Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2009 | May-Jun | (P2-9709/02) | Q#7

Question The diagram shows the curve and its minimum point M.

i.       Find the exact coordinates of M.

ii.       Show that the curve intersects the line y = 20 at the point whose x-coordinate is the root of  the equation iii.       Use the iterative formula with initial value x1 = 1.3, to calculate the root correct to 2 decimal places, giving the result of  each iteration to 4 decimal places.

Solution

i.

We are required to find the coordinates of point M which is minimum point of the curve; A stationary point on the curve is the point where gradient of the curve is equal to zero; Since point M is minimum point, therefore, it is stationary point of the curve and, hence, gradient  of the curve at point M must ZERO.

We can find expression for gradient of the curve at point M and equate it with ZERO to find the x- coordinate of point M.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is: Therefore; If and are functions of , and if , then; If , then; Let and ; Rule for differentiation natural exponential function , or ; Rule for differentiation of is:    Now we need expression for gradient of the curve at point M.

Gradient (slope) of the curve at a particular point can be found by substituting x- coordinates of that point in the expression for gradient of the curve;  Since point M is a minimum point, the gradient of the curve at this point must be equal to ZERO.     Hence, x-coordinate of point M on the curve is .

To find the y-coordinate of point M on the curve, we substitute value of x-coordinate in equation of  the curve.  Hence;  Hence, y-coordinate of the point M is .

Therefore, coordinates of point M ii.

We are required to find the x-coordinate of the point of intersection of the line y=20 and the curve  with equation; 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).

Equation of the line is; Equation of the curve is; Equating both equations;  Taking logarithm of both sides; Since for any ;  Single value of x indicates that there is only one intersection point.

iii.

Iteration method can be used to find the root of the given equation using sequence defined by; If the sequence given by the inductive definition , with some initial value ,  converges to a limit , then is the root of the equation .

Therefore, if , then is a root of .

We use as initial value.   1  2  3  4  5  6  7  8  9  10  It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 1.35.