The point P(1, a) lies on the curve with equation $y = (x + 1)^2(2 - x)$ - Edexcel - A-Level Maths Pure - Question 10 - 2009 - Paper 1
Question 10
The point P(1, a) lies on the curve with equation $y = (x + 1)^2(2 - x)$.
(a) Find the value of a.
(b) On the axes below sketch the curves with the following equat... show full transcript
Worked Solution & Example Answer:The point P(1, a) lies on the curve with equation $y = (x + 1)^2(2 - x)$ - Edexcel - A-Level Maths Pure - Question 10 - 2009 - Paper 1
Step 1
Find the value of a.
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Answer
To find the value of a, substitute x = 1 into the given equation:
y=(1+1)2(2−1)
Calculating this gives:
y = 2^2(1) = 4$$
Thus, a = 4.
Step 2
i) Sketch the curve for $y = (x + 1)^2(2 - x)$.
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Answer
Determine the intercepts:
When x=0: y=(0+1)2(2−0)=12imes2=2
(y-intercept is (0, 2))
When y=0: 0=(x+1)2(2−x)
This occurs when x=−1 and x=2.
(x-intercepts are (-1, 0) and (2, 0))
Shape of the curve:
The curve has a minimum at x=−1 and reaches upward. It extends to the right and approaches the x-axis as x approaches 2.
Step 3
ii) Sketch the curve for $y = \frac{2}{x}$.
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Answer
Determine the intercepts:
When x=0, y is undefined, thus no y-intercept.
When y=0, x=2 gives the point (2, 0).
Shape of the curve:
The curve is a hyperbola with branches in the 1st and 3rd quadrants.
As x approaches zero from the right, y approaches infinity.
As x increases positively, y approaches zero.
Step 4
State the number of real solutions to the equation $(x + 1)^2(2 - x) = \frac{2}{x}$.
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Answer
Referencing the sketch from part (b), the two curves intersect at two distinct points in the first quadrant and none in the third quadrant.
