1
The root of the equation f(x) = 0, where
f(x) = x +ln2x - 4
is to be estimated using the iterative formula
xn1  4  ln2xn , with x0  2.4.
(a) Showing your values of x1, x2, x3,…, obtain the value, to 3
decimal places, of the root.
(4)
(b) By considering the change of sign of f(x) in a suitable interval,
justify the accuracy of your answer to part (a).
(2)
1
The root of the equation f(x) = 0, where
f(x) = x + ln2x - 4
is to be estimated using the iterative formula
xn1  4  ln2xn , with x0  2.4.
(a) Showing your values of x1, x2, x3,…, obtain the value, to 3 decimal
places, of the root.
(4)
x1  4  ln2x0  4  ln2(2.4)  2.4313...
x2  4  ln2(2.431...)
 2.4183...
x3  2.4237...
x4  2.4215...
x  2.422
x5  2.4224...
1
The root of the equation f(x) = 0, where
f(x) = x + ln2x - 4
is to be estimated using the iterative formula
xn1  4  ln2xn , with x0  2.4.
(b) By considering the change of sign of f(x) in a suitable interval,
justify the accuracy of your answer to part (a)
(2)
x  2.422
f(2.4215) = -0.000965...
f(2.4225) = 0.000447...
The function is continuous and there is a change of sign,
therefore x is a root
2
f(x) = x3 + x2  4x  1.
The equation f(x) = 0 has only one positive root, 
(a) Show that f(x) = 0 can be rearranged as
(2)
 4x 1 
x 
 , x  1
 x 1 
The iterative formula xn + 1 =
find an approximation to .
 4xn  1


 xn  1 
is used to
(b) Taking x1 = 1, find, to 2 decimal places, the values of
(3)
x2, x3 and x4.
2
f(x) = x3 + x2  4x  1.
The equation f(x) = 0 has only one positive root, 
(a) Show that f(x) = 0 can be rearranged as
 4x 1 
x 
 , x  1
 x 1 
x  x  4x  1  0
3
2
x3  x2  4 x  1
x2 (x  1)  4 x  1
4x  1
2
x 
(x  1)
(2)
4x  1
x
x 1
2
f(x) = x3 + x2  4x  1.
The equation f(x) = 0 has only one positive root, 
(b) Taking x1 = 1, find, to 2 decimal places, the values of
(3)
x2, x3 and x4.
4 xn  1
xn1 
xn  1
4 1   1
x2 
1   1
 1.58
4 1.58...   1
x3 
1.58...  1
 1.68
x4  1.70
2
f(x) = x3 + x2  4x  1.
The equation f(x) = 0 has only one positive root, 
(c) By choosing values of x in a suitable interval, prove that  = 1.70,
correct to 2 decimal places.
(3)
(d)Write down a value of x1 for which the iteration formula
xn + 1 =  4 x n  1 
 x 1 
 n

does not produce a valid value for x2.
Justify your answer.
(2)
2 4.
2
f(x) = x3 + x2  4x  1.
The equation f(x) = 0 has only one positive root, 
(c) By choosing values of x in a suitable interval, prove that  = 1.70,
correct to 2 decimal places.
(3)
f(1.695) = -0.037…
f(1.705) = 0.0435…
The function is continuous and there is a change of sign,
therefore  is a root
2
f(x) = x3 + x2  4x  1.
The equation f(x) = 0 has only one positive root, 
(d)Write down a value of x1 for which the iteration formula
xn + 1 =  4 x n  1 
 x 1 
 n

does not produce a valid value for x2.
Justify your answer.
(2)
x = -1; Division by zero is not possible
-1 < x < -¼; Cannot find the square root of a negative number
3
(a)
Sketch, on the same set of axes, the graphs of
y = 2 – ex and y = x.
(3)
[It is not necessary to find the coordinates of any points of intersection with the axes.]
Given that f(x) = ex + x – 2, x  0,
(b)
explain how your graphs show that the
equation f(x) = 0 has only one solution, (1)
3
(a)
Sketch, on the same set of axes, the graphs of
y = 2 – ex and y = x.
(3)
[It is not necessary to find the coordinates of any points of intersection with the axes.]
3
Given that f(x) = ex + x – 2, x  0,
(b)
explain how your graphs show that the
equation f(x) = 0 has only one solution, (1)
Where curves meet is
solution to f(x) = 0;
hence there is only one
intersection
3
(c)
show that the solution of f(x) = 0 lies
between x = 3 and x = 4.
(2)
The iterative formula xn + 1 = (2 –e
solve the equation f(x) = 0.
(d)
 xn 2
)
is used to
Taking x0 = 4, write down the values of x1,
x2, x3 and x4, and hence find an
approximation to the solution of f(x) = 0,
giving your answer to 3 decimal places.
(4)
3
(c)
show that the solution of f(x) = 0 lies
between x = 3 and x = 4.
(2)
f(x) = ex + x – 2
f(3) = -0.218…
f(4) = 0.018…
This is a continuous function and there is a change
of sign over this interval, therefore there is a root
within this interval.
The iterative formula xn + 1 = (2 –e
to solve the equation f(x) = 0.
3
(d)
 xn
Taking x0 = 4, write down the values of x1,
x2, x3 and x4, and hence find an
approximation to the solution of f(x) = 0,
giving your answer to 3 decimal places. (4)
x1 =(2 – e–4)2 = 3.92707…
x2
= 3.92158…
x3
= 3.92115…
x4
= 3.92111(9)…
Approx. solution = 3.921 (3 dp)
x0 = 4
)2 is used
4
(a)
The curve with equation y = ln 3x crosses
the x-axis at the point P (p, 0).
Sketch the graph of y = ln 3x, showing the exact value
of p
(2)
The normal to the curve at the point Q, with x-coordinate q,
passes through the origin.
(b)
Show that x = q is a solution of the equation
x2 + ln 3x = 0
5 8.
(4)
4
(a)
The curve with equation y = ln 3x crosses
the x-axis at the point P (p, 0).
Sketch the graph of y = ln 3x, showing the exact value
of p
(2)
ln 1 = 0
ln3x = ln1
3x = 1
x = 1/3
1
3
5 8.
4
The curve with equation y = ln 3x crosses
the x-axis at the point P (p, 0).
The normal to the curve at the point Q, with x-coordinate q, passes
through the origin.
(b)
Show that x = q is a solution of the equation
x2 + ln 3x = 0
y = ln 3x, y’ = 1/x (4)
Q
At Q gradient is 1/q
Gradient of normal is -q
Equation of OQ is
y – 0 = -q(x- 0)
y = -qx
ln 3x = -x2  x2 + ln 3x = 0
5 8.
4
The curve with equation y = ln 3x crosses
the x-axis at the point P (p, 0).
(c) Show that the equation2 in part (b) can be rearranged
in the form x = 13 e  xn
(2)
(d)
 x2
Use the iteration formula xn + 1 = e , with x0 = 13 ,
to find x1, x2, x3 and x4. Hence write down, to 3
decimal places, an approximation for q
(3)
1
3
5 8.
4
The curve with equation y = ln 3x crosses
the x-axis at the point P (p, 0).
(c) Show that the equation2 in part (b) can be rearranged
in the form x = 13 e  xn
(2)
x2 + ln 3x = 0
ln 3x = - x2
 xn2
3x = e
x = 1/3e-x2
5 8.
4
(d)
The curve with equation y = ln 3x crosses
the x-axis at the point P (p, 0).
 x2
Use the iteration formula xn + 1 = e , with x0 = 13 ,
to find x1, x2, x3 and x4. Hence write down, to 3
decimal places, an approximation for q
(3)
1
3
x1 = 0.298280;
x2 = 0.304957, x3 = 0.303731, x4 = 0.303958
Root = 0.304 (3 decimal places)
5 8.