Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2011 | Oct-Nov | (P2-9709/23) | Q#7

Hits: 19

 

  Question

The diagram shows the curve . The curve has a gradient of 3 at the point P.

     i.       Show that the x-coordinate of P satisfies the equation

   ii.       Verify that the equation in part (i) has a root between x = 3.1 and x = 3.3.

  iii.       Use the iterative formula  to determine this root correct to 2 decimal places.  Give the result of each iteration to 4 decimal places.

Solution

     i.
 

We are required to find the equation for x-coordinate of point on the curve where curve has gradient  3.

Equation of the curve is given as;

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 of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation natural exponential function , or ;

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 gradient of the curve at point P is 3.

Hence, x-coordinate of point P on the curve is satisfied by the equation; 

 

   ii.
 

We are required to verify equation   has a root between x=3.1 and x=3.3.

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

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 use as initial value.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 3.21.

Please follow and like us:
error0

Comments