Photo AI
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 do... 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
Write down the x-coordinate of P and the x-coordinate of Q.
Answer
The x-coordinate of P is 1. To find the x-coordinate of Q, we need to solve the equation where the curve cuts the x-axis:
y = (x - 1)(x^2 - 4) = 0.
This gives us:
- (x - 1) = 0 ⟹ x = 1 (for point P)
- (x^2 - 4) = 0 ⟹ x = 2 or x = -2 (for point Q)
Hence, the coordinates for P are (1, 0) and Q are (-2, 0) and (2, 0).
Step 2
Show that \( \frac{dy}{dx} = 3x^2 - 2x - 4 \).
Answer
To find ( \frac{dy}{dx} ), we start with the expression for y:
y = (x - 1)(x^2 - 4).
Applying the product rule: [ \frac{dy}{dx} = (x - 1) \frac{d}{dx}(x^2 - 4) + (x^2 - 4) \frac{d}{dx}(x - 1) ] Computing the derivatives, [ \frac{dy}{dx} = (x - 1)(2x) + (x^2 - 4)(1) = (2x^2 - 2x) + (x^2 - 4) ] Combining the terms results in: [ \frac{dy}{dx} = 3x^2 - 2x - 4. ]
Step 3
Show that \( y = x + 7 \) is an equation of the tangent to C at the point (-1, 6).
Answer
To determine if the line ( y = x + 7 ) is a tangent at the point (-1, 6), we first need to find the slope of the tangent:
-
Evaluate the derivative ( \frac{dy}{dx} ) at x = -1: [ \frac{dy}{dx} = 3(-1)^2 - 2(-1) - 4 = 3 + 2 - 4 = 1. ] Hence, the slope of the tangent at (-1, 6) is 1.
-
The equation of the tangent line with slope 1 at the point (-1, 6): [ y - 6 = 1(x + 1) \implies y = x + 7. ] Thus, ( y = x + 7 ) is indeed the equation of the tangent.
Step 4
The tangent to C at the point R is parallel to the tangent at point (-1, 6).
Answer
Since the slopes of the tangents are equal, we set: [ 3x^2 - 2x - 4 = 1. ]
This simplifies to: [ 3x^2 - 2x - 5 = 0. ]
Using the quadratic formula: [ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 60}}{6} = \frac{2 \pm \sqrt{64}}{6} = \frac{2 \pm 8}{6}. ]
Thus, we find: [ x = \frac{10}{6} = \frac{5}{3} \quad \text{or} \quad x = \frac{-6}{6} = -1. ]
We check the x-coordinate of R which is not -1. Therefore, the x-coordinate is ( \frac{5}{3} ). Now we substitute back to find y: [ \text{Substituting } x = \frac{5}{3} ext{ into } y = (x - 1)(x^2 - 4): ] [ 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} - \frac{36}{9} \right) = \left( \frac{2}{3} \right) \left( \frac{-11}{9} \right) = \frac{-22}{27}. ]
Finally, the coordinates of R are ( \left( \frac{5}{3}, \frac{-22}{27} \right) ).
Step 5