Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 2 (P29709/02)  Year 2011  OctNov  (P29709/23)  Q#7
Hits: 19
Question
The diagram shows the curve . The curve has a gradient of 3 at the point P.
i. Show that the xcoordinate of P satisfies the equation
ii. Verify that the equation in part (i) has a root between x = 3.1 and x = 3.3.
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 xcoordinate of point on the curve where curve has gradient 3.
Equation of the curve is given as;
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 ;
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;
We know that gradient of the curve at point P is 3.
Hence, xcoordinate of point P on the curve is satisfied by the equation;
ii.
We are required to verify equation has a root between x=3.1 and x=3.3.
To use the signchange 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.
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 use as initial value.



1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


11 


12 


13 


14 


15 


It is evident that .
Hence, is a root of .
The root given correct to 2 decimal places is 3.21.
Comments