Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2016 | May-Jun | (P2-9709/23) | Q#5

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Question

The equation of a curve is  . At the point on the curve with x-coordinate p, the gradient of  the curve is .

     i.       Show that .

 

   ii.       Show by calculation that 3.3 < p < 3.5.

 

  iii.       Use an iterative formula based on the equation in part (i) to find the value of p correct to 3

  decimal places. Give the result of each iteration to 5 decimal places.

Solution

     i.
 

We are given that;

First we are required to find the x-coordinate of the point where gradient of the curve is .

Therefore, first find the expression for gradient of this curve.

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

Therefore;

We utilize the Product Rule to differentiate.

If  and  are functions of , and if , then;

If , then;

Let  and , then;

Rule for differentiation of natural exponential function , or ;

Rule for differentiation of  is:

Since at this point the gradient of the curve is 40.

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;

Hence;

Since ;

Taking logarithm of both sides;

 for any

   ii.
 

We are required to show by calculation that the x-coordinate of the point i.e., p lies between 3.3 and  3.5.

We need to use sign-change rule.

To use the sign-change method we need to write the given equation as .

From (i) we have;

Therefore;

Therefore;

If the function  is continuous in an interval  of its domain, and if   and have opposite signs, then  has at least one root between  and .

We can find the signs of at and as follows;

Since and have opposite signs for function , the function has root  between and .

 

  iii.
 

Iteration method can be used to find the root of the given equation using sequence defined by;

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 .

We have already found in (ii) through sign-change rule that root of the given equation lies between
and .

Therefore, for iteration method we use;

We use as initial value.

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It is evident that .

Hence, is a root of .

The root given correct to 3 decimal places is 3.412.

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