Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2006 | January | Q#8

 

Question

The diagram shows the curve with equation  and the line  .

The points  and  have coordinates  and  respectively. The curve touches the x-axis at  the origin  and crosses the x-axis at the point .  The line  cuts the curve at the point   where  and touches the curve at  where .

a.   Find the area of the rectangle .

b.    

                           i.       Find

                          ii.       Hence find the area of the shaded region bounded by the curve and the line .

c.   For the curve above with equation :

                           i.       find

                          ii.       hence find an equation of the tangent at the point on the curve where ;

                        iii.       show that y is decreasing when .

d.   Solve the inequality .

Solution

a.
 

Expression for the area of the rectangle is;

For the given rectangle  it is evident from the diagram that;

Expression to find distance between two given points  and is:

We are given the coordinates of points  so we can find the length of .

Now we need to find length of either  or . The procedure would be same no matter what we  choose to find. We will find here .

It is given that line  cuts the curve at the point  where  which means x-coordinate of point   is -1.

Since point  is the intersection of the given curve and line , it lies on both the curve and the line.  So we can find the y-coordinate of point  by substituting its x-coordinate in the given equation of  the curve.

Hence, coordinates of point  are .

Now we can find length of  .

Expression to find distance between two given points  and is:

We have the coordinates of points  so we can find the length of .

Now we can find the area of rectangle .

b.
          i.
 

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

 

                    ii.
 

It is evident from the diagram that area of the shaded region is given by;

We have area of the rectangle  from (a), so we need to find area under the curve.

To find the area of region under the curve , we need to integrate the curve from point  to   along x-axis.

It is evident from the area under the curve extends from point  to point  .

Therefore;

Rule for integration of  is:

From (b:i) we have

Hence;

Now we can find area of shaded region;

c.
 

We are given that;


i.
 

Rule for differentiation is of  is:

Rule for differentiation is of  is:


ii.
 

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 are required to find the equation of tangent to the curve where . Hence we have neither coordinates of any point on the tangent nor the slope of the tangent. We find them one by  one.

First we find the coordinates of the point  where tangent is drawn to the curve. We are given that  x-coordinates of that point is 1. We can find the y-coordinate of that point from equation of the  curve.

Hence coordinates of point  are .

Now we need slope of the tangent to the curve at point .

The slope of a curve  at a particular point is equal to the slope of the tangent to the curve at the  same point;

Hence we need to find the slope of the curve at point .

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:

We have from c(i);

We need slope of the curve at 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;

Therefore;

Hence;

Finally, we can write the equation of tangent to the curve at point .

Point-Slope form of the equation of the line is;

Hence;


iii.
 

To test whether a function  is increasing or decreasing at a particular point , we  take derivative of a function at that point.

If  , the function  is increasing.

If  , the function  is decreasing.

If  , the test is inconclusive.

 

We are given that;

From c(i) we also have;

 If  is decreasing then;

d.
 

To solve inequality first we need to find the critical points. Hence we solve the equation;

Now we have two options;

Hence the critical points are  and .

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If  (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph.
If
 (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the  graph.

We recognize that given curve  is a parabola opening upwards.

Therefore for the parabola  we must have;

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