Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2010  MayJun  (P19709/13)  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 ycoordinate of vertex represents least value of and xcoordinate 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 oneone function has only one value of against one value of . A function is oneone function if it passes horizontal line test i.e. horizontal line passes only through one point of function. However, if a function is not oneone, 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 oneone 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 oneone function. On both sides of the vertical line given function is a oneone function.
Therefore, for , given function is a oneone 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’;
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