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Numerical Methods: Bisection Method and Newton-Raphson Method Exercises, Quizzes of Numerical Methods in Engineering

DBATU MCQs

Typology: Quizzes

2020/2021

Available from 08/05/2021

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1. Using Bisection method find the root of cos๐‘ฅ โˆ’ ๐‘ฅ๐‘’ ๐‘ฅ= 0 with a = 0 and b = 1.
a) 0.617
b) 0.527
c) 0.517
d) 0.717
Answer: c
2. If a function is real and continuous in the region from a to b and f(a) and f(b) have
opposite signs then there is no real root between a and b.
a) True
b) False
Answer: b
3. The Bisection method is also known as ___________________
a) Binary Chopping
b) Quaternary Chopping
c) Tri region Chopping
d) Hex region Chopping
Answer: a
4. A function is given by ๐‘ฅ โ€“ ๐‘’โˆ’๐‘ฅ = 0. Find the root between a = 0 and b = 1 by using
Bisection method.
a) 0.655
b) 0.665
c) 0.565
d) 0.656
Answer: c
5. Find the root of ๐‘ฅ4โˆ’๐‘ฅโˆ’10 = 0 approximately upto 5 iterations using Bisection
Method. Let a = 1.5 and b = 2.
a) 1.68
b) 1.86
c) 1.88
d) 1.66
Answer: b
6. The Bisection method has which of the following convergences?
a) Linear
b) Quadratic
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff

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  1. Using Bisection method find the root of cos ๐‘ฅ โˆ’ ๐‘ฅ๐‘’๐‘ฅ^ = 0 with a = 0 and b = 1. a) 0. b) 0. c) 0. d) 0.

Answer: c

  1. If a function is real and continuous in the region from a to b and f(a) and f(b) have opposite signs then there is no real root between a and b. a) True b) False

Answer: b

  1. The Bisection method is also known as ___________________ a) Binary Chopping b) Quaternary Chopping c) Tri region Chopping d) Hex region Chopping

Answer: a

  1. A function is given by ๐‘ฅ โ€“ ๐‘’โˆ’๐‘ฅ^ = 0. Find the root between a = 0 and b = 1 by using Bisection method. a) 0. b) 0. c) 0. d) 0.

Answer: c

  1. Find the root of ๐‘ฅ^4 โˆ’ ๐‘ฅ โˆ’ 10 = 0 approximately upto 5 iterations using Bisection Method. Let a = 1.5 and b = 2. a) 1. b) 1. c) 1. d) 1.

Answer: b

6. The Bisection method has which of the following convergences?

a) Linear

b) Quadratic

c) Cubic

d) Quaternary

Answer: a

7. 2 and 4 such that f(2) = 4 and f(4) = 16 are appropriate initial points

for the bisection method.

a)True

b)False

Answer: b

  1. The bisection method of finding roots of nonlinear equations falls under the category of a (an) _________ method. a) open b) bracketing c) random d) graphical

Answer: b

  1. If for a real continuous function ๐‘“(๐‘ฅ), ๐‘“(๐‘Ž) โˆ— ๐‘“(๐‘) < 0 then in the range of [๐‘Ž, ๐‘] ๐‘“๐‘œ๐‘Ÿ ๐‘“(๐‘ฅ) = 0 , there is (are) a) one root b) an undeterminable number of roots c) no root d) at least one root

Answer: d

  1. Rate of convergence of the Newton-Raphson method is generally __________ a) Linear b) Quadratic c) Super-linear d) Cubic

Answer: b

  1. The Iterative formula for Newton Raphson method is given by __________ a) x 1 = x 0 - f(x 0 )/fโ€™(x 0 ) b) x 0 = x 1 - f(x 0 )/fโ€˜(x 0 ) c) x 0 = x 1 +f(x 0 )/fโ€™(x 0 ) d) x 1 = x 0 +f(x 0 )/fโ€˜(x 0 )

Answer: a

c) non-stationary d) stationary

Answer: d

  1. The convergence of which of the following method depends on initial assumed value? a) False position b) Gauss Seidel method c) Newton Raphson method d) Euler method

Answer: c

  1. The equation f(x) is given as x^3 +4x+1=0. Considering the initial approximation at x=1 then the value of x 1 is given as _______________ a) 1. b) 1. c) 1. d) 1.

Answer: c

  1. The Newton-Raphson method of finding roots of nonlinear equations falls under the category of _____________ methods. a) bracketing b) open c) random d) graphical

Answer: b

  1. The Newton-Raphson method formula for finding the square root of a real number R from the equation ๐‘ฅ^2 โˆ’ ๐‘… = 0 is a) ๐‘ฅ๐‘–+ 1 = ๐‘ฅ 2 ๐‘–

b) ๐‘ฅ๐‘–+ 1 = 32 ๐‘ฅ๐‘– c) ๐‘ฅ๐‘–+ 1 = 12 (๐‘ฅ๐‘– + (^) ๐‘ฅ๐‘…๐‘– )

d) ๐‘ฅ๐‘–+ 1 = 12 ( 3 ๐‘ฅ๐‘– โˆ’ (^) ๐‘ฅ๐‘…๐‘– )

Answer: c

  1. The equation f(x) is given as x^2 - 4=0. Considering the initial approximation at x= then the value of x 1 is given as ____________

a) 10/ b) 4/ c) 7/ d) 13/

Answer: a

  1. The equation f(x) is given as x^3 +4x+1=0. Considering the initial approximation at x=1 then the value of x 1 is given as _______________ a) 1. b) 1. c) 1. d) 1.

Answer: c

  1. Cramerโ€™s Rule fails for ___________ a) Determinant > 0 b) Determinant < 0 c) Determinant = 0 d) Determinant = non-real

Answer: c

  1. Cramerโ€™s Rule is not suitable for which type of problems? a) Small systems with 4 unknowns b) Systems with 2 unknowns c) Large systems d) Systems with 3 unknowns

Answer: c

  1. Apply Cramerโ€™s rule to solve the following equations.

a) X = 1, y = 2, z = - 1 b) X = 2, y = 1, z = - 1 c) X = 2, y = - 1, z = 1 d) X = 1, y = - 1, z = 2

Answer: a

  1. Apply Cramerโ€™s rule to solve the following equations.

a) x = 1, y = 2, z = 3 b) x = 2, y = 2, z = 3 c) x = 2, y = 3, z = 7 d) x = 1, y = 3, z = 8 Answer: a

  1. The aim of elimination steps in Gauss elimination method is

to reduce the coefficient matrix to ____________ a) diagonal

b) identity c) lower triangular d) upper triangular

Answer: d

  1. What are the coefficients of the equation obtained during the elimination called?

a) Joints

b) Pivots c) Calculated coefficients d) Operative coefficients

Solution: b

  1. Find the values of x, y, z in the following system of equations

by gauss Elimination Method.

2 ๐‘ฅ + ๐‘ฆ โ€“ 3 ๐‘ง = โˆ’ 10 โˆ’ 2 ๐‘ฆ + ๐‘ง = โˆ’ 2 ๐‘ง = 6

a) 2, 4, 6 b) 2, 7, 6 c) 3, 4, 6 d) 2, 4, 5

Solution: a

  1. For solving the system of equations 5 x + y + 2 z = 34, 4 y โˆ’ 3 z = 12, 10 x โˆ’ 2 y + z = โˆ’ 4 by Gauss elimination method using partial pivoting, the pivots for elimination of x and y are a)10 and 4 b)5 and 4 c)10 and 2 d)5 and โˆ’ 4 Answer: a
  2. For solving the system of equations (^8) y + 2 z = โˆ’7, 3 x + 5 y + 2 z = 8, 6 x + 2 y + 8 z = 26 by Gauss elimination method using partial pivoting, the pivots for elimination of x and y are a) 6 and 3 b) 6 and 8 c) 8 and 5 d) 6 and 4 Answer: b
    1. For solving the system of equations (^8) y + 2 z = โˆ’7, 3 x + 5 y + 2 z = 8, 6 x + 2 y + 8 z = 26 by Gauss elimination method using partial pivoting, the pivots for elimination of x and y are a) 3 and 1 b) 3 and 4 c) 1 and 4/ d) 3 and 13/

Answer: d

  1. The given system of equations is 0 , 4

1 ,^1

x + 1 y + z = x + y + z =

5

1 x + y + z = In Gauss elimination method, on eliminating x from second and third

equations, the system reduces to

Answer: b

  1. The given system of equations is 2 x + y + z = 10 , 3 x + 2 y + 3 z = 18 , x + 4 y + 9 z = 16.

In Gauss elimination method, on eliminating x from second and third equations, the system

reduces to

Answer: a

  1. The given system of equations is 2 x + 2 y + z = 12 , 3 x + 2 y + 2 z = 8 , 2 x + 10 y + z = 12.

In Gauss elimination method, on eliminating x from second and third equations, the system

reduces to

Answer: c

  1. Using Gauss elimination method, the solution of system of equations

y z

y z

x y z is

Answer: b

  1. A curve passes through the set of points (0,1), (1, 3), (2, 7), (3, 13) Value of

3

0

๏ƒฒ y dx by

Trapezoidal rule is given by

A 17 B 15 C 19 D 21

Answer:

46. Value of ๏ฐobtained by evaluating the integral

1 2 0

dx

๏ƒฒ + x

, using Trapezoidal rule with

1 2

h = is given by

1 2 0

given: 1 1 4

dx x

๏ƒง (^) + ๏ƒท ๏ƒจ ๏ƒธ

A 3.15 B 3.1 C 3.2 D3.

  1. Value of^ log 2 e obtained by evaluating the integral

1

0

dx

๏ƒฒ + x

, using Simpsonโ€™s^1 3

rd rule

with^1 2

h = is given by

1

0

given: 1 log 2 1 e

dx x

๏ƒฆ ๏ƒถ ๏ƒง = ๏ƒท

A 0.5934 B 0.6560 C 0.6944 D0.

  1. Fit the straight line to the following data.

a) y = 0.9288x + 7.7815 5 b) y = 7.78155x + 0.928 8 c) y = 0.8288x + 6.7815 5 d) y = 6.78155x + 0.828 8

Answer: a

  1. Fit the straight-line curve to the following data.

a) y = 0.94x + 6. 6 b) y = 6.6x + 0.9 4 c) y = 0.04x + 5. 6 d) y = 5.6x + 0.0 4

Answer: a

  1. Fit a second-degree parabola to the following data.

a) y = - 0.2673x^2 + 3.5232x โ€“ 0.928 6 b) y = 0.2673x^2 + 3.5232x โ€“ 0.928 6 c) y = 0.2673x^2 + 3.5232x + 0.928 6 d) y = - 0.2673x^2 + 3.5232x + 0.928 6

Answer: a

  1. The normal equations for a straight line ๐‘ฆ = ๐‘Ž๐‘ฅ + ๐‘ are:

Answer: a

  1. The normal equations for a second degree parabola y = ax^2 + bx + c are ฮฃy = aฮฃx^2 + bฮฃx + nc, ฮฃxy = aฮฃx^3 + bฮฃx^2 + cฮฃx and ฮฃx^2 y = aฮฃx^4 + bฮฃx^3 + cฮฃx^2 .. Is it true or false? a) True b) False

Answer: a

  1. If the equation y = aebx^ can be written in linear form Y=A + BX, what are Y, X, A, B? a) Y = logy, A = loga, B=b and X=x b) Y = y, A = a, B=b and X=x c) Y = y, A = a, B=logb and X=logx d) Y = logy, A = a, B=logb and X=x Answer: a
  2. If the equation y=axb^ can be written in the linear form Y=A+BX, what are Y, X, A, B? a) Y=logy, A=loga, B=b and X=logx b) Y=y, A=a, B=b and X=x c) Y=y, A=a, B=logb and X=logx d) Y=logy, A=a, B=logb and X=x

Answer: a

Answer : b

  1. To solve the ordinary differential equation

Answer: b

  1. Given

Answer: c

  1. Given that ๐‘‘๐‘ฆ ๐‘‘๐‘ฅ = โˆ’ 2 ๐‘ฅ๐‘ฆ^2 ๐‘ฆ( 0 ) = 1 , โ„Ž = 0. 1 by runge kutta second order method the value of ๐‘˜ 2 for ๐‘ฆ 1 is a) - 0. b) 0. c) 0 d) None

Answer: a

  1. Given that ๐‘‘๐‘ฆ ๐‘‘๐‘ฅ = 3 ๐‘ฅ + 12 ๐‘ฆ ; ๐‘ฆ( 0 ) = 1 ; โ„Ž = 0. 1 by runge kutta second order method the value of ๐‘˜ 2 for ๐‘ฆ 1 is a) - 0. 0825 b) 0. 0825 c) 0 d) None

Answer: b

  1. Given that ๐‘‘๐‘ฆ ๐‘‘๐‘ฅ = 1 + ๐‘ฆ^2 where ๐‘ฆ( 0 ) = 0 then ๐‘ฆ( 0. 2 ) =? a) 0. b) 0. 3 c) 0. d) None

Answer: c

  1. By Runge kutta 4th^ order method 10 ๐‘‘๐‘ฆ ๐‘‘๐‘ฅ = ๐‘ฅ^2 + ๐‘ฆ^2 ; y(0)=1 then ๐‘˜ 3 =? If h=0. a) 0. b) 0. c) 0. d) 0.

Answer: a

  1. By Runge kutta 4th^ order method ๐‘‘๐‘ฆ ๐‘‘๐‘ฅ = ๐‘ฆ^2 โˆ’ ๐‘ฆ ๐‘ฅ; y( 1 )=1 then ๐‘˜ 2 =? If h=0. 2 e) 0. f) 0. 0182 g) 0. h) 0.

Answer: b