Solution of Past Paper of CAIE Math 9709/13 Oct/Nov 2019 Question # 6
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CAIE AS & A Level Mathematics P1 (9709/1)
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CAIE AS & A Level Mathematics P1 (9709/1)
This document presents most carefully worked out and thoroughly explained complete Solution of Past Paper of Cambridge AS/A Level Mathematics (9709), Pure Mathematics 1.
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solution of past paper 9709 13 october november 19
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CIE/GCE/AS/Math/P1/19/Nov/13/Q#6
Question
A line has equation 𝑦 = 3𝑘𝑥 − 2𝑘 and a curve has equation 𝑦 = 𝑥 2 − 𝑘𝑥 + 2, where k is
a constant.
i. Find the set of values of 𝑘 for which the line and curve meet at two distinct points.
i. For each of two particular values of 𝑘, the line is a tangent to the curve. Show
that these two tangents meet on the x-axis.
Solution
i.
We are required to find the values of 𝑘 for which the curve and the line meet at two
distinct points.
If two lines (or a line and a curve) intersect each other at a point then that point lies on
both lines i.e., coordinates of that point have same values on both lines (or on the line
and the curve). Therefore, we can equate 𝑦 coordinates of both lines i.e. equate
equations of both the lines (or the line and the curve).
Equation of the line is;
𝑦 = 𝑥 2 − 𝑘𝑥 + 2
Equation of the curve is;
𝑦 = 3𝑘𝑥 − 2𝑘
Equating both equations;
𝑥 2 − 𝑘𝑥 + 2 = 3𝑘𝑥 − 2𝑘
𝑥 2 − 𝑘𝑥 + 2 − 3𝑘𝑥 + 2𝑘 = 0
𝑥 2 − 4𝑘𝑥 + 2 + 2𝑘 = 0
, CIE/GCE/AS/Math/P1/19/Nov/13/Q#6
𝑥 2 − 4𝑘𝑥 + (2 + 2𝑘) = 0
To find the set of values for which the line and the curve meet at two distinct points, the
above quadratic equation must have two solutions.
Let us find for which set of values of 𝑘 two solution of above quadratic equation exists.
Standard form of quadratic equation is;
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
Expression for discriminant of a quadratic equation is;
𝑏2 − 4𝑎𝑐
If 𝑏2 − 4𝑎𝑐 > 0 ; Quadratic equation has two real roots.
If 𝑏2 − 4𝑎𝑐 < 0 ; Quadratic equation has no real roots.
If 𝑏2 − 4𝑎𝑐 = 0 ; Quadratic equation has one real root/two equal roots.
First, we find the discriminant of the quadratic equation.
𝑏2 − 4𝑎𝑐 = 0
𝑎=1
𝑏 = −4𝑘
𝑐 = (2 + 2𝑘)
Hence;
[−4𝑘]2 − 4(1)(2 + 2𝑘) = 0
16𝑘2 − 4(2 + 2𝑘) = 0
16𝑘2 − 8𝑘 − 8 = 0
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