Exam (elaborations)
COS2633 EXAM PACK 2025 {DETAILED QUESTIONS AND ANSWERS }
- Course
- COS2633
- Institution
- University Of South Africa
COS2633 EXAM PACK 2025 {DETAILED QUESTIONS AND ANSWERS }
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COS2633/LNS2/3/2021
Model Solutions
Question 1
(1.1) The function f(x) is a fourth degree polynomial and by nature is a continuous
everywhere.
When we explore the values, we find that
Roots: f(x) = 0 when x = 0 (multiplicity) and x = 2
Stationary points:′′f′(x) = 42 x3 − 6x2 = 2x2(2x − 3) = 0 when x = 0 and x = 1.5.
Concavity: f (x) = 12x − 12x = 12x(1 − x) = 0 when x = 0 and x = 1. Which implies
inflection at the points when x assumes these values. Thus Option (2) is not true.
The function has stationary points at the points where x = 0 and x = 1.5 The second
derivative at these points has values f′′(0) = 0 and f′′(1.5) = 9 > 0. Thus the function has
a relative minimum at the stationary point where x = 1.5.
Option (1) is therefore not true because we have found a relative minimum point.
At the other stationary point with x = 0, we have f′(0) = f′′(0) = 0. So at this stationary
point the function has neither relative minimum nor a relative maximum point. Thus
Options (4) and (5) are not true since only one of the stationary points is a relative
extremum.
This leaves only Option (3) as true and indeed it is as we have found that there two
stationary points, one of which is a relative minimum and the other an inflection point.
The answer is (3).
(1.2) The question of ’speed’ implies the use of the derivative, and where a function is
increasing, the derivative is positive.
So
f′(x) = 6t2 − 48t + 90 = 6(x2 − 8t + 15),
and f′(x) > 0 when the factors are either both positive or both negative; i.e.
x2 − 8t + 15 = (t − 5)(t − 3) > 0 ⇒ t > 5 and t > 3 OR t < 5 and t < 0.
i.e. t > 5 OR t < 3 or 0 < t < 3 since t ≥ 0
The Answer is (4)
(1.3) A typical code for Muller’s method (using MATLAB/Octave) is as follows:
f=inline(’xˆ4-2*xˆ3-5*xˆ2+12x-5’) kmax=input(’Max iterations :’)
tol=input(’Tolerance :’) x=input(’Define initial value vector x=[x0
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x1 x2]’); y(1)= f(x(1)); y(2)= f(x(2)); y(3)= f(x(3)); c(1)=(y(2)-
y(1))/(x(2)-x(1)); for k=3:kmax c(k-1)=(y(k)-y(k-1))/(x(k)-x(k-1));
d(k-2)=(x(k-1)-c(k-2))/(x(k)-x(k-2)); s=c(k-1)+(x(k)-x(k-1))*d(k-
2); x(k+1)=x(k)-2*y(k)/(s+sign(s)*sqrt(sˆ2*y(k)*d(k-2))); y(k+1)=
f(x(k+1)); if abs(x(k+1)-x(k))<tol disp(’Muller method has
converged’);break; end iter=k; end if iter>=kmax disp(’Zero not
found to desired tolerance’); end
The table below shows the outcome of the different initial values.
p Option Option Option Option (5)
k (2) (3) (4)
x −0.25000000001.50000000003.0000000000 −2.500000000000
0 00 00 00 −1.500000000000
x 1.000000000 2.00000000004.0000000000
1 00 00 −3.500000000000
000
x −1.50000000001.00000000002.0000000000
2 00 00 −2.4437901601945142.4341135
00
x 1.53112887411.949057943348891
3
0.825811859 49 52
x 510 1.51554938411.7709417531 −2.432093256675
4
0.514178203 87 97
x 629 1.52080686121.5752895509 −2.432098956746
5
0.592030467 83 01
1.52077342301.5284179303 −2.432098956719
x 459
6
0.579830728 39 17
1.52077342541.5209219456 −2.432098956719
x 875
0.579810416 92 76
7 −
x 299 1.52077342541.5207735105
8
0.579810415 92 28
x 426 1.5207734254
9
0.579810415 92
426
The answer is (1)
Question 2
The functions y1 and y2 are nice, smooth functions with no discontinuities.
(2.1) Determining the intersection of the two functions requires finding the solution of y1 =
y2. Apart from using the calculus method to determine the behaviour of a function, the
very basic method is to sketch the graph of the function. In this case the easiest way is
2
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COS2633/LNS2/3/2021
to use the aid of a graphing tool such as Matlab/Octave. Here our focus is in using
numerical methods to solve the equation iequation. The answer is option (2).
(2.2) The equation y1(x) = y2(x) is specifically 8sinx = ex + 4.
There different ways to rearrange this equation into the form x = g(x), two of which are
as given by g1(x) and g2(x).
In Option (1) the idea is to apply the Fixed-point theorem to determine convergence of
the the fixed-point formula x = g1(x), by solving the inequality |g′(x)| < 1. We have
from which we deduce that what is suggested in Option 1 is not true.
Option (2) is not true because g2(x) is a valid rearrangement of the equation y1(x) =
y2(x). It’s appropriateness in finding the intersection cannot be disqualified.
The answer is (4)
(2.3) Option 1 is inappropriate because there is no guarantee that the formulas x = g1(x) and
x = g2(x) converge quickly to the desired fixed-point.
Option (2) is not a true statement contrary to the idea that Aiken’s method is aimed at
improving the convergence of fixed-point methods x = g(x) in general.
Option (3) is blatantly not true.
Option (4) is the most appropriate statement regarding Aiken’s acceleration technique
in line with Theorem 2.14 pertaining to convergence of a sequence of approximations
. The conditions for the convergence of Aiken’s technique are that the above
sequence converges linearly and the limit in the error ratio of successive
approximations is zero. These conditions suggest that the technique may not help if
they are not met.
The answer is therefore, (5)
Question 3
Let x and y be the two numbers.
From the statement we have
Substituting y = 20 − x in the latter equation yields
(√ x + x)(√20 − x + 20 − x) = 155.5
which can then be solved to find x from which y can be obtained by substitution.
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