Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2010 | May-Jun | (P1-9709/13) | Q#10

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Question

The function  is defined for .

i.       Find the values of the constant  for which the line  is a tangent to the curve .

ii.       Express  in the form of , where ,  and  are constants.

iii.       Find the range of .

The function g is defined by  is defined for .

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

v.       For this value of , find an expression for  in terms of .

Solution

i.

If the given line is tangent to the given curve, then they intersect at just one point. We need to establish this.

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).

Equationof the line is;

Equation of the curve is;

Equating both equations;

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.

For the given case;

If the given line is tangent to the given curve, then they intersect at just one point i.e. ONLY or two equal solution for  exist. That means;

Now we have two options;

Hence the given line is tangent to the given curve for   and .

ii.

We have the function;

We have the expression;

We use method of “completing square” to obtain the desired form. We take out factor
‘2’ from the terms which involve
;

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 ;

Hence;

iii.

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.

Vertex form of a quadratic equation is;

The given curve , as demonstrated in (i) 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 . Therefore, least value of  is -6 and corresponding value of  is 2.

Hence range of ;

iv.

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.

v.

We have the function;

for

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 (i), we can write the given function as;

Interchanging ‘x’ with ‘y’;