Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2010 | May-Jun | (P2-9709/23) | Q#6

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  Question

     i.       By sketching a suitable pair of graphs, show that the equation

has only one root.

   ii.       Verify by calculation that this root lies between 1.3 and 1.4.

  iii.       Show that, if a sequence of values given by the iterative formula

converges, then it converges to the root of the equation in part (i).

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

Solution

     i.
 

We are required to show that there is only one root of the following equation by sketching.

Root of an equation  is the x-coordinate of a point of intersection of the graphs of   and .

Therefore, first we sketch .

We know that graph of is as shown below. Or we can plot it as follows.

We can calculate a couple of values to be accurate in sketching the graph.

0

0.1

0.5

1.0

1.5

2.0

2.5

3.0

3.5

-2.30

-0.69

0.0

0.41

0.69

0.92

1.10

1.25

We can plot these points to sketch the following graph of  

desmos-graph (8).png

Next, we need to sketch graph of .

It can be seen that it is a quadratic equation which can be rearranged as;

To sketch a quadratic equation, a parabola, we need the coordinates of its vertex and x and y- intercepts, if any.

First we find the coordinates of vertex of this parabola.

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph.
If
(‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the  graph.

We recognize that given curve , is a parabola opening downwards.

Vertex form of a quadratic equation is;

We first write given quadratic equation in vertex form.

We have the equation;

We use method of “completing square” to obtain the desired form.

Next we complete the square for the terms which involve .

We have the algebraic formula;

For the given case we can compare the given terms with the formula as below;

Therefore we can deduce that;

Hence we can write;

To complete the square we can add and subtract the deduced value of ;

For the given case, vertex is .

Next, we need x and y-intercepts of the parabola.

First we find the x-intercept of the parabola.

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).

Therefore, we substitute in equation of the parabola.

Hence, coordinates of x-intercepts of the parabola are  and .

We can sketch the parabola as shown below.

Sketching both graphs on the same axes, we get following.

desmos-graph (11).png

It can be seen that the two graphs of and intersect each other at only a single  point, therefore, the equation has a single roots and at a single value of  .

 

   ii.
 

We are required to verify by calculation that the only root of equation lies between 1.3  and 1.4. We need to use sign-change rule.

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.
 

If we can write the given equation  and transform it to , then we can find the root of  the equation by iteration method using sequence defined as.

We are required to find a root of the equation

We are also given the iterative formula as;

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.

0

1

2

3

4

5

6

7

8

9

It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 1.31.

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