Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2007 | Oct-Nov | (P2-9709/02) | Q#8

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  Question

The diagram shows the curve and its maximum point M.

     i.       Find the x-coordinate of M.

   ii.       Show that the tangent to the curve at the point where x = 1 passes through the origin.

  iii.       Use the trapezium rule with two intervals to estimate the value of

Giving your answer correct to 2 decimal places.

Solution

     i.
 

We are required to find the coordinates of point M which is maximum point of the curve;

A stationary point on the curve is the point where gradient of the curve is equal to zero;

Since point M is maximum point, therefore, it is stationary point of the curve and, hence, gradient of  the curve at point M must ZERO.

We can find expression for gradient of the curve at point M and equate it with ZERO to find the x- coordinate of point M.

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

Therefore;

If  and  are functions of , and if , then;

If , then;

Let  and ;

Rule for differentiation natural exponential function , or ;

Rule for differentiation of  is:

Now we need expression for gradient of the curve at point M.

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;

Since point M is a maximum point, the gradient of the curve at this point must be equal to ZERO.

Hence, x-coordinate of point M on the curve is .

 

   ii.
 

If coordinates of a point satisfies the equation of a curve (or line) then that particular point  lies on that curve (or line).

Therefore, to show whether tangent to the given curve at x=1 passes through the origin we first  need to find equation of the tangent to the curve at point where x=1.

Hence, we are required to find the equation of tangent to the curve at the point where .

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 need coordinates of the point on the curve (and tangent). 

We are given equation of the curve;

We substitute desired value of x-coordinate in equation of curve to find the y-coordinates of  this point on the curve (and tangent).

Therefore, coordinates of the point on the curve (and tangent) are .

Next we need to find slope of tangent at this point 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 then we can find slope of the 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 (i) that;

Therefore;

Hence;

With coordinates of a point on the tangent  and its slope in hand, we can write equation  of the tangent.

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

Now we can substitute coordinates of origin (0,0) in this equation of tangent to see whether it  passes through origin or not.

Hence, equation of tangent is satisfied by coordinates of the origin, therefore, origin lies on the  tangent.

 

  iii.
 

We are required to apply Trapezium Rule to evaluate;

The trapezium rule with  intervals states that;

We are given that there are two intervals, .

We are also given that and .

Hence;

1

2

3

Therefore;

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