# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2018 | May-Jun | (P1-9709/12) | Q#7

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**Question**

The function is defined by for .

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

** ii. **State the coordinates of the stationary point on the curve y = f(x).

The function is defined by for .

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

** iv. **For this value of k, find .

**Solution**

i.

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;

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

ii.

We are given the function;

We are required to state coordinates of the stationary point on the curve.

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 curve , as demonstrated in (i), can be written in vertex form as;

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

For the given case, vertex is .

Vertex of a parabola is a stationary point on the curve.

Hence coordinates of the stationary point on the given curve are;

iii.

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. .

iv.

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

Interchanging ‘x’ with ‘y’;

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