# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2003 | Oct-Nov | (P2-9709/02) | Q#5

Question

i.       By sketching a suitable pair of graphs, for x < 0, show that exactly one root of the equation    is negative.

ii.       Verify by calculation that this root lies between -1.0 and -0.5.

iii.       Use the iterative formula

to determine the root correct to 2 decimal places, showing the result of each iteration.

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 .

It can be seen that it is a quadratic equation;

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;

For the given case, vertex is .

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

Since vertex of parabola is on  there would be no x and y intercepts as is evident from the  sketch of parabola as shown below.

For the case x < 0 , we can restrict the sketch as shown below.

Next we sketch .

It is evident that it is an exponential function.

For an exponential function

Where

Some properties which help to plot/sketch this graph are as follows.

·       The graphs of all exponential functions contain the point .

·       The domain is all real numbers .

·       The range is only the positive .

·       The graph is increasing.

·       The graph is asymptotic to the x-axis as x approaches negative infinity

·       The graph increases without bound as x approaches positive infinity

·       The graph is continuous and smooth.

Therefore the graph of  will look like as shown below.

To be more accurate we may like to calculate coordinates of a couple of points of the graph and then sketch it.

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

 -3 -2 -1 0 1 2 3 4 5 0.125 0.25 0.5 1 2 4 8 16 32

We can plot these points to sketch the following graph of

However, as per given condition, we restrict the graph to desired interval .

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

It can be seen that the two graphs of   and intersect each other at only a single point  and that too on the negative x-axis side, therefore, the equation  has a single roots and at a  single value of   which is negative.

ii.

We are required to verify by calculation that the only root of equation   lies between -1.0 and  -0.5. 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 10 11

It is evident that .

Hence, is a root of .

The root given correct to 2 significant figures is -0.77.