Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2019 | Feb-Mar | (P2-9709/22) | Q#5

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Question

The curve with equation

and the point P on the curve has y-coordinate 10.


i.               
Show that the x-coordinate of P satisfies the equation


ii.               
Use the iterative formula

to find the x-coordinate of P correct to 4 significant figures. Give the  result 

of each iteration to 6 significant figures.


iii.               
Find the gradient of the curve at P, giving the answer correct to 3 significant  figures.

Solution


i.
 

We are given that the curve and the point P on the curve has y-coordinate  10.

We are required to find the x-coordinate of the point P.

We can find the corresponding x-coordinate of point P by substituting y-coordinate of  the same point in equation of the curve.

Taking logarithm of both sides;

 and are inverse functions. The composite function is an identity function, with domain the positive real numbers. Therefore;

   ii.
 

Iteration method can be used to find the root of the given equation using sequence  defined by;

 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 that as initial value.

1

2

3

4

5

6

7

It is evident that .

Hence, is a root of .

The root given correct to 4 significant figures is 2.316.

 

  iii.
 

We are required to find gradient of the curve at the point .

We have demonstrated in (iii) that, .

Gradient (slope) of the curve at the particular point is the derivative of equation of the  curve at that particular point.

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, first we need to find expression for gradient of the curve.

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

We are given equation of the curve;

We use Quotient Rule to differentiate.

If  and  are functions of , and if , then;

If , then;

Let  and , then;

We differentiate  and  one by one.

First we differentiate .

Rule for differentiation of natural exponential function is;

Next, differentiate .

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Hence;

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