Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2012 | May-Jun | (P2-9709/22) | Q#6

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Question

The diagram shows the curve , for . The x-coordinate of the maximum point  M is denoted by .

i.
Find  and show that  satisfies the equation tan 2x = 2x + 4.

ii.       Show by calculation that  lies between 0.6 and 0.7.

iii.       Use the iterative formula to find the value of a correct to 3 decimal  places.

Solution

i.

We are required to find the equation for x-coordinate of point M which is given as maximum point of  the curve with equation;

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 ;

We differentiate and one by one.

First we differentiate ;

If we define , then derivative of is;

If we have  and then derivative of  is;

Let , then , therefore;

Rule for differentiation of  is;

Rule for differentiation of  is:

First we differentiate ;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Finally;

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 ;

ii.

We are required to verify by calculation that   lies between 0.6 and 0.7. We need to use sign- change rule.

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

From (i) we have found;

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 we can write the given equation  and transform it to , then we can find the root of  the equation by iteration method using sequence defined as.

We are also given the iterative formula as;

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 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.

 0 1 2 3 4 5

It is evident that .

Hence, is a root of .

The root given correct to 3 decimal places is .