# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2016 | Feb-Mar | (P2-9709/22) | Q#6

Question

The diagram shows the part of the curve  for , and the stationary point M.

i.       Find the equation of the tangent to the curve at the origin.

ii.       Find the coordinates of M, giving each coordinate correct to 3 decimal places.

Solution

i.

We are given that curve with equation  and we are required to find the equation of  the tangent to the curve at origin (0 , 0).

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We have coordinates of point on the tangent (and curve) as (0,0).

Hence we have coordinates of the point on the curve where we are required to find equation of the  tangent are (0,0).

Next, we need slope of the tangent to write its equation.

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, if we find gradient (slope) of the curve at point (0,0) where tangent  intersects the curve,  then we can find slope of the tangent.

Hence, we need gradient of the curve at point  (0,0).

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 x-

coordinates of that point in the expression for gradient of the curve;

We are given that;

Therefore;

We utilize Product Rule to differentiate .

If  and  are functions of , and if , then;

If , then;

Let  and ;

First we differentiate .

Rule for differentiation of natural exponential function , or ;

Next we differentiate

We utilize Chain Rule to differentiate .

If we define , then derivative of  is;

If we have  and then derivative of  is;

Let ;

Rule for differentiation of  is;

Since ;

Rule for differentiation of  is:

Hence;

We are looking for gradient of the curve at point (0 , 0).

Therefore;

Therefore slope of the curve at point (0 , 0) is;

Hence, slope of the tangent to the curve at this point is;

Now we can write equation of the tangent as follows.

Point-Slope form of the equation of the line is;

ii.

We are required to find the x-coordinate of point M which is a stationary point on the curve.

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

Since point M is stationary point, therefore, gradient of the curve at point M must ZERO.

We can equate expression for gradient of the curve at point M 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:

For the given curve from (i) we have;

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 stationary point, the gradient of the curve at this point must be equal to ZERO.

We need to solve the equation to find the value of  which is given as .

To solve this equation for , we can substitute . Hence,

Since given interval is  , for interval can be found as follows;

Multiplying the entire inequality with 2;

Since ;

Hence the given interval for  is .

To solve  equation for interval ,

Using calculator we can find the value of .

To find all the solutions (roots) of , we utilize the periodic/symmetry property of .

 Properties of Domain Range Periodicity Odd/Even Translation/ Symmetry

Therefore;

It is evident that all solutions obtained from   will be beyond the desired interval .

Therefore,only desired solution is;

Since ;

Hence, x-coordinate of the stationary point M is .

Corresponding values of y coordinate can be found by substituting values of x in equation of the  curve.

Hence coordinates of the stationary point on the curve are;