Nmo MCQ QB PDF

Numerical Methods & Optimization [Multiple Choice Question] UNIT No-01
01 Question: Using Bisection method find the root of 3x2 = 5x+2 in the interval [0,3].
Option A 2.25
Option B 2.52
Option C 2
Option D 2.2
Correct Answer A
02 Question: Find the root of xe-x -0.3 = 0 using Bisection Method in the interval [1,5].
Option A 2
Option B 3
Option C 3.1
Option D 2.5
Correct Answer B
03 Question: Use the bisection method three times on the function f(x) = x^2 − sinx − 1 to
determine where f(x) changes sign on the interval − 2 < x < 0.
Option A f(x) changes sign on the interval − 0.75 ≤ x ≤−0.5
Option B f(x) changes sign on the interval − 0.25 ≤ x ≤ 0
Option C f(x) changes sign on the interval − 1 ≤ x ≤−0.75
Option D We cannot use this method as f(x) does not change sign on this
interval.
Correct Answer A
04 Question: For an equation like x^2 = 0 , a root exists at x = 0. The bisection method cannot
be adopted to solve this equation in spite of the root existing at x = 0 because the function
f(x) = x^2
Option A is a polynomial
Option B has repeated roots at x = 0
Option C is always non-negative
Option D has a slope equal to zero at x = 0
Correct Answer C
05 Question: The equation f(x) is given as x3+4x+1=0. Considering the initial

approximation at x=1 then the value of x1 is given as _______________
Option A 1.67
Option B 1.87
Option C 1.86
Option D 1.85
Correct Answer C
06 Question: The equation f(x) is given as x2-4=0. Considering the initial
approximation at x=6 then the value of next approximation correct upto 2 decimal
places is given as __________
Option A 3.33
Option B 1.33
Option C 2.33
Option D 4.33
Correct Answer A
07 Question: The Newton-Raphson method formula for finding the square root of a real
number R from the equation x^2-R=0 is,
Option A
Option B
Option C
Option D
Correct Answer C
08 Question: The function f(x) = 2x^3 − 2x^2 − 3x + 2 has a root between 0 and 1. Which of
the following conditions fail ?
Option A f(0) and f(1) have opposite signs.
Option B f′(x)≠0 on 0 ≤ x ≤ 1.
Option C f″(x) does not change sign on the interval 0 ≤ x ≤ 1.
Option D The tangents at 0 and 1 cut the axes in the interval 0 ≤ x ≤ 1.
Correct Answer C
09 Question:
◄ QUESTIONS ►
The amount of insulin in microunits per mL in a diabetic patient is given by the

function I(t) = 4.5 + 40.5te−0.26t where t is the number of hours since the last
injection was 7.5. If the next insulin injection must be given after the insulin has
peaked and then fallen to 45 microunits per mL, at what time, to the nearest
minute, must the next insulin injection be given ?
Option A Using Newton-Raphson we find the injection must be given after 1
hour and 23 minutes.
Option B Using Newton-Raphson we find the injection must be given after 8
hours exactly.
Option C Using Newton-Raphson we find the injection must be given after 1
hour and 43 minutes.
Option D Using the Gregory-Dary method, with t = ϕ(t) = lnt we find
the injection must given after 7 hours and 16 minutes.
Correct Answer B
10 Question: The ± relative error in the density of a metal rod is determined to
be ± 0.02. If the calculated value of the density is 8.6321947..., how should
the density be reported?
Option A 8.63
Option B 8.6
Option C 8.632
Option D 9
Correct Answer B
11 Question: If a function is defined at 2 points 3 and 7 as f(3)=8 and f(7)=12, it is
sufficient to find the roots through Bisection Method.
Option A True
Option B False
Option C
Option D
Correct Answer B
12 Question: If it is provided that f(3) = 4 is one of the initial points. What can be the
choice of second point for solving by Bisection Method?
Option A -5 such that f(-5) = -26

Option B 0 such that f(0) = 5
Option C -3 such that f(-3) = -2
Option D 13 such that f(13) = 2
Correct Answer C
13 Question: What is the percentage decrease in an interval containing root after
iteration is applied by Bisection Method?
Option A 20%
Option B 30%
Option C 40%
Option D 50%
Correct Answer D
14 Question: For decreasing the number of iterations in Newton Raphson method:
Option A The value of f’(x) must be increased

Option B The value of f’’(x) must be decreased
Option C The value of f’(x) must be decreased
Option D The value of f’’(x) must be increased
Correct Answer A
15 Question: In Newton Raphson method if the curve f f(x) is constant then
__________
Option A f’’(x)=0
Option B f(x)=0
Option C f’(x)=0
Option D f’(x)=c
Correct Answer C
16 Question: Newton-Raphson method will always converge to a solution for f(x) = 0 on
the interval a ≤ x ≤ b if certain conditions are met. Which of the following is not one of
these conditions ?
Option A f is continuous on the interval a ≤ x ≤ b.
Option B f(a) and f(b) have opposite signs.
Option C f″(x) does not change sign on the interval a ≤ x ≤ b.
Option D f′(x) = 0 on the interval a ≤ x ≤ b.
Correct Answer D
17 Question: The order of convergence of Newton-Raphson iterative algorithm is
Option A First order

Option B Second order
Option C Third order
Option D None of the above
Correct Answer B
Multiple Choice Questions (MCQ)
Unit I Root of equation & Error approximation
Bisection Method
1. Suppose we want to find a root of the polynomial x3 - 5x. Using the Bisection method and
starting boundaries a = 2 and b = 4, what is the third approximation to the root obtained by
the algorithm?
A. 2.875 B. 2.125
B. 2.5 C. 3.0
2. Which method has slow convergence?

(a) false poison (b) Secant
(c) Newton-Raphson (d) Bisection
3. One root of the equation x3 + 3x2- 5x + 2 = 0 lies between:

(a) –5 and –4 (b) –4 and –3
(c) 0 and 1 (d) –1 and +1
4.The root of the equation e power x=4x lies between________.
A. (0, 1) B. (1, 2)
C. (2, 3) D. (3, 4)
5. A root of the equation cos(x) - x * exp(x) = 0 , the first initial guess lies between.
A. (0, 1) B. (-1,-2)
C. (-2, 3) D. (3, 4)
Newton-Raphson methods
5.Solve the equation ex− 4x=0 using Newton-Raphson iteration.

A. x=0.61906 and x=1.51213
B. x=0.35 and x=2.1
C. x=0.35740 and x=2.15329
D. Newton-Raphson iteration cannot be used since the answer oscillates between 2 and −2.
6. Use the Newton-Raphson method to solve 2x3−6x2+6x−1=0 to 4 decimal places.

A. There is no solution since the curve is always increasing.
B. x=0.2063.
C. x=0.7351.
D. Newton-Raphson cannot be used because the tangents to the curve do not cut the axes on the
interval 0≤x≤1.
7. Newton-Raphson method will always converge to a solution for f(x) =0 on the

interval a≤x≤b if certain conditions are met. Which of the following is not one of these
conditions?
A. f is continuous on the interval a≤x≤b.

B. f(a) and f(b) have opposite signs.
C. f′′(x) does not change sign on the interval a≤x≤b.
D. f′(x) =0 on the interval a≤x≤b.
8. The function f(x) =2X3 − 2X2− 3X + 2 has a root between 0 and 1. Which of the following
conditions fail?
A. f(0) and f(1) have opposite signs.

B. f′(x)≠0 on 0≤x≤1.
C. f′′(x) does not change sign on the interval 0≤x≤1.
D. The tangents at 0 and 1 cut the axes in the interval 0≤x≤1.
9. The order of convergence of Newton-Raphson iterative algorithm is
A. First order B. Second order

C. Third order D. None of the above.
10 Newton Raphson method of solution of numerical equation is not preferred when
A. The graph of f(x) is nearly horizontal where it crosses the x-axis.

B. The graph of f(x) is nearly vertical where it crosses the x-axis.
C. Both conditions (A) and (B) above prevail.
D. None of the above.
11. The Newton-Raphson method of finding roots of nonlinear equations falls under the category
of _____________ methods.
(A) Bracketing (B) Open

(C) Random (D) Graphical
12. The next iterative value of the root of X2− 4 = 0 using the Newton-Raphson method, if the
initial guess is 3, is
(A) 1.5 (B) 2.067

(C) 2.167 (D) 3.000
13. Newton Raphson method is also called as

A. Method of chords
B. Interval halving method
C. Method of linear interpolation
D. Method of tangents
14. The Iterative formula for Newton-Raphson method is:
A. Xn+1 = f (Xn) C. Xn+1 = Xn –
B. Xn+1 = Xn- 1 – D. Xn+1 = Xn –
15. Which iterative method requires single initial guess root?

A. Bisection method
B. Secant method
C. Method of false position
D. Newton Raphson Method
16. If initial guess root of the equation x3–5x + 3 = 0 is 1, then first approximation for the root by
Newton Raphson method is:
(a) 0.5 (b) 1.5

(c) 1.0 (d) None of the above
17. Newton-Raphson method is applicable the solution of ______.

A. Both algebraic and transcendental equations
B. Both algebraic and transcendental and also used when the roots are complex
C. Algebraic equations only
D. Transcendental equations only
18. Fourth degree equations are also called _______ equations.

A. quadratic B. cubic
C. linear D. bi-quadratic
19. In which of the following methods proper choice of initial value is very important?
A. Newton Raphson Method
B. Bisection Method
C. Iterative Method
D. Regula Falsi Method
20. In the case of Newton-Raphson method the error at any stage is proportional to______.
A. the error in the previous stage
B. the square of the error in the previous stage
C. the cubic of the error in the previous stage
D. square root of the error in the previous stage
21. The root of x3 - 2x - 5 = 0 correct to three decimal places by using Newton-Raphson method
is
A 2.0946 B. 1.0404
C. 1.7321 D. 0.7011
23.The Newton-Raphson method of finding roots of nonlinear equations falls under the category
of _____________ methods.
(A) bracketing
(B) open
(C) random
(D) graphical
of _____________ methods.
(E) bracketing
(F) open
(G) random
(H) graphical
of _____________ methods.
(I) bracketing
(J) open
(K) random
(L) graphical
26.The Newton-Raphson method formula for finding the square root of a real number R from
the equation x 2 − R = 0 is,
xi
(A) xi +1 =
2
3 xi
(B) xi +1 =
2
1 R
(C) xi +1 =  xi + 
2 xi 
1 R
(D) xi +1 =  3 xi − 
2 xi 
27.The next iterative value of the root of x 2 − 4 = 0 using the Newton-Raphson method, if the
initial guess is 3, is
(A) 1.5
(B) 2.067
(C) 2.167
(D) 3.000
28.The root of the equation f ( x) = 0 is found by using the Newton-Raphson method. The
initial estimate of the root is x0 = 3 , f (3) = 5 . The angle the line tangent to the function f (x)
makes at x = 3 is 57° with respect to the x-axis. The next estimate of the root, x1 most nearly
is
(A) –3.2470
(B) −0.2470
(C) 3.2470
(D) 6.2470
29.The root of x 3 = 4 is found by using the Newton-Raphson method. The successive iterative
values of the root are given in the table below.
Iteration
Value of Root
Number
0 2.0000
1 1.6667
2 1.5911
3 1.5874
4 1.5874
The iteration number at which I would first trust at least two significant digits in the
answer is
(A) 1
(B) 2
(C) 3
(D) 4
30.The ideal gas law is given by

pv = RT
where p is the pressure, v is the specific volume, R is the universal gas constant,
and T is the absolute temperature. This equation is only accurate for a limited range
of pressure and temperature. Vander Waals came up with an equation that was
accurate for larger ranges of pressure and temperature given by
 a
 p + 2 (v − b ) = RT
 v 
where a and b are empirical constants dependent on a particular gas. Given the value of
R = 0.08 , a = 3.592 , b = 0.04267 , p = 10 and T = 300 (assume all units are
consistent), one is going to find the specific volume, v , for the above values. Without
finding the solution from the Vander Waals equation, what would be a good initial guess
for v ?
(A) 0
(B) 1.2
(C) 2.4
(D) 3.6
31 f(a) < 0, f(b) > 0 and if x0∈ (a, b)is first approximation with f(x0) < 0 then in bisection
method,
(a) x0 is to be replaced by a (b) ais to be replaced by x0
(c) bis to be replaced by x0 (d) x0 is to be replaced by b
32 For real root of an equation x3 – 2x – 5 = 0, the root lies between
(a) 0 and 1 (b) 2 and 3 (c) 1 and 2 (d)none of them
33 From the following _______ method is not iterative method.
(a) False position (b) Bisection (c) Lagranges (d)none of them

34 For the function f(x): x3 – 2x – 5 = 0 if the root of equation lies between (2, 3) and if at ith
iteration c= 2.5 then next approximation by bisection method gives c =
3+2.75 2 + 2.5 3+2.5

(a) (b) (c) (d) none of them
2 2 2
35 If in a method of successive approximation, the root of equation lies between 1 and 2,

1
g ( x) = , and initial guess is 1.25 then next approximation is
x −1
2
(a) 0.5625 (b) 1.2177 (c) 1.7777 (d)none of them
36 From the following _______ method is the best method to obtain root of equation f(x) = 0.
(a) False position (b) Bisection (c) Newton’s Raphson (d)none of them
37 Absolute error is defined as
(a) Present Approximation – Previous Approximation
(b) True Value – Approximate Value
(c) abs (True Value – Approximate Value)
(d) abs (Present Approximation – Previous Approximation)
38 The number 0.01850 x 103 has ________ significant digits
(a) 3 (b) 4 (c) 5 (d) 6
39 For an equation like x2 = 0, a root exists at x = 0. The bisection method cannot be adopted
to solve this equation in spite of the root existing at x = 0 because the function f(x) =x2
(a) is a polynomial (b) has repeated roots at x= 0
(c) is always non-negative (d) has a slope equal to zero at x= 0

40 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

Name of Content:
01 Question: Find the values of x, y, z in the following system of equations by gauss

Elimination Method. 2x + y – 3z = -10
-2y + z = -2
z=6
Option A 2, 4, 6
Option B 2, 7, 6
Option C 3, 4, 6
Option D 2, 4, 5
Correct Answer A
02 Question: Solve the given system of equation by Gauss Elimination method.
3x + 4y – z = -6
-2y + 10z = -8
4y – 2z = -2
Option A (-2, -1, -1)

Option B (-1, -2, -1)
Option C (-1, -1, -2)
Option D (-1, -1, -1)
Correct Answer D
03 Question: The following system of equation has:
x–y–z=4
2x – 2y – 2z = 8
5x – 5y – 5z = 20
Option A Unique Solution

Option B No solution
Option C Infinitely many Solutions
Option D Finite solutions
Correct Answer C
04 Question: Solve this system of equations and comment on the nature of the
solution using Gauss Elimination method.
x+y+z=0
-x – y + 3z = 3
-x – y – z = 2
Option A Unique Solution

Option B No solution
Option C Infinitely many Solutions
Option D Finite solutions
Correct Answer B
05 Question: While solving by Gauss Seidal method, which of the following is the first
Iterative solution system; x – 2y = 1 and x + 4y = 4?
Option A (1, 0.75)

Option B (0.25,1)
Option C (0,0)
Option D (1,0.65)
Correct Answer A
06 Question: Solve the following equations by Gauss seidal method.[1 Itr.]
10a - 2b - c - d = 3
- 2a + 10b - c - d = 15
- a - b + 10c - 2d = 27
- a - b - 2c = 10d = -9
Option A a = 0.3, b = 1.56, c = 2.886, d = -0.1368

Option B a = 0.8869, b = 1.9523, c = 2.886, d = -0.1368
Option C a = 0.3, b = 1.56, c = 2.9566, d = -0.1368
Option D a = 0.986, b = 1.9899, c = 2.886, d = -0.1368
Correct Answer A
07 Question: Solve the system of equations by Jacobi’s iteration method.
20x + y – 2z = 17
3x + 20y – z = -18
2x – 3y + 20z = 25
Option A x = 0.85, y = -0.9, z = 1.25
Option B x = 1.02, y = -0.965, z = 1.25
Option C x = 0.85, y = -1.0015, z = 1.00325
Option D x = 0.85, y = -0.9, z = 0.999992
Correct Answer A
08 Question: choose correct hierarchy for reduce set of simultaneous equation in upper
triangular form
i. Make all elements in first column below first row to zero
ii. Similarly make all elements in second column below second row to zero
iii. Write given set of equations in matrix form
iv. Using same procedure make all elements below the diagonal elements to zero
Option A iii, i, ii, iv
Option B i, iv, ii, iii
Option C i, ii, iii, iv
Option D iv, ii, iii, i
Correct Answer A
9 Question: The aim of elimination steps in Gauss elimination method is to reduce
the coefficient matrix to ____________
Option A diagonal
Option B identity
Option C lower triangular
Option D upper triangular
Correct Answer D
10 Question: The Elimination process in Gauss Elimination method is also known as
_____________
Option A Forward Elimination

Option B Backward Elimination
Option C Sideways Elimination
Option D Crossways Elimination
Correct Answer A
11 Question: The reduced form of the Matrix in Gauss Elimination method is also
called ____________
Option A Column Echelon Form

Option B Row-Column Echelon Form
Option C Column-Row Echelon Form
Option D Row Echelon Form
Correct Answer D
12 Question: Gauss seidal method is similar to which of the following methods?
Option A Iteration method

Option B Newton Raphson method
Option C Jacobi’s method
Option D Regula-Falsi method
Correct Answer C
13 Question: What is the main difference between Jacobi’s and Gauss-seidal?
Option A Computations in Jacobi’s can be done in parallel but not in

Gauss-seidal
Option B Convergence in Jacobi’s method is faster
Option C Gauss seidal cannot solve the system of linear equations in
three variables whereas Jacobi cannot
Option D Deviation from the correct answer is more in gauss seidal
Correct Answer A
14 Question: The Gauss-Seidel method is applicable to strictly diagonally dominant
or symmetric________ definite matrices.
Option A Positive
Option B Negative
Option C Zero
Option D Equal
Correct Answer A
15 Question: Gauss seidal requires less number of iterations than Jacobi’s method.
Option A True
Option B False
Option C
Option D
Correct Answer A
16 Question: Which of the following method is employed for solving the system of
linear equations?
Option A Runge Kutta

Option B Newton Raphson
Option C Gauss Seidal
Option D Simpson’s Rule
Correct Answer C
17 Question: What is the limitation of Gauss-seidal method?
Option A It cannot be used for the matrices with non-zero diagonal

elements
Option B It is more complex than Jacobi’s method
Option C It doesn’t guarantees convergence for each and every matrix
Option D It is an iterative technique
Correct Answer C
18 Question: How the transformation of coefficient matrix A to upper triangular matrix
is done?
Option A Elementary row transformations

Option B Elementary column transformations
Option C Successive multiplication
Option D Successive division
Correct Answer A
19 Question: The modified procedure of complete pivoting is called as ____________
Option A Partial
Option B Additional
Option C Reduced
Option D Modified
Correct Answer A
20 Question: Compared to Gauss-Seidel method, Newton-Raphson method takes
Option A Less number of iterations and more time per iteration

Option B Less number of iterations and less time per iteration
Option C More number of iterations and more time per iteration
Option D More number of iterations and less time per iteration
Correct Answer A
21 Question: Which of the following is an assumption of Jacobi’s method?
Option A The coefficient matrix has no zeros on its main diagonal

Option B The rate of convergence is quite slow compared with other
methods
Option C Iteration involved in Jacobi’s method converges
Option D The coefficient matrix has zeroes on its main diagonal
Correct Answer A
2. SIMULTANEOUS EQUATION
2.1 what are different methods to solve simultaneous equations
A. Direct method b. Iterative method

C. All of above d. None of above
2.2 the gauss elimination procedure is one of the several methods to solve
A. Inverse of matrix b. Determinant matrix

C. Procedure matrix d. Eliminated matrix
2.3 apply gauss elimination method to solve the equations as x+3Y+3Z=16, x+4Y+3Z=18, x+3Y+4Z=19
A. X=2, y=1, z=1 b. X=1, y=2, z=3

C. X=1, y=1, z=1 d. X=3, y=2, z=3
2.4 choose correct hierarchy for reduce set of simultaneous equation in upper triangular form
i. Make all elements in first column below first row to zero

ii. Similarly make all elements in second column below second row to zero
iii. Write given set of equations in matrix form
iv. Using same procedure make all elements below the diagonal elements to zero
A. Iii, i, ii, iv b. I, iv, ii, iii
C. I, ii, iii, iv d. Iv, ii, iii, i
2.5 converting square matrix into upper triangular matrix is called
A. Forward substitution method b. Inverse substitution method

C. Backward substitution method d. None of above
2.6 write following matrix in upper triangular matrix form
0 1 2 𝑥1 5
[ 1 2 4 ] [𝑥2] = [ 11 ]
−3 1 −5 𝑥3 −12
0 1 2 𝑥1 5 1 2 4 𝑥1 11
A. [ 1 2 4 ] [𝑥2] = [ 11 ] B. [0 1 2] [𝑥2] = [ 5 ]
−3 1 −5 𝑥3 −12 0 0 0 𝑥3 −14
1 2 4 𝑥1 11 1 1 2 𝑥1 5
C. [0 1 2] [𝑥2] = [ 5 ] D. [0 1 4] [𝑥2] = [ 11 ]
0 0 7 𝑥3 21 0 0 1 𝑥3 −12
1|P ag e
0 1 2
2.7 identify diagonal [ 1 2 4 ]
−3 1 −5
A. Central diagonal, super diagonal, sub diagonal b. Central diagonal, super diagonal, sub
diagonal
C. Central diagonal, super diagonal, sub diagonal d. Central diagonal, super diagonal, sub
diagonal
2.8 using gauss seidel iteration method solve the following equations (2 itr only)
4X1 + 2X3 = 4, 5X2 + 4X2 +10X3 =2
A. X1=1.2, x2=0.6, x3=6.96 b. X1=2.38, x2=0.6, x3=16
C. X1=-2.48, x2=0.12, x3=6.96 d. X1=0.4, x2=0.6, x3=1.2
2.9 Jacobi iteration method is _____substitution method
A. Indirect b. Direct
C. None of above d. Backward
2.10 Jacobi iteration is similar to ______
A. Gauss elimination method b. Tridiagonal method
C. Thomas algorithm method d. Gauss seidel method
2|P ag e
Unit II MCQ
Simultaneous Equation
Gauss Elimination Method, Partial pivoting, Gauss-Seidal method and Thomas algorithm for Tridiagonal
Matrix jacobi
1. Solving following simultaneous equations, 4x - 5y = 17 and x - 5y = 8, we get

2. x = 3, y = -1
3. x = 2, y = 3
4. x = 4, y = 1
5. x = 5, y = 4
Answer A
2. Solve for value of x and y if 5x - y = 5 and 3x + 2y = 29
1. x = 12, y = 3
2. x = 1, y = 4
3. x = -3, y = 24
4. x = 3, y = 10
Answer D
3. Solve simultaneous equations 13x - 6y = 20, 7x + 4y = 18
1. x = 2, y = 1
2. x = 4, y = 8
3. x = 6, y = 1
4. x = 2, y = 4
Answer A
5. Jacobi’s method is also known as

a. Displacement method
b. Simultaneous displacement method
c. Simultaneous method
d. Diagonal method
(Ans:b)
6. In the Gauss elimination method for solving a system of linear algebraic equations, triangularzation
leads to
a. Diagonal matrix
b. Lower triangular matrix
c. Upper triangular matrix
d. Singular matrix
(Ans:c)
7. The goal of forward elimination steps in the Naïve Gauss elimination method is to reduce the
coefficient matrix to a (an) _____________ matrix.
1. Diagonal
2. Identity
3. Lower triangular
4. Upper triangular
8. The following data is given for the velocity of the rocket as a function of time. To find the velocity
at t=21 s, you are asked to use a quadratic polynomial, v(t)=at2+bt+c to approximate the velocity
profile.
t (s) 0 14 15 20 30 35
v(t) m/s 0 227.04 362.78 517.35 602.97 901.67
The correct set of equations that will find a, b and c are
1.
2.
3.
4.
Ans 4
9. Using a computer with four significant digits with chopping, Gauss elimination with partial
pivoting solution to
1. x1 = 26.66; x2 = 1.051
2. x1 = 8.769; x2 = 1.051
3. x1 = 8.800; x2 = 1.000
4. x1 = 8.771; x2 = 1.052
10. Using [x1 x2 x3] = [1 3 5] as the initial guess, the value of [x1 x2 x3] after three iterations in
Gauss-Seidel method for
12 7 3 𝑥𝑥1 2
�1 5 1 � 𝑥𝑥2 = −5
2 7 −11 𝑥𝑥3 6
(A) [-2.8333 -1.4333 -1.9727]

(B) [1.4959 -0.90464 -0.84914]
(C) [0.90666 -1.0115 -1.0242]
(D) [1.2148 -0.72060 -0.82451]
12. Consider the following system of equations
2x1 +x2+x3= 0
x2-x3= 0
x1+x2= 0
This system has
(A) A unique solution
(B) No solution
(C) Infinite number of solutions
(D) Five solutions
Answer: - (C)
UNIT-3
OPTIMIZATION
1. The Maximization or minimization of a quantity is the

a. goal of management science.
b. decision for decision analysis.
c. constraint of operation research.
d. objective of linear programming.
2. Decision variables
a. Represent quantities or product to be manufactured.
b. Represent the values of constraint.
c. Measure the objective function.
d. Must exist for each constraint.
3. When a set of decision variable satisfies all given constraints and non negative restriction
then the solution is called as.
a. Non feasible solution.
b. Feasible solution.
c. Optimal solution.
d. Linear solution.
4. Which variable is added or substracted from inequality constraint to convert to equality
constraint.
a. Artifical variable.
b. slack variable.
c. linear variable.
d. unknown variable.
5. simplex method used to obtain optimum solution is also called as.

a. equality method.
b. Iterative method.
c. graphical method.
d. inequality method.
6. the important condition for graphical method is that it is used to solve problems which involve
a. two unknown or decision variables only.
b. one unknown only.
c. decision variables only.
d. two known or decision variables only.
7. which region refers to area containing all possible solution to the problem.
a. non-Feasible region.
b. feasible region.
c. positive region.
d. negative region.
8. genetic algorithms are also known as.
a. linear algorithms.
b. evolutionary algorithms.
c. simulated algorithms.
d. selection alogrithms.
9. which algorithm is applied to solve optimization problems that does not use any information
gathered during the search.
a. genetic algorithms.
b. evolutionary algorithms.
c. stochastic algorithms.
d. metropolis algorithms.
10. basic requirements of the linear programming problem
a. well defined objective function.
b. limited resources.
c. decision variables.
d. all of the above.
MCQ Unit-III
Optimization
1 Constraints may represent
a) Limitation
b) Requirements
c) Balance conditions
d) All of the above
Solution: d
2 The feasible region of LLP problem is
a) concave
b) convex
c) concave & convex
d) none
Solution: b
3 Distinguishing features of an LP is
a) problem has an objective function & constraints
b) all function in problem are linear
c) optimal values for the decision variables are produced
d) all of the above
Solution: d
Which of the following are not major requirements of a linear programming
4
problem?
a) there must be alternative courses of action among which to decide
b) an objective for the firm must exist
c) the problem must be of the maximization type
d) resources must be limited
Solution: c
Maximize Z=3x1+2x2 subject to

5 4x1+x2≤60, 8x1+x2≤90, 2x1+5x2≤80
X1 ,x2 ≥0 number of corner point feasible solution for above LP model are
a) 3
b) 4
c) 5
d) 0
Solution: b
6 Simplex method has property that
a) at each iteration it gives solution which is at least as good as the earlier
solution
b) at each stage it produces feasible solution
c) it signals that optimal solution has been found
d) none of the above
Solution: a
MCQ Unit-III
Optimization
7 Which of the following is not true of the simlex method
a) at each iteration, the objective valve either says the same or improves
b) it indicates an unbounded or infeasible problem
c) it signals optimality
d) it converges in at most m steps, where m is the number of constraints

Solution: b
8 Artificial variables
a) are used to aid in finding an initial solution
b) are used phase 1 of two method
c) can be used t find optimal dual prices in the final tableau
d) all of the above
Solution: d
Common Data for Question 9 to 11
basic x1 x2 s1 s2 s3 RHS
Z 0 0 0 2 0 48
s1 0 (5/3) 1 (-2/3) 0 14
s3 0 (-1/3) 0 (1/3) 1 5
x1 1 (2/3) 0 (1/3) 0 8
9 the table conclude that
a) solution infeasible
b) solution degenerate
c) unbounded solution exists
d) alternative optimum exists
Solution: d
10 in the above problem
a) S1 leaves& X2 enters basis
b) S3 leaves& X1 enters basis
c) X1 leaves& X2 enters basis
d) X1 leaves& X2 enters basis
Solution: a
11 alternate solution values for the above problem are
a) X1=(12/5) ; X2=(42/5) ; S3=(39/5)
b) X1=14 ; X2=5 ; S3=8
c) X2=8 ; S1=14 ; S3=5
d) X1=(42/5) ; X2= (12/5) S3=0
Solution: a
12 The primal is max model in m equality constraints & n non-negative variable. The
dual has
MCQ Unit-III
Optimization
a) n constraints & m non-negative variables
b) is a min model
c) both a & b
d) none of these
Solution: c
13
when the primal problem is non-optimal , the dual problem is automatically
a) Infeasible
b) Unbounded
c) both a & b
d) none
Solution: c
14 at the optimal solution of maximization problem, the optimal profit must equal to
the worth of
a) used resources
b) unused resources
c) none of the above
d) both a & b
Solution: a
Minimize Z=10x1+x2 +5x3
15 subject to 5x1-7x2+3x3≤50,
X1 ,x2 , x3 ≥0 optimal value of primal is
a) (50/3)
b) (10/3)
c) (250/3)
d) (100/3)
Solution: c
Common Data for Question 16 to 23
Maximize Z=5x1+10x2 +8x3
3x1+5x2+2x3≤60 MATERIAL
4x1+4x2+4x3≤72 M/C HOURS
2x1+4x2+5x3≤100 LABOR
basic x1 x2 X3 S1 s2 s3 RHS
Z (11/3) 0 0 (2/3) (5/3) 0 160
X2 (1/3) 1 0 (1/3) (-1/6) 0 8
X3 (2/3) 0 1 (-1/3) (5/12) 0 10

(-
S3 (-8/3) 0 0 (1/3) 1 18
17/12)
16 if m/c are increased by one unit then the contribution

MCQ Unit-III
Optimization
a) decrease by (11/3)
b) increase by (11/3)
c) increase by (5/3)
d) increase by (2/3)
Solution: c
17 If material increase by 3kgs the increase in contribution is
a) RS 2/-
b) Rs 2/3
c) RS 6/-
d) Rs 8/-
Solution: a
18 if machine hours are decreased by 12 hours the new profit is
a) 140
b) 160
c) 180
d) 200
Solution: a
19 if m/c hours are decreased by 12 hours then the new production of B is
a) 8
b) 10
c) 6
d) 5
Solution: b
20 If material increased by 3kgs then the new production of C is
a) 10
b) 11
c) 9
d) 8
Solution: c
21 For every unit of A produced the decrease in contribution
a) (11/3)
b) (3/11)
c) (2/3)
d) (5/3)
Solution: a
22 if 6 unit of A are to be produced then the new profit is
a) 138
b) 182
c) 160
d) 150
Solution: a
23 if units of A are to be produced the production of B & C
a) decreases by 1 & 2 units
MCQ Unit-III
Optimization
a) increases by 1 & 2 units
a) increases by 1 & decreases by 2
d) decreases by 2 & increases by 1
Solution:a
24
Name of Content: ODE
01 ……………..method is the one of the earliest analytical-numerical algorithms for

approximate solution of initial value problem for ordinary differential equation.
Option A Runge-Kutta fourth Order Method
Option B Taylor’s series
Option C Euler Method
Option D Runge-Kutta second Order Method
Correct Answer B
02 When a differential equation contains all the derivative with respect to single variable,
then it’s called as
Option A An ordinary differential equation

Option B Partial Differential Equation
Option C Numerical Method
Option D Roots of Equation
Correct Answer A
03 In Runge –Kutta fourth Order Method K4=…….
Option A hf(x1+h, y1+k3)

Option B hf(x1+h, y1+k2)
Option C hf(x1+h, y1+k1)
Option D f(x1+h, y1+k3)
Correct Answer A
04 Taylor’s series method is the….. ….for ordinary differential equation.
Option A Boundary value problem

Option B Initial value problem
Option C Valued Problem
Option D None of these
Correct Answer B
05 In which of the following method approximate the curve of solution by the tangent in
each interval.
Option A Trapezoidal Method
Option B Euler Method
Option C Newton’s Method
Option D Runge Kutta Method
Correct Answer B
ODE:- Q.1 to Q.19-1 Marks, Q.20 to Q.37-2 Marks, Q.38 to Q.55 -3 Marks
PDE:- Q.1 to Q.18= Marks, Q.19 to Q.29- 2 Marks, Q.30 to Q.45- 3 Marks
06 …….takes a weighted average of the slopes at more number of points than the……order
R-K method, so it is a little more expensive, but more accurate.
Option A R-K 4th order, lower order

Option B R-K 4th order, higher order
Option C R-K 2nd order, lower order
Option D R-K 2nd order, higher order
Correct Answer A
07 Local truncation error Euler’s method is……
Option A h2
Option B h4
Option C h3
Option D h5
Correct Answer A
08 Local truncation error R-K 4th order method is……
Option A h2
Option B h5
Option C h2
Option D h3
Correct Answer B
09 In Runge –Kutta second Order Method K2=…….
Option A f(x0+h, y0+k)
Option B f(x0+h, y0+k1)
Option C hf(x0+h, y0+k1)
Option D hf(x0+h, y0+k2)
Correct Answer C
10 For small h, error is bound to be quite significant also method is very slow, this
drawback is related to…….
Option A Trapezoidal Method
Option B Euler Method
Option C Newton’s Method
Option D Runge Kutta Method
Correct Answer B
11 The first two steps of the fourth order Runge Kutta method finds the value at which
point?
Option A At the (n+0.5)th point

Option B At the (n+1)th point
Option C At the (n-1)th point
Option D At the nth point
Correct Answer A
12 How many steps does the fourth-order Runge Kutta method use?
Option A Two steps

Option B Five steps
Option C Four steps
Option D Three steps
Correct Answer C
13 The first two steps of the fourth-order Runge Kutta method use………….
Option A Euler method
Option B Forward Euler method
Option C Backward Euler method
Option D Explicit Euler method
Correct Answer A
14 Consider an nth order accurate Runge Kutta method. How many times is the derivative
evaluated at the fourth time-step?
Option A One times
Option B Two times
Option C Four times
Option D n times
Correct Answer D
15 General formula for n iteration become in Runge-Kutta second order method is……..
Option A x0= xn+1, y0= yn+1
Option B x0= xn+1, y0= yn-1
Option C x0= xn-1, y0= yn-1
Option D x0= xn-1, y0= yn+1
Correct Answer C
16 Which of these statements is correct?
Option A When the order of accuracy is the same for two methods, the
accuracy is also the same
Option B Runge Kutta method interpolate at more than one point in a time
interval
Option C Runge Kutta method is not a multipoint method
Option D An nth order Runge-Kutta method is more accurate than the nth
order multipoint method
Correct Answer D
17 How many steps does the second-order Runge Kutta method use?
Option A Two steps

Option B Five steps
Option C Four steps
Option D Three steps
Correct Answer A
18 What is weighted mean value of fourth order R-K method
Option A k=1/6(k1+ 2k2- 2k3 +k4)
Option B k=1/6(k1+ 2k2+ 2k3 +3k4)
Option C k=1/6(k1+ 2k2+ 2k3 +k4)
Option D k=1/2(k1+ 2k2+ 2k3 +k4)
Correct Answer C
19 The weighted mean of second order R-K method k=
Option A =1/2(k1-k2)
Option B =2(k1+k2)
Option C =1/2(k1+ k2)
Option D =(k1+2k2)
Correct Answer C
20 Given y'=x-y2 with h=1 and y(0)=1,find y(1) by Euler’s Method
Option A 0
Option B 1
Option C 0.5
Option D 1.5
Correct Answer A
21 Given y'=x+2y with h=0.1 and x=1, y=1 find y at x=1.1 by using Eulers Method
Option A 1.1
Option B 1.2
Option C 1.3
Option D 1.4
Correct Answer C
22 Using Euler’s method find y(0.2) from dy/dx=x+y, y(0)=1, with h=0.2
Option A 1.2
Option B 1.4
Option C 1.5
Option D 0.5
Correct Answer A
23 Given y’=x+y, y(0)=1 find y(0.1) by Euler’s method. Take h=0.1
Option A 1.3
Option B 1.1
Option C 1.5
Option D 0.5
Correct Answer B
24 Given y'=√𝑥 + 𝑦 with h=0.2 and y(1)=2.2 ,find y(1.2) by Euler’s Method
Correct Answer 2.558
25 Find y(x) if y’=yx2-1.1y, with interval x= 0 to 1, h=1, by applying Euler’s Method where
y(0)=1.
Option A 0.70
Option B 0.9
Option C 0.75
Option D 1
Correct Answer B
26 Find y(0.1) by Euler’s method, given that dy/dx=1-y, y(0)=0 with h=0.1.
Option A 0.0
Option B 0.5
Option C 0.2
Option D 0.1
Correct Answer D
27 dy y2 −x2
Using 2nd order Runge Kutta method solve dx= y2 +x2 with y(0) = 1.0 at h=0.2 and x=0.2,
calculate k1 value.
Option A 0.1
Option B 0.15
Option C 0.2
Option D 0.25
Correct Answer C
28 𝑑𝑦 𝑦 2 −𝑥 2
Using 2nd order Runge Kutta method solve 𝑑𝑥 = 𝑦 2 +𝑥 2 with y(0) = 1.0 at h=0.2 and x=0.2,
calculate k2 value.

29 Given dy/dx+y+xy2=0, y(0)=1. Find k1 when y(0.1) with step size h=0.1 using 2nd order
R-K method.
Option A -0.1
Option B 0.1
Option C -0.2
Option D 0.2
Correct Answer A
30 Find k1 and k2 when y’+xy=2 for y(5)=2 and y(5.1) with h=0.1, K1= -0.8 an k2 =-
0.412using 2nd order R-K method.
Option A k1= -0.8 and k2 = 0.412
Option B k1= 0.8 and k2 = -0.412
Option C k1= -0.8 and k2 =-0.412
Option D k1= 0.8 and k2 =0.412
Correct Answer C
31 Solve differential equation for K1 and l1. x=0.3 R-K 4th order with initial value x=y=0,
dy dz
z=1 dx = (1+xy) and dx = -xy. Take h=0.3.
Option A k1 = 0.3 and l1= 0.1

Option B k1 = 0.3 and l1= 0
Option C k1 = -0.3 and l1= -0.1
Option D k1 = -0.3 and l1= 0
Correct Answer B
32 Find k1 value for given that

𝑑𝑦
=
1
, y(0)=1, for value y(0.5) with step size is 0.5.
𝑑𝑥 𝑥+𝑦
Correct Answer 0.5

33 dy
Find y(0.1) if dx=1+y, y(0)=1 using Taylor Series method. Take step size value is 0.1
Option A 1.2103
Option B 1.5102
Option C 1.4133
Option D 1.001
Correct Answer A
34 Given y’=x+y, y(0)=1. Find y(0)=1 by Taylor series method. Take h=0.01
35 Solve by Taylor series method, y’= xy+y2, y(0)=1 at x=0.1, correct to three decimal
places.
Option A 2.1167
Option B 2.1169
Option C 1.1169
Option D 0.9033
Correct Answer C
36 Calculate up to first iteration dy/dx=-2x3+12x2-20x + 8.5 from x=0 to x=4with step size
of 0.5, the initial condition at x =0 is y = 1. Estimate using Euler’s method.
Option A 5.0
Option B 5.25
Option C 5.5
Option D 5.35
Correct Answer B
37 dy x+y dz
Solve following pair of differential equations dx = z and dx = xy+ z with initial
conditions x0 = 5, y0 = 1.5, z0 = 1 for x = 0.6. Calculate k1 value.
Option A 0.2
Option B 0.3
Option C 0.25
Option D 0.35
Correct Answer A
38 Given y'=x-y2 with h=1 and y(0)=1,find y(2) by Euler’s Method
Option A 0
Option B 1
Option C 2
Option D 1.5
Correct Answer C
39 Given y'=x+2y with h=0.1 and x=1, y=1 find y at x=1.2 by using Eulers Method
Option A 1.5
Option B 1.57
Option C 1.67
Option D 2.12
Correct Answer C
40 dy
Solve dx = log10 (x+y), y(0) =2 by Euler’s method by choosing h=0.2, find y(0.2) and y
(0.4)
Option A 2 and 2.1310
Option B 2.0 and 3.1310
Option C 1.9 and 2.005
Option D 2.0602 and 2.1310
Correct Answer D
41 dy 𝑦2
Solve dx = - 1+𝑥, y(0) =1 by Euler’s method by choosing h=0.1, find y(0.2).

42 Given y'=√𝑥 + 𝑦 with h=0.1 and y(1)=2.2 ,find y(1.2) by Euler’s Method
43 Solve dy/dx = x+y , y(0)=1 Estimate y(1) with h=0.5 by using Euler’s formula method
Option A 3.5
Option B 1.5
Option C 2.5
Option D 0.5
Correct Answer C
44 Using Euler’s method find the solution of the initial value problem y’=y-x2+1, y(0)=0.5
at x= 0.2 and h=0.2
Option A 0.5
Option B 0.3
Option C 0.7
Option D 0.8
Correct Answer D
45 dy
=
x2
given at x=0, y=1.2 find y(0.4) with h=0.4 by R-K second order method.
dx 2y
Option A 1.2133
Option B 1.3541
Option C 0.9034
Option D 1.4891
Correct Answer A
46 Apply Runge Kutta 4th order method to find an approximate value of for x = 0.1 in steps
𝑑𝑦
size is 0.1 if 𝑑𝑥 = x+y2, y(0)=1, correct to four decimal places.
Option A 1.1165
Option B 2.1165
Option C -1.0165
Option D -2.1165
Correct Answer A
47 Using 2nd order Runge-Kutta method solve dy/dx= (y2-x2)/ (y2+x2) with y(0) = 1.0 at
h=0.2 and find y at x=0.2.
48 Given dy/dx+y+xy2=0, y(0)=1. Find y(0.1) with step size h=0.1 using 2nd order R-K
method up to 1st iteration.
49 Use Runge-Kutta method of fourth order to obtained an approximation to y (1.5) for the
𝑑𝑦
solution of𝑑𝑥 = 2xy; y(1)=1 calculate k1 & k2 correct to four decimal places. Take h=0.4
Option A k1= 1 & k2= -1.875

Option B k1= 0 & k2= 1.875
Option C k1= 1 & k2= 1.875
Option D k1= -1 & k2= 1.875
Correct Answer C
50 Using 4th order Runge-Kutta method solve y’ = -y with y(0) = 1.0 at h=0.1 and find y at
x=0.1.
51 Find y(1.1) if y’= x+y, y(1)=0 and h=0.1 by Taylor series.
Option A 0.151
Option B 0.1103
Option C 0.901
Option D 0.16103
Correct Answer B
52 𝑑𝑦
Use Taylor series method 𝑑𝑥 = x2y & y(1)=1, h=0.1 for find y(1.1).
Option A 1
Option B 1.311
Option C 1.016
Option D 1.445
Correct Answer B
53 Define the solution of
dy
= 3x+ y2 , using taylor series method. Given y(0) = 1. Determine
dx
y(0.1)
54 Temperature at one surface of slab of thickness, x=20cm is T = 5000C. Find the
temperature of other surface of slab by taking step size in thickness.
Option A 476.660C
Option B 480.660C
Option C 478.660C
Option D 486.660C
Correct Answer D
55 dy
Using Runge Kutta 4th order method solve - y=0.given y(0) =2, h= 0.1, find k1 & k2
dx
when y(0.1).
Option A k1= -0.2 and k2 = 0.21
Option B k1= 0.2 and k2 = -0.21
Option C k1= 0.2 and k2 =-0.21
Option D k1= 0.2 and k2 = -0.2
Correct Answer C
Name of Content: PDE
01 A partial differential equation requires:
Option A Exactly one independent variable

Option B More than one dependent variable
Option C Two or more independent variable
Option D Equal numbers of dependent variable
Correct Answer C
02 Consider a function u which depends on position x and time t. the partial differential
∂u ∂2 u
equation = is known as the
∂t ∂x2
Option A Wave equation

Option B Heat equation
Option C Laplace equation
Option D Elasticity equation
Correct Answer B
03 Which of these does not come under partial difference equations….
Option A Laplace equation

Option B Equation of motion
Option C 1-D wave equation
Option D Heat equation
Correct Answer B
04 Laplace 2D heat flow method is the…..…….for partial differential equation.
Option A Boundary value problem

Option B Initial value problem
Option C Valued Problem
Option D None of these
Correct Answer A
05 Parabolic equation is also referred as ……………….. Heat equation.
Option A 5 Dimensional
Option B 3 Dimensional
Option C 2 Dimensional
Option D 1 Dimensional
Correct Answer D
06 The Laplace equation comes under the category of ………………differential equation.
Option A Explicate
Option B Elliptical
Option C hyperbolic
Option D Ordinary differential equation.
Correct Answer B
07 The Poisson’s equation comes under the category of elliptical differential equation. The
partial differential equation given form as
Option A ∂u ∂2 u
+ ∂x2 = f (x,y)
∂t
Option B ∂u
=
∂2 u
∂t ∂x2
Option C ∂2 u
+
∂2 u
=f(x,y)
∂x2 ∂x2

Correct Answer C
08 ………..scheme called an implicit scheme because the solution value at any point (i,j+1)
on the (j+1)th level of neighbouring dependent point.

Option B R-K second order
Option C R-K fourth order
Option D Crank Nicolson’s
Correct Answer D
09 Find the order of the continuity equation for steady two- dimensional flow.
Option A 1
Option B 0
Option C 2
Option D 3
Correct Answer A
10 These are essential for solving partial differential equations.
Option A Algebraic equation
Option B Physical principle
Option C Mathematical model
Option D Boundary condition
Correct Answer D
11 Consider a function u which depends on position x and time t. the partial differential
𝜕2 𝑢
equation 𝜕𝑡 2
= c2 Δu is known as the
Option A Wave equation

Option C Laplace equation
Option D Elasticity equation
Correct Answer A
12 Partial differential equation
∂u ∂2 u
= c 2 ∂x2 is called
∂t
Option A Parabolic Heat equation

Option B Hyperbolic Heat equation
Option C Parabolic wave equation
Option D Hyperbolic wave equation
Correct Answer A
13 In one dimensional heat equation
∂u
= α2
∂2 u
, the value of α2 is
∂t ∂x2
Option A 𝑘
𝐶𝜌 2
Option B 𝑘2
𝐶𝜌 2
Option C 𝑘
𝐶𝜌
Option D 𝑘2
𝐶𝜌
Correct Answer C
14 The partial differential equation uxx + uyy = 0 is called

Option C Wave equation
Option D Poisson equation
Correct Answer A
15 What is the value of 𝛾 under which crank Nicolson’s formula
Option A -1
Option B 1
Option C 2
Option D 1/2
Correct Answer B
16 The partial differential equation uxx + uyy = 0 is called
Option A Heat equation
Option B Wave equation
Option C Two Dimensional Heat equation
Option D One Dimensional Heat equation
Correct Answer C
17 ……….equation can be solved by Explicit method or Crank Nicolson Method.
Option A 𝜕𝑢
𝜕𝑦
𝜕𝑢
+ =f(x,y)
𝜕𝑥
Option B 𝜕𝑢
𝜕𝑦
+𝑐
𝜕𝑢
𝜕𝑥
=0
Option C 𝜕𝑢 𝜕2𝑢
= 𝑐2 2
𝜕𝑡 𝜕𝑥
Correct Answer C
18 What is mathematical form of Schmidt Method
Option A ui,j-1 =γui-1,j + (1+2γ)ui,j + γ ui+1,j
Option B ui,j+1 =γui-1,j + (1-2γ)ui,j + γ ui+1,j
Option C ui,j+1 =γui-1,j + (1-2γ)ui,j + γ ui+1,j
Option D None the above
Correct Answer C
19 Obtain the finite difference scheme for the differential equation 2y” + y = 5
Option A 5h2
Option B 6h2
Option C 5h3
Option D 4h2
Correct Answer A
20 Solve Laplace equation with respect to grid as shown in figure. Calculate the temperature
equation for T1 0 40 80 120
20 T4 T3 110
40 T1 T2 180
80 110 160 210

Option A 1
T1 = 4 [T2 + T4 +150]
Option B 1
T1 = 2 [T2 + T4 +150]
Option C 1
T1 = 4 [T2 - T4 +150]
Option D 1
T1 = 4 [T2 + T4 +110]
Correct Answer A
21 Solve Elliptic equation (Laplace equation) with respect to grid as shown in figure.
Compute equation of the temperature T3 0 40 80 135
20 T4 T3 110
40 T1 T2 180
70 110 160 215
Option A 1
T1 = 4 [T2 + T4 +110]
Option B 1
T1 = 4 [T2 + T4 +190]
Option C 1
T1 = [T2 + T4 +80]
4
Option D 1
T1 = 4 [T2 + T4 +135]
Correct Answer B
22 The edges of steel plate of 750 x 750 mm has maintained at temperature as shown in fig.
what will be steady state temperature equation at point T1
500
2000 2000 2000 2000
1000 T1 T2 0
1000 T3 T4 0
500 500 500 500
Option A T1 = (T2 + T3 +3000)/4

Option B T1 = (T2 + T4 +3000)/4
Option C T1 = (T4 + T3 -3000)/4
Option D T1 = (T2 + T3 +1000)/4
Correct Answer A
23 Solve Elliptic equation (Laplace equation) with respect to grid as shown in figure
Compute the temperature equation for T4. 0 10 20 30
20 T1 T2 40
40 T3 T4 50
60 60 60 60
Option A 1
T4 = 4 [T1 + T4 +110]
Option B 1
T4 = 4 [T2 + T4 -110]
Option C 1
T4 = 4 [T2 + T4 +110]
Option D 1
T4 = 4 [T3 + T4 +110]
Correct Answer D
24 Solve Laplace equation (2D heat Flow) with respect to grid as shown in figure Compute
the equation for U1. 25
U4 U3
60 10
U1 U2
80
Option A 4U1 = [U2 + U4 +60-80]

Option B U1/4 = [U2 + U4 +60+80]
Option C 4U1 = [U2 + U4 +60+80]
Option D 4U1 = [U2 + U4 - 60+80]
Correct Answer C
25 Solve Laplace equation (2D heat Flow) with respect to grid as shown in figure Compute
the temperature equation for T1. 0 10 20 30
20 T4 T3 40
40 T1 T2 50
60 60 60 60
Option A 1
T1 = 4 [T2 + T4 +100]
Option B 1
T1 = 4 [T2 + T4 +110]
Option C 1
T1 = 4 [T2 + T4 +120]
Option D 1
T1 = 4 [T2 + T4 -100]
Correct Answer A
26 The edges of steel plate of 750 x 750 mm has maintained at temperature as shown in fig.
what will be steady state temperature equation at point T3.
500
2000 2000 2000 2000
T1 T2
1000 0
1000 0
T4
T3
500 500 500 500
Option A T3 = (T1 + T4 +3000)/4

Option B T3 = (T1 + T4 +1500)/4
Option C T3 = (T1 + T4 +500)/4
Option D T3 = (T1 + T4 +2500)/4
Correct Answer B
27 𝑑2 𝑦 𝑑2 𝑦 1
Solve the poissons equation 𝑑𝑥 2 + 𝑑𝑦 2 = 𝑥𝑦 corresponding to grid mesh size, h =2 shown
in fig, calculate equation form for u2. 0
0 y 0
0
0 1 2
x
4 3 0
0
0
(0,0)
Option A 1
u2 = [u1 + u3 -4]
4
Option B 1
u2 = 4 [u1 + u3 + 4]
Option C 1
u2 = 4 [u1 - u3 -4]
Option D 1
u2 = 4 [u1 - u3 +4]
Correct Answer A
28 𝑑2 𝑦 𝑑2 𝑦 1
Solve the equation 𝑑𝑥 2 + 𝑑𝑦 2 = 𝑥𝑦 corresponding to grid mesh size, h =2 shown in fig,
calculate equation form for u4. 0
0 y 0
0
0 1 2
x
4 3 0
0
0
(0,0)
Option A 1
u2 = 4 [u1 + u3 -4]
Option B 1
u2 = 4 [u1 + u3 + 4]
Option C 1
u2 = 4 [u1 - u3 -4]
Option D 1
u2 = 4 [u1 - u3 +4]
Correct Answer A
𝜕3 𝑈
29 What is equation of T1 using Laplace equation >1 for the square mesh a shown in fig.
𝜕𝑦 3
500
60 100 60 20
T4 T3
80 40
100 80
T1 T2
40
50 10 15
Option A T1 = (T3 + T4 +50)/4

Option B T1 = (T2 + T4 +180)/4
Option C T1 = (T2 + T4 +150)/4
Option D T1 = (T2 + T4 +90)/4
Correct Answer C
30 Consider the following partial differential equation for u(x,y) with the constant c>1:
∂u ∂u
+ c = 0 solution of this equation is
∂y ∂x
Option A u(x,y) = f(x+cy)
Option B u(x,y) = f(x-y)
Option C u(x,y) = f(cx+y)
Option D u(x,y) = f(cx-cy)
Correct Answer B
31 Solve by crank Nicolson’s implicit method, ut = uxx, 0<x<1,t>0 with u(x,0)=100(x-x2),
u(0,t)=0, u(1,t)=0. Compute u for one time step with h=0.25
Option A 8.82, 14.29
Option B 9.82, 14.29
Option C 9.82, 12.29
Option D 9.82, 10.29
Correct Answer B
32 Solve the poissons equation uxx + uyy = -81xy, 0<x<1, 0 < y < 1 and u(0,y)=u(x,0)=0,
u(x,1) = u(1,y) =100 with the square mesh, each of length h=1/3.
Option A 51.08, 76.54, 25.79
Option B -51.08, -76.54, 25.79
Option C -51.08, 76.54, -25.79
Option D 51.08, -76.54, 25.79
Correct Answer A
33 Solve the boundary value problem y”- 64+10=0 with y(0) = y(1) = 0 by the finite
difference method. h=0.25.
Option A y(0.5)= 0.129

Option B y(0.5)= 0.147
Option C y(0.5)= 0.157
Option D y(0.5)= 0.111
Correct Answer B
34 2nd order differential equation is x2 y’’(x-2) y’-3y = 10x, subjected to consideration y (0)
=0, y (0.3) = 10, h= 0.1. Write equation by using finite difference method.
Option A 10y1+17y2= -2, 26y1 - 22y2= 104
Option B 10y1- 17y2= -2, 26y1 - 22y2= 104
Option C 10y1+17y2= -2, 26y1 + 22y2= 104
Option D 10y1+17y2= 2, 26y1 + 22y2= 104
Correct Answer A
35 Solve y”+ y=0 with boundary condition y(0)=0, y(1)=0. Find y(0.5), using finite
difference method.
Option A 0.16285
Option B 0.14285
Option C 0.15285
Option D 0.13285
Correct Answer B
36 Using the finite difference solve y’ – y = 0 of y(0)=0, y(1)=1, n=2.
Option A 0.4444
Option B 0.5555
Option C 0.3333
Option D 0.6666
Correct Answer A
37 Solve ∇2 u = 8x2y2 over the square with x=-2, x=2,y=-2,y=2 with u=0 on the boundary
sand mesh length 1.what is value of u1, u3, u7 and u9
Option A 3
Option B 2
Option C -2
Option D -3
Correct Answer D
38 Calculate y1 & y2 value equation by using finite difference method of given differential
equation is x2 y’’(x-2) y’-3y = 10x, subjected to consideration y (0) =0, y (0.3) = 10, h=
0.1.
Option A y1=-3.73 , y2= -1.6495
Option B y1=3.73 , y2= 1.6495
Option C y1=3.73 , y2= -1.6495
Option D y1=-3.73 , y2= 1.6495
Correct Answer C
39 Solve ∇2 𝑢 = 8x2y2 over the square with x=-2, x=2,y=-2, y=2 with u=0 on the boundary
sand mesh length 1.what is value of u5
Option A -1
Option B -2
Option C -3
Option D -4
Correct Answer B
40 Solve ∇2 𝑢 = 8x2y2 over the square with x=-2, x=2,y=-2,y=2 with u=0 on the boundary
sand mesh length 1.what is value of u2, u4, u6 and u8
Option A -4
Option B -3
Option C -2
Option D -1
Correct Answer C
41 Find u4 value by using the Laplace equation ∇2 𝑢= 0 for the given boundary condition
shown in fig. 25
u4 u3
60 10
u1 u2
80
42 Find u4 value by using the Laplace equation ∇2 𝑢= 0 for the given boundary condition
shown in fig.
0 40 80 120
20 u1 u2 110
40 u3 u4 180
80 110 160 210
Correct Answer 60
43 Solve the equation ∇2 u = -10(x2 + y2 +10) over the square mesh length =1, with sides
x=0=y, x=3=y with u=0. y
2 3
1 2
x
(0,0)
Option A u1= 60.5, u2 = 75 u3 =82.5

Option B u1= 60, u2 = 75 u3 =80
Option C u1= 65, u2 = 75 u3 =80
Option D u1= 67.5, u2 = 75 u3 =82.5
Correct Answer D
44 Given
∂u
=
∂2 u
using explicit finite difference scheme at t=0, u= sinπx (0 < x <1) at x=0
∂t ∂x2
& x=1 , u= 0 for all values of t. take increment for t is 0.002 (0 to 0.006) and for x as 0.2
(0 to 1). Find u1 value

45 Given
𝜕𝑢 𝜕2 𝑢
= 𝜕𝑥 2 using crank Nicolson’s method. At x=0 and x=3, u=0 (for all value of t)
𝜕𝑡
& u=0.3. At t=0, u=x2 for 0 < x < 3. Take increment in x as 1 and t as 0.1, find value of
u1.
Name of Content: Curve Fitting [UNIT V]
01 Question:Interpolation is done by
Option A Curve fitting

Option B Regression analysis
Option C Curve fitting & Regression analysis
Option D None of the mentioned
Correct Answer Answer: c
Explanation: Interpolating the value requires or is done by
curve fitting and regression analysis.
02 Question:Interpolation provides a mean for estimating functions
Option A At the beginning points

Option B At the ending points
Option C At the intermediate points
Explanation: Interpolation provides a mean for estimating the
function at the intermediate points.
03 Question: Interpolation methods are
Option A Linear interpolation

Option B Piecewise constant interpolation
Option C Polynomial interpolation
Option D All of the mentioned
Correct Answer Answer: d
Explanation: Some of the interpolation techniques are linear
interpolation, piecewise constant interpolation, polynomial
interpolation, spline interpolation etc.
04 Question:Linear interpolation is
Option A Easy
Option B Precise
Option C Easy & Precise
Correct Answer Answer: a
Explanation: Linear interpolation is quick and easy but not
precise.
05 Question:Error is equal to
Option A Distance between the data points

Option B Square of the distance between the data points
Option C Half the distance between the data points
Correct Answer Answer: b
Explanation: Error is equal to square of the distance between
the data points.
06 Question:Which produces smoother interpolants?
Option A Polynomial interpolation

Option B Spline interpolation
Option C Polynomial & Spline interpolation
Explanation: Polynomial interpolation and spline interpolation
produces smoother interpolants.
07 Question:Which is more expensive?
Option A Polynomial interpolation

Option B Linear interpolation
Option C Polynomial & Linear interpolation
Explanation: Polynomial interpolation is more expensive than
linear interpolation.
08 Question:Interpolation means
Option A Adding new data points

Option B Only aligning old data points
Option C Only removing old data points
Explanation: Interpolation is a method of adding new data
points within the range of a discrete set of known data points.
09 Question:Interpolation is a method of
Option A Interrelating
Option B Estimating
Option C Integrating
Option D Combining
Explanation: Interpolation is a method of estimating the value
of the function.
10 Question:The process of finding the values inside the interval (X0, Xn) is called
Option A Interpolation
Option B Extrapolation
Option C Iterative
Option D Polynomial equation
Correct Answer Ans- A

11 Question:The Delta of power two is called the ____order difference operator.
Option A First
Option B Second
Option C Third
Option D Fourth
Correct Answer Ans- B

12 Question:Newton forward interpolation formula is used for _________ intervals.
Option A open
Option B unequal
Option C equal
Option D closed
Correct Answer Ans- C

13 Question:For the given distributed data find the value of Δ 3y0 is?
3.60 3.65 3.70 3.75

x
36.598 38.475 40.447 42.521
y
Option A 0.095
Option B 0.007
Option C 1.872
Option D 0.123
Correct Answer Ans- B

14 Question:Fit a straight line into the following data.
x: 0 1 2 3 4 5
y: 3 6 8 11 13 14
Option A y=3.52+2.26x
Option B y=3.52
Option C y=2.26x
Option D y=4+3x
Explanation: Here, N=6
Calculations of ∑x and ∑x2
x y x2 xy
0 3 0 0
1 6 1 6
2 8 4 16
3 11 9 33
4 13 16 52
5 14 25 70
∑x=15 ∑y=55 ∑x2=55 ∑xy=177
We know that,
∑y=Na+b∑x
∑xy=a∑x+b∑x2
Substituting the values from the table into the equations-
55=(6)a+b(15) – (1)
177=(a)15+b(55) – (2)
Solving equations (1) and (2) simultaneously
a=3.52 and b=2.26
Thus the equation of the line is given by y=a+bx
Thus, the equation of the line is y=3.52+2.26x.
15 Question:Fit a straight line y=a+bx into the given data:

(x,y):(5,12)(10,13)(15,14)(20,15)(25,16).
Option A y=11
Option B y=0.2x
Option C y=11+0.2x
Option D y=1.1+0.2x
x y x2 xy
5 12 25 60
10 13 100 130
15 14 225 210
20 15 400 300
25 16 625 400
∑x=75 ∑y=70 ∑x2=1375 ∑xy=1100
We know that,
∑y=Na+b∑x
∑xy=a∑x+b∑x2
70=(5)a+b(75) – (1)
1100=(a)75+b(1375) – (2)
a=11 and b=0.2
Thus, the equation of the line is y=11+0.2x.
16 Question:Fit a straight line y=a+bx into the given data. What is the value of y
when x=8 ?
x: 1 2 3 4 5 6
y: 20 21 22 23 24 25
Option A 45.2
Option B 26
Option C 28
Option D 37

x y x2 xy
1 20 1 20
2 21 4 42
3 22 9 66
4 23 16 92
5 24 25 125
6 25 36 216
∑x=21 ∑y=135 ∑x2=91 ∑xy=561
We know that,
∑y=Na+b∑x
∑xy=a∑x+b∑x2
135=(6)a+b(21) – (1)
561=(a)21+b(91) – (2)

a=4.8 and b=5.05
Thus, the equation of the line is y=4.8+5.05x.
Putting x=8,
y=4.8+(5.05)×(8)
y=45.2.
Unit V
curve Fitting
1) Least square technique is use...........
a) To minimize sum of residual error
b) To minimize sum of absolute value of
Residual error
c) maximize sum of square of error
d) both a& b.
Ans : d) both a&b.
2) Using least square method , the value of Y(22)=
X 0 2 4 6
Y 10 12 18 22
a) 40.02. b) 45.00. c) 55.4. d) 60.20
Ans: c) 55.4
3) Equation of straight line in fitting curve of the form y=ab^x
a) b'£X+na'=£Y , b'£X^2+a'£X=£XY
b) a'£X+nb'=£X , a'£x^2+b'£x=£XY
c) a£X^2+b£X+nc=£Y , a£X^3+b£X^2+c£X=£XY , a£X^4+b£X^3+c£X^2=£X^2Y
d) b'£X+na'=£XY , b'£X+a'£X^2=£Y
Ans: a) b'£X+na'=£Y , b'£X^2+a'£X=£XY
4) If y=ae^bx then what is the value of a' & b' respectively.
a) a'=b , b'=log a
b) a'=b , b'= ln a
c) a'=ln a , b'= b
1
d) a'= log a , b'= b
Ans: b) a'=b , b'= ln a
5) Simultaneous equation for 2nd degree polynomial curve or parabola is.......
a) a£X^2+b£X+nC=£Y , a£X^3+b£X^2+c£X=£XY , a£X^4+b£X^3+c£X^2=£X^2Y
b) a£X+b+nc=£Y , a£X^2+b£X+nc=£XY , a£X^4+b£X^3+c£X^2=£X^2Y
c) b'£X+na'=£Y , b'£X^2+a'£X=£XY
d) a'£X+nb'=£Y , a'£X^2+b'£X=£XY
Ans: a) a£X^2+b£X+nc=£Y , a£X^3+b£X^2+c£X=£XY , a£X^4+b£X^3+c£X^2=£X^2Y
6) Fit the curve of the form N=ab^t & estimate N when t=7.
t 0 1 2 3
N 32 47 65 92
a) 3.099×10^-3.
b) 4.088×10^-3
c) 3.099
d) 30.99
Ans: a) 3.099×10^-3
7) Fit a straight line y=a+bx into a given data:
(X,Y) : (5,12) (10,13) (15,14) (20,15) (25,16)
a) y=11
b) y=0.2x
c) y=11+0.2x
d) y=1.1+0.2x
Ans: c) y=11+0.2x
8) The method of ..............is the most systematic procedure to fit a unique curve from given data
2
a) least square
b) least cube
c) square
d) none of these
Ans: a) least square
9) What type of equation is f(x)=ae^bx ?
a) logarithmic
b) exponential
c) power equation
d) polynomial
Ans: b) exponential
10) Principal of least square state that......
a) The sum of square of all points from curve is minimum
b) The sum of square of root of all points from curve is minimum
c) The sum of square of all points from curve is maximum
d) The sum of square of root of all points from curve is maximum
Ans: a) The sum of square of all points from curve is minimum
Regression Analysis
1.Let h be the finite difference, then forward difference operator is defined by……
a) f(x)=f(x+h)-f(x) b) f(x)=f(x-h)-f(x) c) f(x)=f(x*h) d) f(x)=f(x)
Ans: a) f(x)=f(x+h)-f(x)
2. Yg= Y0*L0 + Y1*L1 + Y2*L2 +……+Yn-1*Ln-1 given equation represents ….
a) inverse interpolation b) Newton’s interpolation c) Lagrange’s interpolation d) Hermit

interpolation
Ans: c) Lagrange's Interpolation
3
3. From following data calculate line of regression
∑x=89 , ∑y=283 , ∑xy=5071 , ∑x^2=1611 , ∑y^2=16089
Estimate value of Y when X=25
a) 283 b) 96.352 c) 65.629 d) 0
Ans: c) 65.629
4. Find the equation of the lines of regression based on following data:
∑x=15 , ∑y=15 ,∑xy=44 ,∑x^2=49
a) 2x-8y-15=0 b) 2x+8y-15=0 c) 2x+8y+15=0 d) 2x-8y+15=0
Ans: b) 2x+8y-15=0
5. Using Ladrange’ s formula , find a unique polynomial P(x) of degree 2 such that P(1) =1,P(3)=27 ,
P(4) =64 and evaluate P(1.5)
a) 1.5 b) 0.625 c) 0 d) -1.5
Ans : a) 1.5
6. From following data , calculate value of u
x 45 50 55 60 65
y 2.871 2.404 2.083 1.862 1.712
a) 1 b) 0.5 c) 0.1 d) 0.2
Ans: d) 0.2
4
Unit V- Curve fitting and Interpolation
1. Use least square regression to fit a straight line y=ax+b to the data given below
x 1 2 3 4 5 6 7
Y 0.5 2.5 2.0 4.0 3.5 6.0 5.5
The value of a is
a) a = 0.83 b) a = 0.65 c) a = 0.39 d) a = 0. 07
2. Use least square regression to fit a straight line y=ax+b to the data given below
x 1 2 3 4 5 6 7
Y 0.5 2.5 2.0 4.0 3.5 6.0 5.5
The value of b is
b) a = 0.83 b) a = 0.65 c) a = 0.39 d) b = 0. 07
3. The table of points is given below
x 0 2 4 6 8 12 20
Y 10 12 18 22 20 30 30
Use least square regression to fit a straight line y=ax+b, the value of a is
c) 1.055 b) 12.444 c) 17.564 d) 19.714
4. The table of points is given below
x 0 2 4 6 8 12 20
Y 10 12 18 22 20 30 30
Use least square regression to fit a straight line y=ax+b, the value of a is
a) 1.055 b) 12.444 c) 17.564 d) 19.714
5. The equation of best fit curve is of the type y=abx find the value of a if
x 2 3 4 5 6
y 144 172.8 207.4 248.8 298.5
a) 73.45 b) 50.32 c) 99.66 d) 1.2
6. The equation of best fit curve is of the type y=abx find the value of b if
x 2 3 4 5 6
y 144 172.8 207.4 248.8 298.5
b) 73.45 b) 50.32 c) 99.66 d) 1.2
7. fit a straight line to the data given below
x 1 2 3 4 5 6 7
Y 0.5 2.5 2.0 4.0 3.5 6.0 5.5
a) y= 0.83x+0.07 b) y= 0.43x+0.47 c) y= 1.93x+4.08 d) y= 9.43x+0.12
8. A set of values of x and f(x) are given below using Lagrange’s interpolation formula find
f(9)
x 5 7 11 13 17
Y=F(x) 150 392 1452 2366 52010
a) 1258 b) 420 c) 1029 d) 810
9. Use the following points to fit the polynomial using Lagrange’s method and find the
value of y at x = 2.7, (2.10, 5.14) (2.15, 6.78) (3.10, 10.29) (3.50, 13.58)
a) 7.78 b) 8.95 c) 12.76 d) 15.55
10. Given the two points [a, f (a )], [b, f (b )] , the linear Lagrange polynomial f1 ( x ) that passes
through these two points is given by
x−b x−a
(A) f1 (x ) = f (a ) + f (b )
a −b a−b
f1 (x ) = f (a ) + f (b )
x x
(B)
b−a b−a
f (b ) − f (a )
(C) f1 (x ) = f (a ) + (b − a )
b−a
x−b x−a
(D) f1 (x ) = f (a ) + f (b ) -----------------------ANS
a −b b−a
11. The Lagrange polynomial that passes through the 3 data points is given by
x 15 18 22
y 24 37 25
f 2 ( x ) = L0 ( x )(24 ) + L1 (x )(37 ) + L2 (x )(25)

The value of L1 ( x ) at x = 16 is most nearly
(A) –0.071430
(B) 0.50000
(C) 0.57143
(D) 4.3333
12. The following data of the velocity of a body is given as a function of time.
Time ( s ) 10 15 18 22 24
Velocity ( m s ) 22 24 37 25 123
A quadratic Lagrange interpolant is found using three data points, t = 15 , 18 and 22.
From this information, at what of the times given in seconds is the velocity of the
body 26 m/s during the time interval of t = 15 to t = 22 seconds.
(A) 20.173
(B) 21.858
(C) 21.667
(D) 22.020
13. The path that a robot is following on a x, y plane is found by interpolating four data
points as
x 2 4.5 5.5 7
y 7.5 7.5 6 5
y (x ) = 0.15238 x 3 − 2.2571x 2 + 9.6048 x − 3.9000

The length of the path from x = 2 to x = 7 is
(A) (7.5 − 7.5)2 + (4.5 − 2)2 + (6 − 7.5)2 + (5.5 − 4.5)2 + (5 − 6)2 + (7 − 5.5)2
7
(B) ∫
2
1 + (0.15238 x 3 − 2.2571x 2 + 9.6048 x − 3.9000) 2 dx
7
(C) ∫
2
1 + (0.45714 x 2 − 4.5142 x + 9.6048) 2 dx ---------ANS
7
(D) ∫ (0.15238x − 2.2571x 2 + 9.6048 x − 3.9000)dx
3
2
Time (s) 0 15 18 22 24
Velocity (m/s) 22 24 37 25 123
If you were going to use quadratic interpolation to find the value of the velocity at
t = 14.9 seconds, what three data points of time would you choose for interpolation?
(A) 0, 15, 18
(B) 15, 18, 22
(C) 0, 15, 22
(D) 0, 18, 24
15. When using the linearized data model to find the constants of the regression model
y = ae bx to best fit ( x1 , y1 ), ( x 2 , y 2 ),........( x n , y n ), the sum is the square of the residuals that
is minimized is
( )
n
(A) ∑ y i −ae bxi
2
i =1
n
(B) ∑ (ln( y i ) − ln a − bxi ) -------------------ANS
2
i =1
n
(C) ∑ ( y − ln a − bx )
2
i i
i =1
n
(D) ∑ (ln( y i ) − ln a − b ln( xi ) )
2
i =1
16. It is suspected from theoretical considerations that the rate of flow from a firehouse is
proportional to some power of the nozzle pressure. Assume pressure data is more
accurate. You are linearizing the data.
Flow rate, F 96 129 135 145 168 235

(gallons/min)
Pressure, p (psi) 11 17 20 25 40 55
The exponent of the power of the nozzle pressure in the regression model,F=apb
most nearly is
(A) 0.497
(B) 0.556
(C) 0.578
(D) 0.678
17. The linearized data model for the stress-strain curve σ = k1εe − k 2ε for concrete in
compression, where σ is the stress and ε is the strain is
(A) ln σ = ln k1 + ln ε − k 2 ε
σ
(B) ln = ln k1 − k 2 ε --------------ANS
ε
σ
(C) ln = ln k1 + k 2 ε
ε
(D) ln σ = ln(k1ε ) − k 2 ε
18. In nonlinear regression, finding the constants of the model requires solution of
simultaneous nonlinear equations. However in the exponential model, y = ae bx that is
best fit to ( x1 , y1 ), ( x 2 , y 2 ),........( x n , y n ), the value of b can be found as a solution of a
sample nonlinear equation. That equation is given by
n n n
(A) ∑ y i xi e bxi − ∑ y i e bxi ∑ xi = 0
i =1 i =1 i =1
n
n ∑ye i
bxi
n
(B) ∑ y i xi e bxi − i =1
n ∑x e i
2 bxi
=0
i =1
∑e
i =1
2 bxi i =1 ------------------ANS
n ∑ye i
bxi
n
(C) ∑ yi xi e bxi − i =1
n ∑e bxi
=0
i =1
∑e
i =1
2 bxi i =1
n ∑ye i
bxi
n
(D) ∑ y i e bxi − i =1
n ∑x e i
2 bxi
=0
i =1
∑e
i =1
2 bxi i =1
19. There is a functional relationship between the mass density p of air and altitude h
above the sea level
Altitude above sea level, 0.32 0.64 1.28 1.60

h (km)
Mass Density, ρ (kg/m3) 1.15 1.10 1.05 0.95
In the regression model ρ = k1e .− k 2 h , the constant k 2 is found as k 2 = 0.1315 .
Assuming the mass density of air at the top of the atmosphere is 1 / 1000 th of the
mass density of air at sea level. The altitude in km of the top of the atmosphere most
nearly is
(A) 46.2
(B) 46.6
(C) 49.7
(D) 52.5
20. A steel cylinder at 80oF of length 12" is placed in a liquid nitrogen bath (−315 o F ) . If
thermal expansion coefficient of steel behaves as a second order polynomial of
temperature and the polynomial is found by regressing the data below,
Temperature Thermal expansion

(oF) coefficient
( µ in/in/oF)
-320 2.76
-240 3.83
-160 4.72
-80 5.43
0 6.00
80 6.47
the reduction in the length of cylinder most nearly is

(A) 0.0219
(B) 0.0231
(C) 0.0235
(D) 0.0307
Sr. No Question Option 1 Option 2 Option 3 Option 4 Correct Ans Marks
All the formulaae of interpolation are based on the
1 fundamental assumption that the given data can be Polynomial Equation Algorithm None of the Above Polynomial 1
expressed as a________
to estimate the value of dependent variable x for given value
2 Extrapolation Inverse Interolation Interpolation Polynomial Inverse Interolation 1
of independent variable y, the process known as…….
3 …… is called the forward difference operator. ∆ α λ ∆ 1
The amain disadvanatages of Lagrangian interpolation is tha it
4 Polynomial Equation Algorithm None of the Above Polynomial 1
is difficult to find the ordr of the ……… to be fitted.
5 _____is not a type of interpolation method. Forward difference Backward difference Newton divided difference Moving Average method Moving Average method 1
The formula for inverse interpolation is obtained
6 from________interpolation formula by changing the variable Forward difference Backward difference Newton divided difference Lagrangian Lagrangian 1
x and y=f(x).
7 the process of fitting function to data is known as….. Regression Data fitting Curve fiting Interpolation Curve fitting 1
When we predict values that fall within the range of our data,
8 Regression Extrapolation Inverse Interpolation Interpolation Interolation 1
this is known as_____
We can find out the equation of the regression line by using
9 Least square Method Power Equation Exponential Function Method Interpolation Least Square Method 1
an algebric method called the…….
When we predict valuesof a variable beyond the range of our
10 Regression Extrapolation Inverse Interpolation Interpolation Extrapolation 1
data, this is known as_____
A method of fitting a Parabola …………….. In short apolynomial 〖 = 〗^ ^2+ ^2= ^ = ^2+ + = ^2+ +
11 through the given set of point is called as polynomial y=ax+b 1
regression.
12 Power equation is mathematical form of ……. Regression Polynomial Linear Non-Linear None of the Above Non-Linear 1
The………. Is used to model relationship in which constant
13 change in the independent variable gives the same power function exponential Function Quadratic function All of the Above exponential Function 1
proportional change in the dependent variable.
14 ∆ is defined as any functions says f(x) it gives difference……. ∆f(x)=f(x+h)-f(x) ∆f(x)=f(x+h)-f(x+ih) ∆f(x)=f(x+h)-f(x1) ∆f(x)=f(x3)-f(x1) ∆f(x)=f(x+h)-f(x) 1
15 ……….is 2nd degree of polynomial regession form. 〖 = 〗^ y=ax+b = ^2+ + ^2+ ^2= ^2 = ^2+ + 1
16 if ∆f(x)=f(x+h)-f(x), then a constant k, ∆k equals 1 0 f(k)-f(0) f(x+k)-f(x) 0 2
Name of Content: Trapezoidal Rule
01 Trapezoidal rule for integration gives exact result when the integrate is a polynomial of
degree
Option A 0 but not 1

Option B 1 but not 0
Option C 0 or 1
Option D 2
Correct Answer C
02 1.5
The value of ∫−1 |𝑥| 𝑑𝑥 Computed using trapezoidal rule if the interval of integration is
divided into 3 equal intervals is____________
Correct Answer 1.11
03 4
The value of ∫2.5 ln 𝑥 𝑑𝑥 computed using trapezoidal rule if the interval of integration is
divided into 5 equal intervals is_______________

04 Integration using trapezoidal rule gives the best result for a single variable function, which
is ________
Option A Linear
Option B Parabolic
Option C Logarithmic
Option D Hyperbolic
Correct Answer D
05 П
The error in numerically computing the integral ∫𝟎 (𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙)𝒅𝒙 using the trapezoidal
rule with three intervals of equal length between 0 and П is __________

𝑥
06 The integral ∫𝑥 2 𝑥 2 𝑑𝑥 with is evaluated analytically as well as numerically using a single
1
application of trapezoidal rule. If capital I is the exact value of the integral obtained
analytically and J approximate value of obtained using the trapezoidal rule, which of the
following statements is correct about their relationship?
Option A J>I
Option B J<I
Option C J=I
Option D Insufficient data to determine the relationship
Correct Answer A
Dr. Sharad S. Mulik, Professor, RMDSSOE, Warje, Pune

07 The value of function f(x) at 5 discrete point are given below.
X 0 0.1 0.2 0.3 0.4
F(x) 0 10 40 90 160
0.4
Using Trapezoidal rule with step size of 0.1, the value of ∫0 𝑓(𝑥) 𝑑𝑥 is _______
Correct Answer 22
08 2
Using a step size of ∫𝑥 𝑥 ln 𝑥 𝑑𝑥 by trapezoidal rule is____________
09 A river is 80 metre wide. Its depth d metre and corresponding distance x metre from
when bank is given below in the table:
x 0 10 20 30 40 50 60 70 80
f(x) 0 4 7 9 12 15 14 8 3
Approximate area a cross section of river by trapezoidal rule is
Option A 705 m2
Option B 710 m2
Option C 730 m2
Option D 750 m2
Correct Answer A
10 The following table, using trapezoidal rule area bounded by the curve, x axis and the line
x =7.47, x = 7.52 is
x 7.47 7.48 7.49 7.50 7.51 7.52

f(x) 1.93 1.95 1.98 2.01 2.03 2.06
.
Option A 0.0776
Option B 0.1096
Option C 0.0896
Option D 0.0996
Correct Answer D
11 The definite integral ∫1
31
𝑑𝑥 , is evaluated using Trapezoidal rule with a step size 1. The
𝑥
correct answer is _______________
Correct Answer 1.16

12 A Passing through the points given by the following table
X 1 2 3 4 5
f (x) 10 50 70 80 100
By Trapezoidal rule, the area bounded by the curve, the x 12axis and the lines x =1 and
x =5
Option A 255
Option B 275
Option C 305
Option D 310
Correct Answer A
13 1
The value of ∫0 𝑥 3 𝑑𝑥 by Trapezoidal rule taking five sub-intervals is
Option A 0.21
Option B 0.23
Option C 0.24
Option D 0.26
Correct Answer D
14 A Second-degree polynomial f(x) has values of 1,4 and 15 at x= 0,1 and 2 respectively.
2
The Integral ∫0 𝑓(𝑥) 𝑑𝑥 used to be estimated by applying the trapezoidal rule to this
data. What is error define h True Value- approximate value in the statement?
Option A −
4
3
Option B −
2
3
Option C 0
Option D 2
3
Correct Answer A
15 A Calculator has accuracy up to 8 digits after decimal place. The Value of
2П
∫0 sin 𝑥 𝑑𝑥 When evaluated using this calculator by trapezoidal method with 8 equal
intervals, to 5 significant digits is
Option A 0.00000
Option B 1.0000
Option C 0.00500
Option D 0.00025
Correct Answer A

16 The minimum number of equal length of sub intervals needed to approximate
2
∫1 𝑥𝑒 𝑥 𝑑𝑥 to an accuracy of at least 1/3 X 10-6 using the trapezoidal rule is
Option A 1000e
Option B 1000
Option C 100e
Option D 100
Correct Answer A
17 Using the Trapezoidal rule and dividing the intervals of integration into three equal
+1
subintervals, the definite integrals ∫−1 |𝑥| 𝑑𝑥 is___________
Correct Answer 1.11

18 1
The numerical value of the definite integral ∫0 𝑒 −𝑥 𝑑𝑥 using Trapezoidal rule with function
evaluation at points x =0, 0.5 and 1 is _____________ (round off to three decimal places)
19 Evaluation of
4
∫0 𝑥 3 𝑑𝑥 using two equal segment, Trapezoidal rule gives a value of ______
Correct Answer 63

Name of Content: Simpson’s 1/3rd Rule
01 3
By Simpson’s rule, the value of ∫−3 𝑥 4 𝑑𝑥 by taking 6 sub-intervals is _________
Option A 96
Option B 98
Option C 99
Option D 100
Correct Answer B
02 By Simpson’s rule, the value of ∫1
2 𝑑𝑥
dividing the interval (1,2) into 4 equal parts is
𝑥
Option A 0.6932
Option B 0.6753
Option C 0.6692
Option D 0.6319
Correct Answer A
03 3
By Simpson’s 1/3 rule, the value of ∫1 𝑓(𝑥)𝑑𝑥 for the following data is
x 1 1.5 2 2.5 3
f(x) 2.1 2.4 2.2 2.8 3
Option A 4.975
Option B 5.05
Option C 11.1
Option D 55.5
Correct Answer B
04 If 𝑒 0 = 1 ; 𝑒 1 = 2.72; 𝑒 2 = 7.39; 𝑒 3 = 20.09; 𝑎𝑛𝑑 𝑒 4 = 54.60 By Simpson’s rule, value of
4
∫0 𝑒 𝑥 𝑑𝑥 is
Option A 5.387
Option B 52.78
Option C 53.17
Option D 53.87
Correct Answer D
05 Simpson's rule for integration gives exact results when f(x) is a polynomial function of
degree less than or equal to___________
Correct Answer 3

06 If by Simpson’s rule, ∫0
1 1
𝑑𝑥 =
1
[3.1 + 4(𝑎 + 𝑏)] when the interval [0,1] is
1.+𝑥 2 12
1
divided into 4 subintervals and a & b are the values of at two of its division
1+𝑥 2
points, then a &b are
Option A a=
1
; b = 125
1
1.0625
Option B a=
1
; b = 1.5625
1
1.0625
Option C a=
1
;b=1
1.25
Option D a=
1
; b = 1.25
1
1.5625
Correct Answer A
07 Taking 4 subintervals, the value of ∫0
1 1
𝑑𝑥 by Simpson’s rule is
1.+𝑥
Option A 0.6035
Option B 0.6945
Option C 0.6145
Option D 0.5945
Correct Answer B
08 𝑖𝑓 ℎ = 1 𝑖𝑛 Simpson’s rule, the value of ∫0
5 𝑑𝑥
is
𝑥
Option A 1.43
Option B 1.48
Option C 1.56
Option D 1.62
Correct Answer D
09 For Step size ∆𝑥 = 0.4 the value of following integral using Simpson's 1/3rd rule
𝟎.𝟖
is__________ ∫𝟎 (𝟎. 𝟐 + 𝟐𝟓𝒙 − 𝟐𝟎𝟎𝒙𝟐 + 𝟔𝟕𝟓𝒙𝟑 − 𝟗𝟎𝟎𝒙𝟒 + 𝟒𝟎𝟎𝒙𝟓 )𝒅𝒙

𝛑
10 Taking the step size
𝛑
the value of ∫02 √1 − 0.162𝑠𝑖𝑛2 𝑥𝑑𝑥 by Simpson’s 1/3rd rule is
12
Option A 1.5058
Option B 1.5759
Option C 2.5056
Option D 1.5056
Correct Answer D
11 The value of ∫1
21
𝑑𝑥 computed using Simpsons rule with a step size of h = 0.25
𝑥
is______________

12 3
The value of ∫−1 ln 𝑥 𝑑𝑥 computed using Simpson's rule if the interval of integration is
divided into two equal intervals of width one is __________________
Correct Answer 1.29
13 The Estimate of ∫0.5
1.5 𝑑𝑥
obtained using Simpsons rule with three-point evolution exceeds
𝑥
the exact value by
Option A 0 but not 1

Option B 1 but not 0
Option C 0 or 1
Option D 2
Correct Answer D
14 The table below gives value of a function f(x) obtained for values of X at interval
of 0.25
X 0 0.25 0.5 0.75 1
F(x) 1 0.9412 0.8 0.64 0.5
The value of integral of the function between limits 0 and 1, using Simpson's rule
is ______________
15 The velocity v (in kilometre per minute) of a motorbike which starts from rest is given
at fixed interval of time t (in minutes) as follows
1 2 4 6 8 10 12 14 16 18 20
v 10 18 25 29 32 20 11 5 2 0
The approximate distance in kilometre rounded to two places covered in using Simpson's
1/3rd rule is _____________
16 Simpson's 1/3rd rule is used to integrate the function 𝑓(𝑥) =
3
𝑥2 −
9
between
5 5
x = 0 and x = 1 using the least number of equal sub-internal. The value of integral is __
Correct Answer 2
17 In numerical integration using Simpson's rule the function in the interval is a ________
Option A Constant
Option B straight line
Option C cubic B spline
Option D parabola
Correct Answer D

18 P(0,3), Q(0.5,4) and R(1,5) are defined by f(x). Numerical integration is carried out using
trapezoidal rule and Simpson's rule within limits x=0 and x=1 for the curve. The
difference between the two results will be
Option A 0
Option B 0.25
Option C 0.5
Option D 1
Correct Answer D
19 The accuracy of Simpson’s rule quadrature for a step size h is
Option A O (h2 )
Option B O (h3 )
Option C O (h4 )
Option D O (h5 )
Correct Answer C
20 31
The integral ∫1 𝑑𝑥 , when evaluated by using Simpson’s 1/3rd rule on two equal sub
𝑥
intervals each of length 1, equals
Option A 1.0000
Option B 1.098
Option C 1.111
Option D 1.120
Correct Answer C
20 The estimate of
1.5 1
∫0.5 𝑑𝑥 , Obtained using Simpson’s rule with three point function
𝑥
evaluation exceeds the exact value by
Option A 0.235
Option B 0.068
Option C 0.024
Option D 0.012
Correct Answer D
21 The magnitude of the error (correct to two decimal places) in the estimation of following
integral using Simpson’s 1/3rd rule. Take the step length as 1.
4
∫ (𝑥 4 + 10)𝑑𝑥
0
Correct Answer 0.5

22 Function f is known at the following points.
x 0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3.0
f(x) 0 0.09 0.36 0.81 1.44 2.25 3.24 4.41 5.76 7.29 9.00
3
The value of ∫0 f(x)𝑑𝑥 computed using the continuous at x = 3?
Option A 8.983
Option B 9.003
Option C 9.017
Option D 9.045
Correct Answer D
23 Using Simpson’s 1/3rd rule of Numerical Integration, the consecutive points are joined
by a______
Option A Line
Option B Parabola
Option C Polynomial with power 3
Option D Polynomial with power 1/3
Correct Answer B

6. Numerical integration
2
1. Using a unit step size, the value of integral∫1 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑by trapezoidal rule is ________
a. 0.68
b. 0.69
c. 0.62
d. 0.70
3 9
2. Simpson’s 𝟏𝟏/𝟑𝟑rd rule is used to integrate the function (𝑥𝑥)= 𝑥𝑥2+ between x = 0 and x = 1 using
5 5
the least number of equal sub -intervals. The value of the integral is ______________
a. 4
b. 2
c. 3
d. 5
3. The values of function f (x) at 5 discrete points are given below:

x 0 0.1 0.2 0.3 0.4
F(x) 0 10 40 90 160
0.4
Using Trapezoidal rule with step size of 0.1, the value of ∫0 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 is…..
a. 40
b. 22
c. 30
d. 4
2
4. Using a unit step size, the value of integral∫1 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑by trapezoidal rule is ________
e. 0.68
f. 0.69
g. 0.62
h. 0.70
1
5. Using a three steps the definite value of integral∫−1 |𝑥𝑥| 𝑑𝑑𝑑𝑑 by trapezoidal rule is ________
a. 1.1189
b. 2.1189
c. 1.3452
d. 2.3891
31
6. The definite integral ∫1 𝑑𝑑𝑑𝑑 is evaluated using Trapezoidal rule with a step size of 1. The
𝑥𝑥
correct answer is _______
a. 2.27
b. 1.17
c. 1.18
d. 1.20
7. The best estimate of the distance in meters covered by the body from t=4 to t=15 using combined
Simpson’s 1/3rd rule and the trapezoidal rule would be
a. 354.70
b. 362.50
c. 368.00
d. 378.80
19
8. The value of ∫3 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 by using two-segment Simpson’s 1/3 rule is estimated as 702.039. The
estimate of the same integral using four-segment Simpson’s 1/3 rule most nearly is
a. 702.39 + 8/3 [2f(7)-f(11)+2f(15)]
b. 702.39/2 + 8/3 [2f(7)-f(11)+2f(15)]
c. 702.39 + 8/3 [2f(7)+2f(15)]
d. 702.39/2 + 8/3 [2f(7)+2f(15)]

Time (s) 4 7 10 15
Velocity (m/s) 22 24 37 46
The best estimate of the distance in meters covered by the body from t=4 to t=15 using combined
Simpson’s 1/3rd rule and the trapezoidal rule would be
a. 354.70
b. 362.50
c. 368.00
d. 378.80
10. The velocity of a body is given by
V(t)= 2t 1≤t≤5
2
V(t)= 5t + 3 5 ≤ t ≤ 14
where t is given in seconds, and v is given in m/s. Using two-segment Simpson's 1/3 rule, the
distance covered in meters by the body from t=2 to t=9 seconds most nearly is
a. 949.33
b. 1039.7
c. 1200.5
d. 1442.0
2
11. The value of∫0.2 𝑒𝑒^𝑥𝑥 𝑑𝑑𝑑𝑑 by using four-segment Simpson's 1/3 rule is most nearly
a. 7.8036
b. 7.8062
c. 7.8423
d. 7.9655
12. The highest order of polynomial integrand for which Simpson’s 1/3 rule of integration is exact is
a. First
b. Second
c. Third
d. Fourth
13. The two-segment trapezoidal rule of integration is exact for integrating at most ______ order
polynomials.
a. First
b. Second
c. Third
d. Fourth
14. The following data of the velocity of a body as a function of time is given as follows.
Time (s) 0 15 18 22 24
Velocity (m/s) 22 24 37 25 123
The distance in meters covered by the body from t=12 s to t=18 s calculated using using
Trapezoidal Rule with unequal segments most nearly is
a. 162.9
b. 166.0
c. 181.7
d. 436.5
2.2
15. The value of ∫0.2 𝑥𝑥𝑥𝑥^𝑥𝑥 𝑑𝑑𝑑𝑑 by using the three-segment trapezoidal rule is most nearly
a. 11.672
b. 11.807
c. 12.811
d. 14.633
2.2
16. The value of ∫0.2 𝑥𝑥𝑥𝑥^𝑥𝑥 𝑑𝑑𝑑𝑑 by using the one-segment trapezoidal rule is most nearly
a. 11.672
b. 11.807
c. 20.099
d. 24.119
17. The shaded area shows a plot of land available for sale. The numbers are given in meters
measured from the origin. Your best estimate of the area of the land in square meters is most
nearly
4
18. Evaluate- ∫1 𝑥𝑥^ − 0.5 𝑑𝑑𝑑𝑑 by suitable method
a. -2
b. -7/16
c. ½
d. 2
19. The number of strips required in simpsons 3/8th rule is a multiple of
a. 1
b. 2
c. 3
d. 6
20. The error involved in simpson’s 1/3rd rule is
ℎ3
a. − 𝑓𝑓"(𝑥𝑥)
12
ℎ5
b. − 𝑓𝑓′′′′(𝑥𝑥)
19
3ℎ 5
c. − 𝑓𝑓′′′′(𝑥𝑥)
80
8ℎ 7
d. − 𝑓𝑓′′′′′′(𝑥𝑥)
345
1 1
21. The value of ∫0 𝑑𝑑𝑑𝑑 by using Simpson’s rule is
1+𝑥𝑥
a. 0.96315
b. 0.63915
c. 0.69315
d. 0.69351
Sr. No Question Option 1 Option 2 Option 3 Option 4 Correct Ans Marks
According to newtons Cotes formula degree of polynomial is one , such
1 Simpson's 1/3 method Simpson's 3/8 method Trapezoidal method Guass Legendres Formula Trapezoidal method 1
method called as…….
2 Area enclosed by the curve known as….. Integration Differentiation Partial Derivative Total Area Integration 1
The order of error's the Simpson's 1/3 rule for numerical integration with ℎ^2 ℎ^3 ℎ^4 ℎ^4
3 h 1
a step size h is
Trapezoidal method calculate area under the curve is divided into n
4 Even Multiple of 3 Any Multiple of 4 Any 1
number of strips, then is number of interval is
In…… method integral within the limits x0 to xn can be converted to -1 to
5 Guass Legendres 2 point Formula
Simpson's 3/8 method Trapezoidal method Simpson's 1/3 methodGuass Legendres 2 point Formula1
1 limits.
Simpson's 1/3 method calculate area under the curve is divided into n
6 Any Even Multiple of 3 Multiple of 4 Even 1
The order of error's the trapezoidal rule for numerical integration with a ℎ^2 ℎ^3 ℎ^4 ℎ^2
7 h 1
step size h is
when Degree of polynomial of Newtons Cotes formula is three is known
8 Trapezoidal method Simpson's 1/3 method Simpson's 3/8 method 2 point Quadreture formula Simpson's 3/8 Rule 1
as….
The order of error's the Simpson's 3/8 rule for numerical integration with ℎ^2 ℎ^3 ℎ^4 ℎ^5 ℎ^5
9 1
a step size h is
Simpson's 3/8 method calculate area under the curve is divided into n
10 Any Even Multiple of 3 Multiple of 4 Multiple of 3 1

Nmo MCQ QB PDF

Uploaded by

Copyright:

Available Formats

Nmo MCQ QB PDF

Uploaded by

Document Information

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Nmo MCQ QB PDF

Uploaded by

Copyright:

Available Formats

Numerical Methods & Optimization [Multiple Choice Question] UNIT No-01

Option A f(x) changes sign on the interval − 0.75 ≤ x ≤−0.5

Option B f(x) changes sign on the interval − 0.25 ≤ x ≤ 0

Option C f(x) changes sign on the interval − 1 ≤ x ≤−0.75

Option B has repeated roots at x = 0

Option C is always non-negative

Option D has a slope equal to zero at x = 0

05 Question: The equation f(x) is given as x3+4x+1=0. Considering the initial

Option A f(0) and f(1) have opposite signs.

Option C f″(x) does not change sign on the interval 0 ≤ x ≤ 1.

Option D The tangents at 0 and 1 cut the axes in the interval 0 ≤ x ≤ 1.

The amount of insulin in microunits per mL in a diabetic patient is given by the

Option A -5 such that f(-5) = -26

Option A The value of f’(x) must be increased

Option A f is continuous on the interval a ≤ x ≤ b.

Option B f(a) and f(b) have opposite signs.

Option C f″(x) does not change sign on the interval a ≤ x ≤ b.

Option D f′(x) = 0 on the interval a ≤ x ≤ b.

Option A First order

2. Which method has slow convergence?

3. One root of the equation x3 + 3x2- 5x + 2 = 0 lies between:

4.The root of the equation e power x=4x lies between________.

5.Solve the equation ex− 4x=0 using Newton-Raphson iteration.

6. Use the Newton-Raphson method to solve 2x3−6x2+6x−1=0 to 4 decimal places.

7. Newton-Raphson method will always converge to a solution for f(x) =0 on the

A. f is continuous on the interval a≤x≤b.

A. f(0) and f(1) have opposite signs.

A. First order B. Second order

10 Newton Raphson method of solution of numerical equation is not preferred when

A. The graph of f(x) is nearly horizontal where it crosses the x-axis.

(A) Bracketing (B) Open

(A) 1.5 (B) 2.067

13. Newton Raphson method is also called as

B. Xn+1 = Xn- 1 – D. Xn+1 = Xn –

15. Which iterative method requires single initial guess root?

(a) 0.5 (b) 1.5

17. Newton-Raphson method is applicable the solution of ______.

18. Fourth degree equations are also called _______ equations.

30.The ideal gas law is given by

(a) x0 is to be replaced by a (b) ais to be replaced by x0

(c) bis to be replaced by x0 (d) x0 is to be replaced by b

32 For real root of an equation x3 – 2x – 5 = 0, the root lies between

(a) 0 and 1 (b) 2 and 3 (c) 1 and 2 (d)none of them

33 From the following _______ method is not iterative method.

(a) False position (b) Bisection (c) Lagranges (d)none of them

3+2.75 2 + 2.5 3+2.5

35 If in a method of successive approximation, the root of equation lies between 1 and 2,

(a) 0.5625 (b) 1.2177 (c) 1.7777 (d)none of them

37 Absolute error is defined as

(a) Present Approximation – Previous Approximation

(b) True Value – Approximate Value

(c) abs (True Value – Approximate Value)

(d) abs (Present Approximation – Previous Approximation)

38 The number 0.01850 x 103 has ________ significant digits

(a) 3 (b) 4 (c) 5 (d) 6

(a) is a polynomial (b) has repeated roots at x= 0

(c) is always non-negative (d) has a slope equal to zero at x= 0

(a) one root (b) an undeterminable number of roots

(c) no root (d) at least one root