Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2018 | May-Jun | (P2-9709/21) | Q#4

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Question

The diagram shows the curve .

The curve crosses the x-axis at the point P and has a maximum point M.

     i.       Find the gradient of the curve at the point P.

   ii.       Show that the x-coordinate of the point M satisfies the equation

  iii.       Use an iterative formula based on the equation in part (ii) to find the x-coordinate of M correct  to 4 significant figures. Show the result of each iteration to 6 significant figures.

Solution

     i.
 

We are required to the gradient of the curve at P which is x-intercept of the curve.

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 equation of the curve;

Therefore first we find from given equation of the curve.

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 ;

First we differentiate

Rule for differentiation natural logarithmic function , for  is;

Next we differentiate .

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Hence;

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

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 know that point P is the x-intercept of the curve.

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).

Hence, to find the x-coordinate of point, we substitute in given equation of the curve.

 and are inverse functions. The composite function is an identity function, with  domain the positive real numbers. Therefore; 

Taking anti-logarithm of both sides;

Therefore, coordinates of point P are (1, 0).

Hence;

 

   ii.
 

We are required to find the x-coordinates of point M which is given as maximum point of the curve  with equation;

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, 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.

From (i) we have;

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 stationary point, the gradient of the curve at this point must be equal to ZERO.

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

 

  iii.
 

We are required to apply an iterative formula based on the equation in part (ii) to find the value of m  correct to 4 significant figures.

We have found in (i) that;

We can write an iterative formula based on this equation.

If the sequence given by the inductive definition , with some initial value , converges  to a limit , then  is the root of the equation .

Therefore, if , then  is a root of .

It is evident from the diagram that root of the curve lies where .

We use , as initial value since will yield , whereas  will yield zero in denominator.

1

2

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9

It is evident that .

Hence, is a root of .

The root given correct to 4 significant figures is 3.181.

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