Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2006  January  Q#9
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Question
Figure 2 shows part of the curve C with equation
The curve cuts the xaxis at the points P, (1, 0) and Q, as shown in Figure 2.
a. Write down the xcoordinate of P, and the xcoordinate of Q.
b. Show that .
c. Show that y=x+7 is an equation of the tangent to C at the point (–1, 6).
The tangent to C at the point R is parallel to the tangent at the point (–1, 6).
d. Find the exact coordinates of R.
Solution
a.
It is evident from the diagram that point P, (1, 0) and Q are xintercepts fo the given curve C.
The point at which curve (or line) intercepts xaxis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore, for the given curve C;
Now we have two options.









It is evident that xintercepts of the given curve C are located at xcoorindates 1, 2 and 2. It is evident from the figure that P has xcoordinate 2 and Q has xcoordinate 2.
b.
We are required to show that;
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
We are given that;
Therefore;
Rule for differentiation is of is:
Rule for differentiation is of is:
Rule for differentiation is of is:
c.
We are required to find the equation of tangent to the curve C at point (1,6).
To find the equation of the line either we need coordinates of the two points on the line (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope form of Equation of Line).
We already have coordinates of a point on the tangent as P(1,6).
We are given equation of the curve as;
We need to find slope of tangent at (1,6) in order to write its equation.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, if we can find slope of the curve C at point (1,6) then we can find slope of the tangent to the curve at this point.
We need to find the gradient of the curve C at point (1,6).
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
We have found in (b) expression for gradient of the given curve as;
For gradient of the curve at point (1,6), substitute in derivative of the equation of the curve.
Therefore;
With coordinates of a point on the tangent (1,6) and its slope in hand, we can write equation of the tangent.
PointSlope form of the equation of the line is;
d.
The tangent to C at the point R is parallel to the tangent at the point (–1, 6).
If two lines are parallel to each other, then their slopes and are equal;
Therefore, both tangents to curve C at points R and (–1, 6) have same slope.
We have found slope of tangent to curve C at (–1, 6) as .
Hence;
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, gradient of the curve C at point R must be equal to slope of tangent to the curve at this point.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x coordinates of that point in the expression for gradient of the curve;
We have found in (b) expression for gradient of the given curve as;
For gradient of the curve at point R, substitute in derivative of the equation of the curve.
Therefore;
Now we have two options.









Two values of x indicate two points where the curve C has gradient 1.
We already know, as demonstrated in (c) that one of them is (1,6), therefore, second must be R. So we have xcoordinate of point R as .
Corresponding values of y coordinate can be found by substituting value of x in equation of the curve C because point R lies on curve C.
We are given equation of the curve as;
Therefore;
Hence, coordinates of point R on the curve C are .
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