# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2017 | Oct-Nov | (P2-9709/21) | Q#7

Question

The diagram shows the curve

The curve crosses the y-axis at the point P and the gradient of the curve at P is m. The point Q on  the curve has x-coordinate q and the gradient of the curve at Q is m.

i.       Find the value of m and hence show that q satisfies the equation

where the values of the constants a and b are to be determined.

ii.       Show by calculation that 4.5 < q < 4.0.

iii.       Use an iterative formula based on the equation in part (i) to find the value of q correct to 3  significant figures. Give the result of each iteration to 5 significant figures.

Solution

i.

We are required to find the value of m which is gradient of the curve at P which is given as y- intercept point of the curve with equation;

We can find expression for gradient of the curve and then gradient at point P.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Therefore;

Rule for differentiation of  is:

We utilize Chain Rule to differentiate .

If we define , then derivative of  is;

If we have  and  then derivative of  is;

Let , then;

Hence;

Rule for differentiation of  is;

Since ;

Hence;

Hence, it is expression for gradient of the curve.

We need gradient of the curve at point P which is y-intercept of the curve.

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, we need value of to substitute in expression for gradient of the curve.

Similarly gradient of the curve at point Q;

We are given that point Q on the curve has x-coordinate q, therefore;

We are given that the gradient of the curve at point Q is .

Therefore;

Hence;

Hence, x-coordinate of the point Q on the curve;

ii.

We are required to show by calculation that 4.5 < q < 4.0.

Since “q”, the x-coordinate of point Q on the curve and its expression as obtained in (i) is;

Therefore;

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.

We are required to apply an iterative formula based on the equation in part (i) to find the value of m correct to 4 significant figures.

We have found in (i) that;

We can write an iterative formula based on this equation.

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 .

As demonstrated in (ii) that and have opposite signs for function  , the function has root between and .

We use as initial value.

 1 2 3 4 5 6 7 8 9 10 11 12 13

It is evident that .

Hence, is a root of .

The root given correct to 3 significant figures is -4.11.