Past Papers’ Solutions  Assessment & Qualification Alliance (AQA)  AS & A level  Mathematics 6360  Pure Core 1 (6360MPC1)  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 xaxis at the origin and crosses the xaxis 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 xcoordinate 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 ycoordinate of point by substituting its xcoordinate 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 xaxis.
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 (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 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 xcoordinates of that point is 1. We can find the ycoordinate 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 .
PointSlope 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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