# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/12) | Q#8

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Question

The function f is defined, for , by , where a and b are constants.

i.       In the case where a = 6 and b = 8, find the range of f.

ii.       In the case where a = 5, the roots of the equation f(x)=0 0 are k and 2k, where k is a constant. Find the values of b and k.

iii.       Show that if the equation f(x+a)=a has no real roots, then .

Solution

i.

We are given the function as;

We can write it as;

We are also given that a = 6 and b = 8,therefore;

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 upwards. Its range will be  always greater than or equal to its minimum value. Let’s find the minimum value of the given function.

Vertex form of a quadratic equation is;

The given curve , can be written in vertex form by method of completing square;

First 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 ;

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, least value of  is -17 and corresponding value  of  is -2.

Hence, range of ;

ii.

In the case where a = 5, we can write the given function as;

We are also given that the roots of the equation f(x)=0 are k and 2k. If;

If K and -2K are roots of the equation  then according to factor theorem we can write;

We can expand the left hand side as;

Hence, equating both equations and comparing the coefficients of x on both sides;

Substituting ;

iii.

We have;

For ;

It is evident that it is a quadratic equation.

Standard form of quadratic equation is;

We can write it as;

We are given that  has no real roots.

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 given case;