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=(4x)eand 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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