# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2016 | June | Q#7

Question

The diagram shows the sketch of a curve and the tangent to the curve at P.

The curve has equation  and the point P(-2,24) lies on the curve. The tangent at P  crosses the x-axis at Q.

a.

i.               Find the equation of the tangent to the curve at the point P, giving your answer in the  form .

ii.               Hence find the x-coordinate of Q.

b.

i.               Find

ii.               The point R(1,0) lies on the curve. Calculate the area of the shaded region bounded by  the curve and the lines PQ and QR.

Solution

a.

i.

We are required to find the equation of the tangent to the curve at the point P(-2,24).

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 P(-2,24). Therefore, we need slope of the tangent 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 have gradient for the curve at point P, we can find the slope of tangent to the curve  at point P.

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 are given the equation of the curve as;

Therefore;

Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:

Now we need gradient of the curve at point P(-2,24). Therefore, we substitute ;

Hence,

Now we can find slope of tangent to the curve at point P(-2,24).

Now we can write equation of the tangent.

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

ii.

We are required to find coordinates of x-intercept (point Q) of the tangent to the curve at point P.

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).

From (a:i) we know the equation of the tangent to the curve at point P as;

Therefore;

Hence coordinates of x-intercept (point Q) of the tangent to the curve at point P are;

b.

i.

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

ii.

Consider the diagram below.

It is evident from the diagram that;

First we find area under curve.

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

We are given equation of the curve as;

It is evident that shaded region extends from point P (x=-2) to point R (x=1) along x-axis.

Therefore;

We have found in (b:i) that;

Therefore;

Next we find area of triangle PQS.

Expression for the area of the triangle is;

For triangle PQS;

It is evident that SQ and PS are distances between point S&Q and P&S, respectively.

To find these distance we need coordinates of all three points. We are already given coordinates of  point P(-2,24) and we have found coordinates of  in (a:ii).

It is evident from the diagram above that x-coordinate of S is -2 (as that of point P) and y-coordinate  0 (as it is x-intercept of PS). Therefore, S(-2,0).

Expression to find distance between two given points  and is:

Therefore, for PS;

Hence;

Next we find QS.

Hence;

Now we can find area of triangle PSQ.

Finally;