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CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
APR2016
ASSESSMENT_CODE BCA3010_APR2016
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
10676
QUESTION_TEXT
Define algebraic and transcendental equation. List out any three basic
properties of an algebraic equation.
SCHEME OF
EVALUATION
An equation f(x)=0 is called an algebraic equation if it is purely a
polynomial in x (2 Marks)
An equation f(x)=0 is called an transcendental equation if f(x)
contains trigonometric, exponential or logarithmic functions. (2
Marks)
Properties:
1.Every algebraic equation of nth degree, has ne only n roots (2
Marks)
2.Complex roots occur in pairs (2 Marks)
3.(x-a) is a factor of f(x) (2 Marks)
Or
4.Descartes rule of signs (2 Marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
10681
QUESTION_TEXT
Explain Inherent errors and numerical errors with their
component
SCHEME OF
EVALUATION
Inherent errors (2 Marks)
Data error (2 Marks)
Conversion error (2 Marks)
Numerical errors (2 Marks)
Truncation numerical error (2 Marks)
QUESTION_TYPE DESCRIPTIVE_QUESTION
QUESTION_ID
73417
Using the given figure explain Regula–Falsi method.
QUESTION_TEXT
Choose two points xo and x1 such that f(x1) and f(x2) are of opposite
signs. Since the graph of y=f(x) crosses the X–axis between these two
points.
This indicates that a root lies between these two points x1 and x2.
SCHEME OF
EVALUATION
Equation of the chord joining the points A(x1, f(x1)) and B(x2, f(x2)) is
y–f(x1) = f(x2)–f(x1) divided by x2–x1 Whole multiplied by (x–x1)-------(i) (3.5 marks)
Where f(x2)–f(x1) divided by x2–x1 is the slope of the line AB.
The method consists in replacing the curve AB by means of the Chord
AB and taking the point of intersection of the chord with the X–axis as
an approximation to the root. The point of intersection in the present
case is given by putting y=0 in (i).
Thus we obtain
0–f(x1)=f(x2)–f(x1) divided by x2–x1 whole multiplied by (x–x1). Solve
for x,
We get x=x1–f(x1)(x2–x1) divided by f(x2)–f(x1)-------(ii) (3.5 marks)
Hence the second approximation to the root of f(x)=0 is given by
x3=x=x1–f(x1)(x2–x1) divided by f(x2)–f(x1)------(iii)
If f(x3) and f(x1) are of opposite signs, then the root lies between x1
and x3, and we replace x2 by x3 in (iii), and obtain the next
approximation. Otherwise, f(x3) and f(x1) are of same sign; we replace
x1 by x3 and generate the next approximation. The procedure is
replaced till the root is obtained to the desired accuracy. (3 marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
125706
QUESTION_TEXT
If
where a and b are real constants, calculate
.
Solution: We have
SCHEME OF
EVALUATION
=
=
=
=
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
125707
QUESTION_TEXT
Explain the bisection method?
Step 1: Consider two trail points which by enclose the roots. Two
points a and b enclose a root if f(a) < 0 (negative) and f(b) > 0
(positive) are of opposite signs.
Step 2: Bisect the interval (a, b) and denote the mid-point by x1, so
that
x1 =
equation.
SCHEME OF
EVALUATION
. If f(x1) = 0, we conclude that x1 is a root of the
Otherwise
Step 3: The root lies either between x1 and b if f(x1) < 0, or the root lies
between x1 and a if f(x1) > 0.
If f(x1) > 0 (positive), then
Step 4: Replace b by x1 and search for the root in this new interval
which is half the previous interval. Then the second approximation to
the root is
x2 =
. If f(x2) = 0 then x2 is a root of f(x) = 0.
If f(x2) < 0 (negative), then
Step 5: The root lies between x1 and x2. Then the third approximation
to the root is x3
=
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
164972
i.
QUESTION_TEXT
.
Mention 3 situations where we need integration technique
ii.
Define linear difference equation and write 2 elementary properties
of linear difference equation
We need numerical integration techniques in the following situations: 1.
Functions do not possess closed form solutions. 2. Closed form solutions
exist but these solutions are complex and difficult to use for calculations.
3. Data for variables are available in the form of a table, but no
mathematical relationship between them is know, as is often the case
with experimental data. (5 marks)
A linear difference equation is that in which Yn+1, Yn+2 etc. occur to the
first degree only and are not multiplied together. A linear difference
equation with constant coefficients is of the form Yn+1 + a1Yn+r-1 +
a1Yn+r-2 +….+ arYn = f(n) --- (1) where a1, a2,….ar are constants. (1
mark)
SCHEME OF
EVALUATION
Elementary Properties: i. If U1(n), U2(n),…Ur(n) be r independent
solutions of the equation Yn+r + a1Yn+r-1……+ arYn=0 ---(2)
Then its complete solution is Un = C1U1(n) + C2U2(n) + ………+ CrUr(n)
where C1,C2,……., Cr are arbitrary constants. (2 marks)
ii. If Vn is a particular solution of (1), then the complete solution of (1)
is The part Un is called complementary function (C.F.) and the Vn is
called the particular integral (P.I) of (1). This complete solution (denoted
as C.S.) of (1) is Yn = C.F. + P.I. (2 marks)