# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2017 | May-Jun | (P2-9709/21) | Q#8

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Question

The diagram shows the curve with equation

The curve crosses the x-axis at the point P and has a minimum point M.

i. Find the gradient of the curve at the point P.

ii. Find the exact coordinates of the point M.

Solution

i.

We are required to find the gradient of the curve at the point P which is given as x-intercept of the  curve.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to  is:

Therefore, first we find expression for gradient of the curve.

We are given that;

Therefore;

If  and  are functions of , and if , then;

If , then;

Let  and ;

First we solve .

Rule for differentiation of  is:

Next we solve .

We utilize Chain Rule to differentiate .

If we define , then derivative of  is;

If we have  and then derivative of  is;

Let , then;

Rule for differentiation natural logarithmic function , for  is;

Rule for differentiation of  is:

Since ;

Hence;

We need gradient of the curve at point P.

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;

Therefore, we need value of  at point P.

The point at which curve (or line) intercepts x-axis, the value of . So we can find the  value of coordinate by substituting in the equation of the curve (or line).

We are given equation of the curve;

Therefore, we substitute to find the value of .

Now we have two options.

 Taking anti-logarithm of both sides. and are inverse functions. The  composite function is an identity function, with domain the positive real  numbers. Therefore;

Since denoted origin whereas point P is not origin, therefore, for point P .

Hence, gradient of the curve at point P is;

Since

Hence gradient of the curve at point P is 18.

ii.

We are required to find the exact coordinates of point M which is minimum point of the curve.

We know that minimum point of the curve is a stationary point.

A stationary point on the curve is the point where gradient of the curve is equal to zero;

Since it is stationary point of the curve and, hence, gradient of the curve at this point must be ZERO.

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

Therefore;

Taking anti-logarithm of both sides

and are inverse functions. The composite function is an identity function, with

domain the positive real numbers. Therefore;

Hence .

To find the y-coordinate of point M we can substitute in equation of the curve.

for any

Hence coordinates of minimum point M are