Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | May-Jun | (P1-9709/12) | Q#11

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Question

The function  is defined by  for .

     i.       Find the set of values of x for which f(x) 3.

   ii.       Given that the line y=mx+c is a tangent to the curve y = f(x), show that

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

  iii.       Express   in the form , where a and b are constants.

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

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

Solution


i.
 

To find the set of values of x for which ;

We solve the following equation to find critical values of

Now we ave two options;

Hence the critical points on the curve for the given condition are 4 & -2.

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.

Therefore conditions for  are; 


ii.

We are given that line with equation  is tangent to the curve  .

Tangent indicates that line and the curve intersect at only single point.

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 line is;

Equation of the curve is;

Equating both equations;

If we solve this equation for x, we will get x-coordinates of the point(s) of intersection of line and  curve.

Single value of x indicates that there is only one intersection point.

Since, in the given case, line is tangent the curve, there must be single value of x in the solution of  above equation. 

We recognize that  is a quadratic equation.

Standard form of quadratic equation is;

Expression for discriminant of a quadratic equation is;

If   ; Quadratic equation has two real roots.

If   ; Quadratic equation has no real roots.

If   ; Quadratic equation has one real root/two equal roots.

Since line is tangent to the curve, only single root must yield from solution of above equation, therefore;


iii.
 

We have the expression;

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

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 ;


iv.
 

We are given the function;

For a given function  the inverse function  exists only if  is one-one function.

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.

We recognize that  is a quadratic equation which results in parabola,  graphically.

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 form of a quadratic equation is;

The given curve , as demonstrated in (iii) can be written in vertex form as; 

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 .

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.

Since we are given that  is defined for , we choose that given function is a  one-one function for  although it is also one-one function for .

   v.

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 .

As demonstrated in (iii), we can write the given function as;

Interchanging ‘x’ with ‘y’;

 

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