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#3


A curve has equation .

    i.       Find the set of values of  for which .

   ii.       Find the value of the constant  for which the line  is a tangent to the curve.



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

We are given that;


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 -1 & 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.
 (‘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 upwards.

Therefore conditions for  are;


If any given line is tangent to the curve, then line and the curve intersect at only a 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;

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

Therefore, solution of above equation, which is a quadratic equation, must yield equal or repeated roots.

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.

Therefore, for the above equation;