Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2018  MayJun  (P19709/13)  Q#8
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Question
i. The tangent to the curve y = x^{3} − 9x^{2} + 24x − 12 at a point A is parallel to the line y = 2 − 3x. Find the equation of the tangent at A.
ii. The function f is defined by f(x) = x^{3} − 9x^{2 }+ 24x − 12 for x > k, where k is a constant. Find the smallest value of k for f(x) to be an increasing function.
Solution
i.
We are required to find the equation of tangent to the curve at point A.
To find the equation of the line either we need coordinates of the two points on the line (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope form of Equation of Line).
First we find slope of the tangent to the curve.
We are given that tangent to the curve y = x^{3} − 9x^{2} + 24x – 12 at point A is parallel to the line y = 2 − 3x.
If two lines are parallel to each other, then their slopes and are equal;
Hence, slope of the tangent to the curve y = x^{3} − 9x^{2} + 24x – 12 at point A is equal to slope of the line y = 2 − 3x.
Therefore, we find slope of the line y = 2 − 3x to find slope of the tangent to the curve.
SlopeIntercept form of the equation of the line;
Where is the slope of the line.
Therefore, we can rearrange the equation of the line in slopeintercept form as;
y = − 3x + 2
Hence;
Next we find coordinates of a point on the tangent to the curve.
We are given that tangent is drawn to the curve at point A. Hence, point A lies both on the curve and the tangent.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, gradient of the curve at point A, where tangent is drawn to the curve, is same as slope of the tangent.
Therefore, we find expression for gradient of the curve at point A and equate it with;
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting xcoordinates of that point in the expression for gradient of the curve;
Therefore, we find of the curve and then equate it with slope of the tangent to the curve .
We are given that;
Therefore;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Now gradient of the curve at point A;
Since gradient of the curve at point is .
To find the ycoordinate of a point on a given line or curve we simply substitute the x coordinate of the point in the equation of the curve or line.
Therefore we substitute in;
Hence, coordinates of point A, on the tangent and the curve, are (3, 6).
Now we can write equation of the tangent to the curve at point A.
PointSlope form of the equation of the line is;
ii.
We are given function;
To test whether a function is increasing or decreasing at a particular point , we take derivative of a function at that point.
If , the function is increasing.
If , the function is decreasing.
If , the test is inconclusive.
Therefore, we need of the given function.
As demonstrated in (i);
Since for an increasing function;
Therefore;
To find the set of values of x for which , 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 2 & 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.
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.
Therefore conditions for are;
Hence function is increasing, only if we consider the parabola when or .
We are given that function is defined for and we are required to find the smallest value of k for f(x) to be an increasing function.
Therefore, as ensures the function is increasing.
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