CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
SMUAPR15
ASSESSMENT_CODE BCA3010_SMUAPR15
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
10677
QUESTION_TEXT
Define curve fitting. What are the methods of curve fitting.
SCHEME OF
EVALUATION
The process of finding the equation of the curve of best fit which
may be suitable for predicting the unknown values in known is
curve fitting. (2 Marks)
Methods:
i.Graphic method (1 Mark)
ii.Method of group average (1 Mark)
iii.Method of moments (1 Mark)
iv.Method of least squares (1 Mark)
Graphical method explanation (4 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
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)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
164973
i.
Why Runge Kutta method is better than Taylor’s series
method and Euler’s method of solving differential equations.
QUESTION_TEXT
ii.
What are the merits and demerits of Taylor’s series method of
solving differential equation?
SCHEME OF
EVALUATION
Runge-Kutta method The Taylor’s series method of solving
differential equations is restricted by the labour involved in the
determination of higher order derivatives. Euler’s method is less
efficient in practical problems since it requires h to be small for
obtaining reasonable accuracy. A class of method known as
Runge-Kutta method does not require the calculations of higher
order derivatives and they are designed to give greater accuracy
with the advantage of requiring only the function values at some
selected points on the sub-interval. (5 marks)
Merits: i. The method of numerical solution by using Taylor series is
of the single-step untruncated type.
ii. The method is very powerful if we can calculate the successive
derivatives of y in an easy manner.
iii. If there is a simple expression for the higher derivatives in terms
of the previous derivatives of y, Taylor’s method will work very
well.
Demerits: The differential equation dy/dx= f(x, y), the function f(x, y)
may have a complicated algebraic structure. Then the evaluation
of higher order derivatives may become tedious and so this
method has little application for computer programmes. (5
marks)