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 :

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;

Adding  in the entire inequality;

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 ;