Download Numerical Methods: Bisection Method and Newton-Raphson Method Exercises and more Quizzes Numerical Methods in Engineering in PDF only on Docsity!
- Using Bisection method find the root of cos 𝑥 − 𝑥𝑒𝑥^ = 0 with a = 0 and b = 1. a) 0. b) 0. c) 0. d) 0.
Answer: c
- 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
- The Bisection method is also known as ___________________ a) Binary Chopping b) Quaternary Chopping c) Tri region Chopping d) Hex region Chopping
Answer: a
- A function is given by 𝑥 – 𝑒−𝑥^ = 0. Find the root between a = 0 and b = 1 by using Bisection method. a) 0. b) 0. c) 0. d) 0.
Answer: c
- Find the root of 𝑥^4 − 𝑥 − 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
- 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
- If for a real continuous function 𝑓(𝑥), 𝑓(𝑎) ∗ 𝑓(𝑏) < 0 then in the range of [𝑎, 𝑏] 𝑓𝑜𝑟 𝑓(𝑥) = 0 , there is (are) a) one root b) an undeterminable number of roots c) no root d) at least one root
Answer: d
- Rate of convergence of the Newton-Raphson method is generally __________ a) Linear b) Quadratic c) Super-linear d) Cubic
Answer: b
- The Iterative formula for Newton Raphson method is given by __________ a) x 1 = x 0 - f(x 0 )/f’(x 0 ) b) x 0 = x 1 - f(x 0 )/f‘(x 0 ) c) x 0 = x 1 +f(x 0 )/f’(x 0 ) d) x 1 = x 0 +f(x 0 )/f‘(x 0 )
Answer: a
c) non-stationary d) stationary
Answer: d
- 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
- 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
- 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
- The Newton-Raphson method formula for finding the square root of a real number R from the equation 𝑥^2 − 𝑅 = 0 is a) 𝑥𝑖+ 1 = 𝑥 2 𝑖
b) 𝑥𝑖+ 1 = 32 𝑥𝑖 c) 𝑥𝑖+ 1 = 12 (𝑥𝑖 + (^) 𝑥𝑅𝑖 )
d) 𝑥𝑖+ 1 = 12 ( 3 𝑥𝑖 − (^) 𝑥𝑅𝑖 )
Answer: c
- 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
- 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
- Cramer’s Rule fails for ___________ a) Determinant > 0 b) Determinant < 0 c) Determinant = 0 d) Determinant = non-real
Answer: c
- Cramer’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
- Apply Cramer’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
- Apply Cramer’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
- 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
- What are the coefficients of the equation obtained during the elimination called?
a) Joints
b) Pivots c) Calculated coefficients d) Operative coefficients
Solution: b
- Find the values of x, y, z in the following system of equations
by gauss Elimination Method.
2 𝑥 + 𝑦 – 3 𝑧 = − 10 − 2 𝑦 + 𝑧 = − 2 𝑧 = 6
a) 2, 4, 6 b) 2, 7, 6 c) 3, 4, 6 d) 2, 4, 5
Solution: a
- For solving the system of equations 5 x + y + 2 z = 34, 4 y − 3 z = 12, 10 x − 2 y + z = − 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 − 4 Answer: a
- For solving the system of equations (^8) y + 2 z = −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
- For solving the system of equations (^8) y + 2 z = −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
- 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
- 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
- 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
- Using Gauss elimination method, the solution of system of equations
y z
y z
x y z is
Answer: b
- A curve passes through the set of points (0,1), (1, 3), (2, 7), (3, 13) Value of
3
0
y dx by
Trapezoidal rule is given by
A 17 B 15 C 19 D 21
Answer:
46. Value of obtained by evaluating the integral
1 2 0
dx
+ x
, using Trapezoidal rule with
1 2
h = is given by
1 2 0
given: 1 1 4
dx x
(^) +
A 3.15 B 3.1 C 3.2 D3.
- Value of^ log 2 e obtained by evaluating the integral
1
0
dx
+ x
, using Simpson’s^1 3
rd rule
with^1 2
h = is given by
1
0
given: 1 log 2 1 e
dx x
=
A 0.5934 B 0.6560 C 0.6944 D0.
- 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
- 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
- Fit a second-degree parabola to the following data.
a) y = - 0.2673x^2 + 3.5232x – 0.928 6 b) y = 0.2673x^2 + 3.5232x – 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
- The normal equations for a straight line 𝑦 = 𝑎𝑥 + 𝑏 are:
Answer: a
- The normal equations for a second degree parabola y = ax^2 + bx + c are Σy = aΣx^2 + bΣx + nc, Σxy = aΣx^3 + bΣx^2 + cΣx and Σx^2 y = aΣx^4 + bΣx^3 + cΣx^2 .. Is it true or false? a) True b) False
Answer: a
- 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
- 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
- To solve the ordinary differential equation
Answer: b
- Given
Answer: c
- Given that 𝑑𝑦 𝑑𝑥 = − 2 𝑥𝑦^2 𝑦( 0 ) = 1 , ℎ = 0. 1 by runge kutta second order method the value of 𝑘 2 for 𝑦 1 is a) - 0. b) 0. c) 0 d) None
Answer: a
- Given that 𝑑𝑦 𝑑𝑥 = 3 𝑥 + 12 𝑦 ; 𝑦( 0 ) = 1 ; ℎ = 0. 1 by runge kutta second order method the value of 𝑘 2 for 𝑦 1 is a) - 0. 0825 b) 0. 0825 c) 0 d) None
Answer: b
- Given that 𝑑𝑦 𝑑𝑥 = 1 + 𝑦^2 where 𝑦( 0 ) = 0 then 𝑦( 0. 2 ) =? a) 0. b) 0. 3 c) 0. d) None
Answer: c
- By Runge kutta 4th^ order method 10 𝑑𝑦 𝑑𝑥 = 𝑥^2 + 𝑦^2 ; y(0)=1 then 𝑘 3 =? If h=0. a) 0. b) 0. c) 0. d) 0.
Answer: a
- By Runge kutta 4th^ order method 𝑑𝑦 𝑑𝑥 = 𝑦^2 − 𝑦 𝑥; y( 1 )=1 then 𝑘 2 =? If h=0. 2 e) 0. f) 0. 0182 g) 0. h) 0.
Answer: b
