# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Oct-Nov | (P1-9709/12) | Q#10

Question

A function f is defined by  for .

i.       Find the range of f.

ii.       Sketch the graph of

iii.       Solve the equation , giving answers in terms of .

The function  is defined by  for , where k is a constant.

iv.       State the largest value of k for which g has an inverse.

v.       For this value of k, find an expression for .

Solution

i.

We are given the function with its domain.

The function is;

The domain of the given function is;

We are required to find the range of function.

The set of values a function  can take against its domain is called range of the function.

Finding range of a function :

·       Substitute various values of  from given domain into the function to see what is happening to y.

·       Make sure you look for minimum and maximum values of y by substituting extreme values of   from given domain.

We know following properties of .

 Properties of Domain Range

We can deduce that for the given function;

We just utilize the extreme values of  to find the extreme values of range.

 For For

Therefore range of  can be expressed as;

ii.

We are required to sketch    for .

We can find the points of the graph as follows.

Now we can sketch required graph from these points as shown below.

iii.

We are required to solve .

We are given that    for .

Therefore;

We are required to solve this for .

To solve this equation for ., we can substitute . Hence,

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 solve  equation for interval ,

Using calculator we can find the value of .

We have following properties of .

 Properties of Domain Range Odd/Even Periodicity Translation/ Symmetry

We utilize the symmetry property of   to find other solutions (roots) of

 Translation/ Symmetry

Therefore;

Hence;

Hence, we have following two solutions of ;

To find all the solutions (roots) of  within  interval, we utilize the periodic property of   for both these values of .

 Periodicity

For the given case,

 For For

Now;

 For

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

Since ;

Therefore;

iv.

The function  is defined by  for , where k is a constant.

We are required to state the largest value of k for which g has an inverse.

A one-one function has only one value of  against one value of . A function is one-one function if  it passes horizontal line test i.e. horizontal line passes only through one point of function. However,  if a function is not one-one, we can make it so by restricting its domain.

The given function is a sinusoidal as observed in (ii) and sinusoidal does not pass horizontal line test, so it is not a one-one function. However, we can make it so by restricting its domain.

We can see from the graph obtained in (ii) that starting from  the graph remains one-one  function till  where .

After , the graph of function starts repeating itself and hence does not remain on-one function.

Hence;

v.

As found in (iv), the function  is defined by  for .

We have found in (iv) that function is one-one for this domain and hence its inverse exists for this domain only.

We have;

We write it as;

To find the inverse of a given function  we need to write it in terms of  rather than in terms of .

Interchanging ‘x’ with ‘y’;