Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2003 | Oct-Nov | (P2-9709/02) | Q#6
Question
The diagram shows the curve y=(4−x)ex and its maximum point M. The curve cuts the x-axis at A and the y-axis at B.
i. Write down the coordinates of A and B.
ii. Find the x-coordinate of M.
iii. The point P on the curve has x-coordinate p. The tangent to the curve at P passes through the origin O. Calculate the value of p.
Solution
i.
We are required to find coordinates of points A and B.
It is evident that point A and B are x and y intercepts, respectively, of the curve with equation;
First we find the coordinates of the point A which is x-intercept of the curve.
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).
Therefore we substitute y=0 in equation of the curve;
Now we have two options. First is;
The second option is;
It is evident that is exponential function.
Properties of an exponential function are as follows.
Where
Some properties, which help to plot/sketch, of this graph are as follows.
· The graphs of all exponential functions contain the point .
· The domain is all real numbers .
· The range is only the positive .
· The graph is increasing.
· The graph is asymptotic to the x-axis as x approaches negative infinity
· The graph increases without bound as x approaches positive infinity
· The graph is continuous and smooth.
As per following two properties .
· The graph is asymptotic to the x-axis as x approaches negative infinity
· The graph increases without bound as x approaches positive infinity
Therefore, only option is .
Hence, coordinates of point A are (4,0).
Next we find the coordinates of the point B which is y-intercept of the curve.
The point at which curve (or line) intercepts y-axis, the value of
. So we can find the value of
coordinate by substituting
in the equation of the curve (or line).
Therefore we substitute x=0 in equation of the curve;
Hence, coordinates of point B are (0,4).
ii.
We are required to find the x-coordinate of point M which is given as maximum point of the curve with equation;
The maximum or minimum point of a curve is a stationary point.
A stationary point on the curve
is the point where gradient of the curve is equal to zero;
Since point M is maximum 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 of is:
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation natural exponential function , or
;
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 maximum point, the gradient of the curve at this point must be equal to ZERO.
Hence, x-coordinate of the point M is 3.
iii.
We are given a point on the curve as P(p,y) and the tangent to the curve at P passes through the origin O.
We are required to find the value of p.
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, slope of the tangent OP (to the curve at point P) and gradient of the curve at point P are equal.
We have found in (ii) expression for gradient of the curve;
Now we need expression for 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;
Next we need slope of the tangent OP (to the curve at point P).
Expression for slope of a line joining points and
;
We have coordinates of O(0,0) and P(p,y), therefore, we need y-coordinates of the point P at curve.
If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve (or the line).
Therefore, we can substitute the coordinates of point P in equation of the curve;
Now we can find slope of the tangent OP;
We can equate the slope of tangent OP and gradient of the curve at point P;
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