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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)