# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2013 | May-Jun | (P2-9709/23) | Q#7

Question

i.       Express in the form , where  and , giving the  exact value of R and the value of a correct to 2 decimal places.

ii.      Hence solve the equation

Giving all solutions in the interval  correct to 1 decimal place.

iii.       Determine the least value of  as  varies.

Solution

i.

We are given the expression;

We are required to write it in the form;

If  and are positive, then;

can be written in the form

can be written in the form

where,

and , , with

Considering the given equation, we have following case at hand;

can be written in the for

Comparing it with given equation Therefore

Therefore;

Finally, we can find , utilizing the equation;

Using calculator we can find that;

Therefore;

ii.

We are required to solve the equation;

As demonstrated in (i), we can write;

Therefore, we need to solve;

Let , then:

Using calculator we can find that;

We utilize the symmetry property of   to find another solution (root) of :

 Properties of Domain Range Odd/Even Periodicity Translation/ Symmetry

Hence;

Therefore;

We have found two solutions of equation .

Since given interval is  , for interval can be found as follows;

Multiplying the entire inequality with 2;

Since ;

Hence the given interval for  is .

To find all other solutions of the equation in the given interval  we  utilize the periodicity property of :

For the given case,

 For For

Now;

 For

Only following solutions (roots) of the equation  are within  interval;

Since ;

iii.

We are required the find the least value of

As demonstrated in (i), we can write;

Hence;

To find the least value of we need to find the greatest value of because it is in the denominator.

We know that , hence the greatest value of will occur when and therefore;

Hence the least value of will be ;