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The curve C has equation $y = x(5 - x)$ and the line L has equation $2y = 5x + 4$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 1
Question 6
The curve C has equation $y = x(5 - x)$ and the line L has equation $2y = 5x + 4$. a) Use algebra to show that C and L do not intersect. b) In the space on page 11... show full transcript
Worked Solution & Example Answer:The curve C has equation $y = x(5 - x)$ and the line L has equation $2y = 5x + 4$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 1
Step 1
a) Use algebra to show that C and L do not intersect.
Answer
To determine whether the curve C and the line L intersect, we first express both equations in a standard format.
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Equation of Curve C: [ y = x(5 - x) \implies y = -x^2 + 5x \quad (1) ]
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Equation of Line L: [ 2y = 5x + 4 \implies y = \frac{5}{2}x + 2 \quad (2) ]
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Setting Equations Equal: To find the intersection points, we set (1) equal to (2):
[ -x^2 + 5x = \frac{5}{2}x + 2 \quad (3) ] -
Rearranging Equation: Rearranging gives us: [ -x^2 + 5x - \frac{5}{2}x - 2 = 0 \implies -x^2 + \left(5 - \frac{5}{2}\right)x - 2 = 0 ] [ -x^2 + \frac{5}{2}x - 2 = 0 \quad (4) ]
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Using the Quadratic Formula: To solve for x in equation (4), we apply the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) where ( a = -1, b = \frac{5}{2}, c = -2 ). Thus: [ b^2 - 4ac = \left(\frac{5}{2}\right)^2 - 4(-1)(-2) = \frac{25}{4} - 8 = \frac{25}{4} - \frac{32}{4} = -\frac{7}{4} ]
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Determining Intersection: Since the discriminant is negative (( -\frac{7}{4} )), the quadratic equation (4) has no real roots. Therefore, curves C and L do not intersect.
Step 2
b) In the space on page 11, sketch C and L on the same diagram, showing the coordinates of the points at which C and L meet the axes.
Answer
To sketch the curve C and line L, we first find their x- and y-intercepts.
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Finding Intercepts of Curve C:
- x-intercept: Set ( y = 0 ):
[ x(5 - x) = 0 \implies x = 0 \text{ or } x = 5. ] Thus, the x-intercepts are (0, 0) and (5, 0). - y-intercept: Set ( x = 0 ):
[ y = 0(5 - 0) = 0. ]
Thus, the y-intercept is (0, 0).
- x-intercept: Set ( y = 0 ):
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Finding Intercepts of Line L:
- x-intercept: Set ( y = 0 ):
[ 0 = \frac{5}{2}x + 2 \implies x = -\frac{4}{5}. ]
Thus, the x-intercept is (-0.8, 0). - y-intercept: Set ( x = 0 ):
[ y = 2. ]
Thus, the y-intercept is (0, 2).
- x-intercept: Set ( y = 0 ):
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Sketching the Diagram:
In the diagram:- Curve C is a downward-opening parabola passing through (0, 0) and (5, 0).
- Line L is a straight line passing through (-0.8, 0) and (0, 2) with a positive gradient. The coordinates of the intercepts should be clearly marked.