Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2011 | May-Jun | (P2-9709/21) | Q#7

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  Question

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

has exactly two real roots.

   ii.       Show by calculation that this root lies between 1.2 and 1.3.

  iii.       Show that this root also satisfies the equation

  iv.       Use an iteration process based on the equation in part (iii), with a suitable starting value, to find  the root correct to 2 decimal places. Give the result of each step of the process to 4 decimal  places.

Solution

     i.
 

We are required to show that there are exactly two roots 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 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.

Next, we need to sketch graph of .

Where  

Some properties, which help to plot/sketch, of 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 graph of is as shown below.

desmos-graph (10).png

Therefore graph of is as shown below.

desmos-graph (12).png

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 two points;  therefore, the equation has exactly two roots.

 

   ii.
 

We are required to verify by calculation that the only root of equation   lies between  1.2 and 1.3. 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.
 

We are required to show that root of equation  is also a root of the equation;

If we can write the given equation  and then transform it to , then both will have the  same root.

We can show the desired result if we can show that given equation  can be written  as;

Therefor , if the given equation can be rewritten as  , it is evident  that roots of both will be same.

Taking logarithm of both sides;

Since  for any ;

Since given equation   can be rewritten as    , the root of will also be root of  .

 

  iv.
 

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.

As demonstrated in (iii) given equation   can be rewritten as   , therefore, 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.

1

2

3

4

5

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7

8

It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 1.26.

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