Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2010 | May-Jun | (P2-9709/23) | Q#5

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  Question

The equation of a curve y=x3e-x.

     i.       Show that the curve has a stationary point where x = 3.

   ii.       Find the equation of the tangent to the curve at the point where x = 1.

Solution

     i.
 

We are required to show that the curve has a stationary point where x=3.

We are given equation of the curve as;

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

Hence, gradient of the curve at point where x=3 must be ZERO.

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

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 of  is:

Rule for differentiation natural exponential function , or ;

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

Hence, x-coordinate of the stationary point is 3.

 

   ii.
 

We are given that curve with equation  and we are required to find the equation of the  tangent to the curve at point where .

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 (1,y).

Therefore, first we need to find the y-coordinate of the curve.

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

Hence coordinates of the point on the curve where we are required to find equation of the tangent  are (1,)

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 (1,) where tangent  intersects the curve,  then we can find slope of the tangent. 

Hence, we need gradient of the curve at point  (1,).

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 have found in (i) that;

We are looking for gradient of the curve at point (1,).

Therefore;

Therefore slope of the curve at point (1,) 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;

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