# 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

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** Question**

The diagram shows the curve y=(4−x)e^{x }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

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

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

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

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

Therefore;

If

If

Let

Rule for differentiation of

Rule for differentiation of

Rule for differentiation of

Rule for differentiation natural exponential function

Now we need expression for gradient of the curve at point M.

Gradient (slope)

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

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)

Therefore;

Next we need slope of the tangent OP (to the curve at point P).

Expression for slope of a line joining points

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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