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

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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;

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 ;

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