# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2011 | Oct-Nov | (P1-9709/12) | Q#5

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Question

i.    Sketch, on the same diagram, the graphs of  and  for .

ii.       Verify that  is a root of the equation , and state the other root of this equation for which .

iii.       Hence state the set of values of , for , for which

Solution

i.

We are required to sketch  and  for .

First we sketch  for .

We can sketch the graph of  for  as follows.

We can find the points of the graph as follows.

Now we sketch  for .

We can sketch the graph of  for  as follows.

We can find the points of the graph as follows.

Red curve is for  while orange curve is for .

ii.

We have sketched both sides of the equation  in (i) and we can see that both curves intersect at points. Therefore, there are two roots of the equation.

If we substitute  in the given equation;

Using calculator we can find that;

Hence, both sides of the equation  are equal for , therefore,  is the root of the equation.

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

 Symmetry Property or

Therefore;

If we substitute  in the above equation;

Using calculator we can see that;

Hence other root for   is .

We can also check this root  for .

Using calculator we can see that;

Hence, roots of equation  for  are  and .

iii.

We do not need to do tedious calculations for this part, we are required to “state”.

By closely observing the sketch graphed in (i) we can see that two points of intersection of the two curves are roots of the equation as found and discussed in (ii). Values of  at these intersection points are are  and .

By mere observation we can tell that curve  remains below the curve of  when  and .

Therefore range of values of  for the given condition is;