Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2006 | January | Q#9

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Question

Figure 2 shows part of the curve C with equation

The curve cuts the x-axis at the points P, (1, 0) and Q, as shown in Figure 2.

a.   Write down the x-coordinate of P, and the x-coordinate 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 x-intercepts fo the given curve C.

The point  at which curve (or line) intercepts x-axis, 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 x-intercepts of the given curve C are located at x-coorindates 1, 2 and -2. It is  evident from the figure that P has x-coordinate -2 and Q has x-coordinate 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 (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  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.

Point-Slope 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 x-coordinate 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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