Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2009 | Oct-Nov | (P2-9709/22) | Q#7

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Question

The diagram shows the curve y = x2 cos x, for , and its maximum point M.

    i.       Show by differentiation that the x-coordinate of M satisfies the equation

   ii.       Verify by calculation that this equation has a root (in radians) between 1 and 1.2.

  iii.       Use the iterative formula to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

Solution

     i.
 We are required to find the equation for x-coordinates of point M which is maximum point of the curve;

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;

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.

Since ;

Hence, x-coordinate of point M is found from .

   ii.       


 

We are required to verify by calculation that the root of equation lies between 1 and 1.2  radians. We need to use sign-change rule.

To use the sign-change method we need to write the given equation as .

Therefore;

If the function  is continuous in an interval  of its domain, and if   and have opposite signs, then  has at least one root between  and .

We can find the signs of at and as follows;

Since and  have
opposite signs for function
, the function has root between and .

 

  iii.
 

If the sequence given by the inductive definition , with some initial value , converges  to a limit , then  is the root of the equation .

Therefore, if , then  is a root of .

We are given the iterative formula as;

We have already found in (ii) through sign-change rule that root of the given equation lies between
and .

Therefore, for iteration method we use;

We use as initial value.

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It is evident that .

Hence, is a root of .

The root given correct to 2 decimal places is 1.08.

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