CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
JULY15
ASSESSMENT_CODE BCA3010_JULY15
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
10679
QUESTION_TEXT
Define:
i. Difference equation
ii. Order of difference equation
iii. Degree of difference equation
iv. General solution
v. Particular solution of difference equation
SCHEME OF
EVALUATION
i.Which involves relationship between independent variables,
dependent variables and successive difference of the dependent
variables. (2 Marks)
ii.difference between the largest and the smallest arguments divided by
the unit of increment (2 Marks)
iii.Highest power of Y (2 Marks)
iv.That in which number of arbitrary constants is equal to the order of
the difference equation (2 Marks)
v.That solution which is obtained from the general solution by giving
particular values to the constants (2 Marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
10680
QUESTION_TEXT
Define significant digits. State the rules and describe the notion of
significant digits
SCHEME OF
EVALUATION
The digits that are used to express a number (1 Mark)
Rule 1: (Numbers without decimal point) (2 Marks)
Rule 2: (Numbers with decimal point) (2 Marks)
Note:
1.All non zero digits are significant (1 Mark)
2.All zeros occurring between non zero digits are significant digits
(1 Mark)
3.Trailing zeros following a decimal point (1 Mark)
4.Zeros between decimal point and preceding a non zero digit are
not significant (1 Mark)
5.When the decimal point is not written trailing zeros are not
considered to be significant. (1 Mark)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
73416
QUESTION_TEXT
Explain six steps to apply Cramer’s Rule
SCHEME OF
EVALUATION
Step i. Write the given equations in order so
that
constant terms all on the right side (1.5
Marks)
Step ii. Take = the determinant formed by
the
coefficients of x, y, z (1.5 Marks)
Step iii. Replace the first column of  by constant
terms of
the equations and denote as x
(1.5 Marks)
Step iv. Replace the second column of by
constant
terms of the equations and denote
as y (1.5 Marks)
Step v. Replace the third column of by
constant
terms of the equations and denote
as z (1.5 Marks)
Step vi. Write the solution
(3 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
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)
SCHEME OF
EVALUATION
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)
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)