IT3201 Mathematics for ComputingII mc2ict'cmb'ac'lk

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PREPARING FOR THE
IT3201 Mathematics for Computing-II
mc2_at_ict.cmb.ac.lk
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It will be very helpful if the following three
point are noted

• Some questions are purely based on definitions of
  certain mathematical entities. If the definition
  is known, the correct answers can be identified
  directly.
• If the definition is known and some other simple
  and elementary properties of the entity are also
  known, then, sometimes, certain obvious indirect
  methods can be used for easier identification of
  the correct options.

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• In either case, without understanding the entire
  question, a suitable strategy cannot be worked
  out. We start with Part1 and the following
  questions have be chosen for discussion because,
  identification of certain special features, makes
  the strategy more evident.

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Q8)
The given data is direct but the alternative
answers are of different types. Here, each
alternative has to be considered independently.
(a), (e) can be eliminated by using simple
counter examples. For example for (a), take A1
and B-1 of the same order. Then, the given data
is satisfied and AB0 which has no inverse. Thus
(a) can be eliminated.
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As regards (e), take ABI,and eliminate (e).
Now (AB)B-1A-1I. Therefore (AB)-1B-1A-1.
(Some students would know this). Thus (b) is
false but (c),(d) are true.
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Q12)
In (iii), there is an undefined A-AI which has
got in as a printing error. Once, this is wiped
out, all the transformations are evidently,
linear.
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Q13)
It is useful to write the transformation as
xI1.x0.y(-3)z yI1.x(-2)y4.z zI0.x(2)y1.z
Then, the matrix of the transformation is
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Q14)
The question is, for which values of x,y, is
This can be done directly by solving two
simultaneous equation in x,y. However, if it is
noted that
, then
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we have to solve

i.e xy0 3x-y0 implying that x0,
y-0. OR. This is of the form
Since A is non-singular, it has an inverse.
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Q26)
Here A is given. Therefore AI is known. Hence 2A
and AAI can be worked out, and therefore B. This
is the direct method and this involves several
matrix operations. However, if is noted that 2B
AAI - 2A AAI - A2
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y
Q34)
Reflexion on the line xy transforms to
Hence, the matrix is
Stretch by a factor of 4 in the direction Oy
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transforms
which corresponds to the matrix
Reflection is done first which takes
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This is followed by the next operation which
finally results in
Note that the successive linear transformations
have to be read from the right to the left. That
is AB means B followed by A.
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Q36)
Whenever considered useful, it is best that a
diagram is drawn
y
gt
The direct method is to take
so that OA OB 1. Then icos ?isin ?j and j
(-sin ?)icos ?j
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Otherwise, note that if ?0, ii and jj. Only
(b) satisfies this conduction
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Q37)
A linear transformation is given and the question
is on its inverse. It is best that we deal with
the inverse transformation directly. If it is
represented by A then
Determination of A involves that of 4 constants
which means the solution of 4 simultaneous
equations. Since this takes a lot of time, an
indirect method coupled with some properties of
linear transformations will have to be used.
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Let us write
(3,5)? (2,6) ----- (1) (1,1)?
(6,2)-------(2) ?(3,3)?(18,6)------(3)
3x(2) ?(0,2)?(-16,0)-----(4)
(1)-(3) ?(0,1)?(-8,0)------(5)
(1/2)x(4) Sim (5,5)?(30,10)(6)
5x(2) ?(2,0)?(28,4)--- (7)
(6)-(1) ?(1,0)?(14,2)----(8)
(1/2)x(7) But (0,1) ? (-8,0) ---(5) ?(2,0)?(28,4
)---(9) 2x(8)
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and (0,3)?(-24,0)---(10) 3x(5) ?(2,3)?(4,4)
(9)(10) In solving the problem, the
explanations are not needed. ?(c) is the correct
answer.
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Q38)
There is an error in the question. x-axis at the
origin should be replaced by y-axis A rough
sketch of the two graphs will reveal (c) as the
correct choice. Note (a),(b),(d),(e) can all be
eliminated because the tangents to the curves y
sinx and y cosx will never be perpendicular to
the x- axis.
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Q39)
,the messy
In the integral
looking term to handle is lnx. The obvious step
is to change the variable using ylnx
Hence, the integral is
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Care must be exercised in selecting the correct
choice because sin2 can be written in many
different ways, one way being 2(sin1)(cos1).
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Q40)
There is a error in the question. In the 3rd
line, y-axis(end of the last sentence) should
be replaced by x-axis.
y
gt
A
C
gt
x
B
o
(2,0)
(1,0)
(3,0)
As far as B is concerned, y is negative in this
region.
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because area is positive.
By symmetry CA
Thus (i) and (ii) are correct and (iii) is
incorrect. Thus (a)is the only correct
choice. Note Even if the error is not corrected,
C becomes infinite and therefore (iii) is
incorrect. (i) and (ii) are unaffected.
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Now, let us get to part 2
Q1(a)
One way is to use the fact that R3-R1 and R2-R1
leave the determinant unchanged where R3 stands,
for example, for Row3. Then, it is equal to
R2 has a factor (b-a) and R3 has a factor c-a
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OR One can use the Remainder Theorem. If ab,
then ?0 ?(a-b) is a factor of ? By symmetry
(b-c),(c-a) are also factors. ?(a-b)(b-c)(c-a) is
a factor. If a,b,c are replaced by ?a, ?b, ?c
respectively, ? become ?4 ?. (? is homogeneous of
degree 4 in a,b,c). If a is replaced by b, b by c
and c by a, then ? remains unchanged.? is
unchanged for a cyclic change where a ? b ? c ?
a.
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• ??(a-b)(b-c)(c-a)(abc) where ? is independent
  of a,b,c.
• The coefficient of bc3 in ? is 1
• ?1

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Q2(b)(i)

• ?Y sin2x
• Y(1)2cos2x 2 sin
• Note that sinx becomes cosx if is added to x
  in sinx.
• Y(2)22cos 22sin
• Now, a pattern emerges. With each
  differentiation, a factor 2 appears and the angle
  increases by .
• By a very intelligent guess,
• Y(n)2nsin

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Q2(c)
From first principles, y2 and y?y2
It is clear that, if ?x0, the R.H.S. reduces to
0. ?getting rid of the square root signs is
vital. This can be achieved by multiplying the
numerator and denominator by
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