^2 + 2 and the graph of the line k intersect at (0, p) and (1, q). Which one of the following is the smallest possible slope of line k ?
(A) \(-\sqrt{2} -1\)
(B) \(-\sqrt{2} +1\)
(C) \(\sqrt{2} - 1\)
(D) \(\sqrt{2} + 1\)
(E) \(\sqrt{2} + 2\)
Are You Up For the Challenge: 700 Level Questions
(A) \(-\sqrt{2} -1\)
(B) \(-\sqrt{2} +1\)
(C) \(\sqrt{2} - 1\)
(D) \(\sqrt{2} + 1\)
(E) \(\sqrt{2} + 2\)
Are You Up For the Challenge: 700 Level Questions
11 Jun 2022, 16:51
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We need to find lowest value of possible slopes of line k. Line k must pass through points (0, p) and (1, q).
=> Slope of line k = \(\frac{(q-p) }{ (1-0)}\) = q-p
Let's find value of p & q. Line (or curve) x = -y^2 + 2 passes through points (0, p) and (1, q).
Therefore, substituting values of x & y with value of point (0,p) we get,
0 = -p^2 + 2
=> p^2 = 2
=> p = +√2 or p =-√2
Substituting values of x & y with value of point (1,q) we get,
1 = -q^2 + 2
=> q^2 = 1
=> q = +1 or q =-1
Therefore, Slope of line k = (q-p) / (1-0) = q-p can be any of the 4 values:
(-√2 - 1) ; (-√2 + 1) ; (√2 - 1) ; (√2 + 1)
The lowest of these is (-√2 - 1)
Answer: A
=> Slope of line k = \(\frac{(q-p) }{ (1-0)}\) = q-p
Let's find value of p & q. Line (or curve) x = -y^2 + 2 passes through points (0, p) and (1, q).
Therefore, substituting values of x & y with value of point (0,p) we get,
0 = -p^2 + 2
=> p^2 = 2
=> p = +√2 or p =-√2
Substituting values of x & y with value of point (1,q) we get,
1 = -q^2 + 2
=> q^2 = 1
=> q = +1 or q =-1
Therefore, Slope of line k = (q-p) / (1-0) = q-p can be any of the 4 values:
(-√2 - 1) ; (-√2 + 1) ; (√2 - 1) ; (√2 + 1)
The lowest of these is (-√2 - 1)
Answer: A
11 Jun 2022, 20:07
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Given: The graph of x = -y^2 + 2 and the graph of the line k intersect at (0, p) and (1, q).
Asked: Which one of the following is the smallest possible slope of line k ?
Let the equation of the line be : y = mx + c ; where m is the slope of the line and c is y-intercept.
At point (0,p):
0 = - p^2 + 2; p^2 = 2; \(p = \sqrt{2}\) or \(p = -\sqrt{2}\)
p = c
At point (1, q)
1 = - q^2 + 2
q^2 = 1; q = 1 or q = -1
q = m + c = m + p
m = q - p
[m]m = {\sqrt{2}-1, \sqrt{2}+1,-\sqrt{2}-1, - \sqrt{2}+1}
m = -\sqrt{2}-1; is the smallest slope.
IMO A
Asked: Which one of the following is the smallest possible slope of line k ?
Let the equation of the line be : y = mx + c ; where m is the slope of the line and c is y-intercept.
At point (0,p):
0 = - p^2 + 2; p^2 = 2; \(p = \sqrt{2}\) or \(p = -\sqrt{2}\)
p = c
At point (1, q)
1 = - q^2 + 2
q^2 = 1; q = 1 or q = -1
q = m + c = m + p
m = q - p
[m]m = {\sqrt{2}-1, \sqrt{2}+1,-\sqrt{2}-1, - \sqrt{2}+1}
m = -\sqrt{2}-1; is the smallest slope.
IMO A
