Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2016  FebMar  (P29709/22)  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 (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).
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.
PointSlope form of the equation of the line is;
ii.
We are required to find the xcoordinate 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 xcoordinate 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, xcoordinate 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;
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