# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | May-Jun | (P1-9709/11) | Q#9

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Question

Functions f and g are defined for  by;

i.Find the points of intersection of the graphs of y = f(x) and y = g(x).

ii.Find the set of values of x for which f(x) > g(x).

iii.Find an expression for fg(x) and deduce the range of fg.

The function h is defined by  for .

iv.Find the smallest value of k for which h has an inverse.

Solution

i.

We are required to find the points of intersection of the graphs of y = f(x) and y = g(x)

We are given that;

If two lines (or a line and a curve) intersect each other at a point then that point lies  on both lines i.e. coordinates of that point have same values on both lines (or on the  line and the curve). Therefore, we can equate coordinates of both lines i.e. equate  equations of both the lines (or the line and the curve).

Equation of the one curve is;

Equation of the other curve is;

Equating both equations;

Multiplying entire equation with ‘2’;

Now we have two options.

Two values of x indicate that there are two intersection points.

Corresponding values of y coordinate can be found by substituting values of x in any  of the two equations of the curves.

We choose the equation;

Therefore, coordinates of the points of intersection are;

ii.

We are required to find the set of values of for which .

We can solve the inequality  algebraically;

We are given that;

Therefore;

Multiplying entire inequality with ‘2’;

We solve the following equation to find critical values of ;

Now we have two options.

Hence the critical points on the curve for the given condition are -3 & 4.

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola  opens upwards and its vertex is the minimum point on the graph.

If (‘a’ is negative) then parabola opens downwards and its vertex is the  maximum point on the graph.

Hence is an upwards opening parabola.

Therefore,  when;

iii.

Next we are required to find an expression for fg(x) and deduce the range of fg.

We are given that;

We are required to find expression for ;

We are also required to find range of .

It is evident that  is a quadratic equation.

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola  opens upwards and its vertex is the minimum point on the graph.
If
(‘a’ is negative) then parabola opens downwards and its vertex is the  maximum point on the graph.

We recognize that given curve , is a parabola opening downwards.

Vertex form of a quadratic equation is;

The given function can be rearranged to write in vertex form

so that coordinates of vertex can be found.

We utilize method of “completing square” to obtain the desired form.

We have the function;

We have the expression;

Next we complete the square for the terms which involve .

We have the algebraic formula;

For the given case we can compare the given terms with the formula as below;

Therefore we can deduce that;

Hence we can write;

To complete the square we can add and subtract the deduced value of ;

Comparing with , results in;

Hence;

Coordinates of the vertex are .Since this is a parabola opening upwards the  vertex is the minimum point on the graph.
Here y-coordinate of vertex represents least value of
and x-coordinate of vertex  represents corresponding value of .

For the given case, vertex is . Therefore, maximum value, since it is a downward opening parabola, of  is and corresponding value of  is 1.

Hence range of ;

iv.

We are given that;

for

We can write it as;

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 parabola and parabola does not pass horizontal line test, so it  is not a one-one function. However, we can make it so by restricting its domain  around its line of symmetry.

The line of symmetry of a parabola is a vertical line passing through its vertex.

Vertex of the given parabola, as demonstrated in (iii) is . Hence the vertical line  passing through the vertex is . By restricting the domain of the given function to  , we can make it one-one function. On both sides of the vertical line given  function is a one-one function.

Therefore, for , given function is a one-one function i.e. .