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Figure 2 shows part of the curve C with equation y = (x - 1)(x^2 - 4) - Edexcel - A-Level Maths Pure - Question 10 - 2006 - Paper 1
Question 10
Figure 2 shows part of the curve C with equation y = (x - 1)(x^2 - 4). The curve cuts the x-axis at the points P, (1, 0) and Q, as shown in Figure 2. a) Write dow... show full transcript
Worked Solution & Example Answer:Figure 2 shows part of the curve C with equation y = (x - 1)(x^2 - 4) - Edexcel - A-Level Maths Pure - Question 10 - 2006 - Paper 1
Step 1
a) Write down the x-coordinate of P and the x-coordinate of Q.
Answer
The x-coordinate of P is found by setting the equation to zero:
[(x - 1)(x^2 - 4) = 0]
This gives us:
- (x - 1 = 0) → (x = 1) (Point P)
- For (x^2 - 4 = 0), we have (x^2 = 4) → (x = -2) or (x = 2) (Point Q)
Thus, the x-coordinates are: P = -2, Q = 2.
Step 2
Step 3
c) Show that y = x + 7 is an equation of the tangent to C at the point (-1, 6).
Answer
To show this, we first find the slope at the point (-1, 6):
[ \frac{dy}{dx} \text{ at } x = -1 ]
Substituting (x = -1):
[ \frac{dy}{dx} = 3(-1)^2 - 2(-1) - 4 = 3 + 2 - 4 = 1. ]
Thus, the slope is 1.
Using the point-slope form of a line:
[ y - y_1 = m(x - x_1) ]
[ y - 6 = 1(x + 1) ]
This simplifies to:
[ y = x + 7. ]
Step 4
d) The tangent to C at the point R is parallel to the tangent at point (-1, 6).
Answer
Since the tangent at (-1, 6) has a slope of 1, we need to find where the slope of the curve is also 1:
Set ( \frac{dy}{dx} = 1 ):
[ 3x^2 - 2x - 4 = 1 ]
This simplifies to:
[ 3x^2 - 2x - 5 = 0. ]
Using the quadratic formula:
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-5)}}{2(3)} = \frac{2 \pm \sqrt{4 + 60}}{6} = \frac{2 \pm \sqrt{64}}{6} = \frac{2 \pm 8}{6}. ]
This gives:
[ x = \frac{10}{6} = \frac{5}{3} \text{ or } x = -1. ]
Step 5
e) Find the exact coordinates of R.
Answer
Now, substituting ( x = \frac{5}{3} ) to find ( y ):
[ y = \left( \frac{5}{3} - 1 \right) \left( \left( \frac{5}{3} \right)^2 - 4 \right) = \left( \frac{2}{3} \right) \left( \frac{25}{9} - 4 \right) ]
Calculating further:
[ y = \left( \frac{2}{3} \right) \left( \frac{25}{9} - \frac{36}{9} \right) = \left( \frac{2}{3} \right) \left( -\frac{11}{9} \right) = -\frac{22}{27}. ]
Thus, the exact coordinates of R are ( \left( \frac{5}{3}, -\frac{22}{27} \right) ).