Matoshri College of Engineering & Research Center, Nashik
NMCP
Test No. 1 4. The Bisection method can be used to solve
______.
A. Algebraic equation only
Name: __________________________ B. Transcendental equation only
Div: _______ Roll No.: ______ C. Both Algebraic and Transcendental equation
D. Linear equation only
Bisection Method
5. Which is not correct statement?
1. Which of the following method has
guaranteed convergence? Bisection method requires less number of
A.
steps for desired accuracy.
A. Secant method
B. Bisection method Bisection method is based on intermediate
B.
value theorem.
C. Newton Raphson Method
D. Regula falsi method Bisection method is only suitable to
C.
determine real positive and negative root.
2. The Bisection method of finding root of D. Bisection method always coverage.
nonlinear equation falls under the category
of a (an) ______ method.
6. Let f(x) = 0 and an initial approximation is
A. Open [0, 2]. The number of iteration required is
_____, if it is to be solved by Bisection
B. Bracketing method with accuracy 0.001.
C. Graphical
D. Random A. 14
B. 10
3. In the Bisection method, the minimum C. 11
number of iterations required are ____ if x0 D. 8
and x1 are initial approximations with
required accuracy as
7. In the Bisection method, an initial
[ x1 x0 ] approximation [2, 3] can be used for ____
log e
A. log e error
n A. f ( x) x 2 5.0625
log e 2
[x x ] B. f ( x) x 2 3.6
log e 1 0
B. log e 2
n C. f ( x) x 2 2.5
log e 2
[x x ] D. None of these
log e 1 0
C. log e 10
n
log e 2
D. None of these
Numerical Method & Computer Programming MCQ on Unit 3 Page 1
Matoshri College of Engineering & Research Center, Nashik
8. In the Bisection method, an initial 12. If f ( x) x 2 5x 6 0 is to be solved by
approximation [5, 8] can be used for ____ Bisection method, correct interval is?
A. f ( x) x 2 16 A. [ 2.5, 3.5]
B. f ( x) x 36
2 B. [2.5, 3.5]
C. f ( x) x 2 25 C. [ 1.5, 0]
D. [1.5, 0]
D. None of these
13. In Bisection method, an initial
approximation [0, 1] can be used for ___ in
9. Which is not correct statement x0 and x1 are
A. f ( x) sin( x)
initial approximation to solve f ( x) 0 used B. f ( x) cos( x)
in Bisection method is suitable, if _______. f ( x) sinh( x)
C.
D. None of these
A. f ( x0 ) is positive and f ( x1 ) is negative
B. f ( x0 ) is negative and f ( x1 ) is positive
C. f ( x0 ) is negative and f ( x1 ) is negative Regula - falsi Method
D. f ( x0 ) * f ( x1 ) < 0 14. In false position method of finding roots of
equations, generalize formula is?
xn xn 1
A. xn 2
10. A positive root of equation is x 3 4 x 9 , at f ( xn 1 ) f ( xn )
the end of 2nd iteration using Bisection
method is ____ if initial interval is [2, 3] xn f ( xn 1 ) xn 1 f ( xn )
B. xn 2
f ( xn 1 ) f ( xn )
A. 2.5
xn f ( xn 1 ) xn 1 f ( xn )
B. 2.75 C. xn 2
f ( xn 1 ) f ( xn )
C. 3.5
D. 1.5 D. None of these
11. Let f ( x) 3x cos( x) 1 and initial 15. Find x3 by false position method, if x1 4
approximate interval is [0, 1] the next
approximate interval is ___ in Bisection and x2 5 for method fx x 2 log 10 ( x) 3
method.
A. 3.747
A. [0.5, 1] B. 5.253
B. [0, 0.5] C. 4.253
C. [0.5, 1.5] D. 4.747
D. None of these
Numerical Method & Computer Programming MCQ on Unit 3 Page 2
Matoshri College of Engineering & Research Center, Nashik
16. Find x3 by false position method, if x1 2 Secant Method
and x2 3 for equation f ( x) x3 3x 5
20. In secant method, if f ( x0 ) is positive, f ( x1 )
A. 2.187 is negative and f ( x2 ) is positive then next
B. 2.718 interval is _______.
C. 2.178
A. [x1, x2]
D. 2.817
B. [x0, x2]
C. [x0, x1]
17. Find x3 by false position method, if
D. [x0, x1, x2]
x1 1.5 and x2 2.7 for equation
f ( x) x 3 4 x 4
21. In secant method of finding roots of
A. 0.809 equations, generalization formula is ___.
B. 2.009
C. 3.091 xn xn 1
D. 2.191 A. xn 2
f ( xn 1 ) f ( xn )
xn . f ( xn 1 ) xn 1. f ( xn )
18. Regula falsi method of finding roots of B. xn 2
equation, initial interval will be [x0, x1]. f ( xn 1 ) f ( xn )
If _____. xn . f ( xn 1 ) xn 1. f ( xn )
C. xn 2
f ( xn 1 ) f ( xn )
A. f ( x0 ) f ( x1 )
B. f ( x0 ) f ( x1 ) ( xn 1 xn )
D. xn 2 xn 1 . f ( xn 1 )
C. f ( x0 ) * f ( x1 ) 0 f ( xn 1 ) f ( xn )
D. Both C and D
22. The secant method to finding roots of
equation falls under the category of ____
19. Find x3 by false position method, if x1 3
methods.
and x2 5 for equation f ( x) x3 5 x 2 4
A. Bracketing
A. 1.4555
B. Graphical
B. 4.5525
C. Open end
C. 3.2222
D. 6.5555 D. Random
Numerical Method & Computer Programming MCQ on Unit 3 Page 3
Matoshri College of Engineering & Research Center, Nashik
23. Find x3 by Secant method, if x1 0.5 and Newton’s Rahpsons Method
x2 0.85 for the equation f ( x) x3 x 1
26. Which statement is incorrect?
A. 2.187
B. 2.718 A. N.R. method is fast convergent method.
C. -2.178
D. B. N.R. method is known as method of tangent.
2.817
N.R. method also works when the y f (x)
C.
is oscillating near maximum and minimum.
Chebyshev Method N.R. method fails if f ( x) 0
D.
24. In Chebyshev method of finding roots of
equations, generalize formula is?
27. f ( x0 )
If the ratio of is high N.R. method
f ( xn ) f ( x0 )
A. xn 1 xn
f ( xn )
A. The next approximation is near to root
2
f ( xn ) 1 [ f ( xn )] B. The next approximation is root
B. xn 1 xn f ( x)
f ( xn ) 2 [ f ( xn )]3 C. The next approximation is away from root
D. Can’t say
f ( xn ) 1 [ f ( xn )]2
C. xn 1 xn f ( x)
f ( xn ) 2 [ f ( xn )]3
28. To solve the given system by Newton’s
D. None of these Rahpson method f ( x) x3 x 2 5 with
f ( x) 3x 2 2 x and values at the end of
1st iteration is 2.125. the approximation
25. To solve the given system by chebyshev after 2nd iteration is
method f ( x) 5x3 20 x 3 with
f ( x) 15x 2 20 and f ( x) 30 x . Initial A. 3.11638
approximation as x0= 0. The approx. value B. 2.11639
after 1st iteration is? C. 2.7189
D. 1.96445
A. 0.15
B. 0.25
29. Use Newton’s Rahpson method find the
C. 0.35 root of equation f ( x) x 0.8 0.2 sin x
D. 0.158 with f ( x) 1 0.2 cos x . Approximate
value at the end of 1st iteration is x1 = 1.00
The approximation value after 2nd iteration
is?
A. 0.9887
B. 1.0456
Numerical Method & Computer Programming MCQ on Unit 3 Page 4
Matoshri College of Engineering & Research Center, Nashik
C. 0.8554 Curve Fitting
D. 0.96445
33. The best representative curve to the given
set of point for which the sum of the
30. Use Newton’s Rahpson method to find the squares of the residual is a minimum is
root of equation f ( x) x cos x . With known as _______.
initial approximation x0 = 1.00. The
approximation value after 1st iteration is? A. Curve fitting
B. Least Square method
A. 0.75036
B. 0.73911 C. Regration
C. 0.55036 D. None of these
D. 0.25987
34. The curve obtained by the method of least
N.R. Method by two equations square is known as the curve of ___
31. To solve the given system by Newton’s
A. Straight line
Rahpson method f ( x, y) x 2 y 2 11 is
f g B. Second degree equation
g ( x, y) y 2 x3 7 the values and
y y C. Best fit
at x0 3.5 and y0 1.8
D. Polynomial equation
A. -3.6 and 7
B. 7 and -3.6 35. Which method gives a unique set of values
to the constants in the case of curve fitting?
C. 7 and 3.6
D. -7 and 3.6 A. Honer’s method
B. Least squares method
32. To solve the given system by Newton’s C. Interpolation
Rahpson method f ( x, y ) sin xy x y
D. Newton’s method
f
is g ( x, y ) y cos xy 1 the values and
y
g 36. In curve fitting by method of least square,
y straight line (first order curve) fitting
A. y cos xy 1 and y 2 sin xy respectively equation are as follows except
B. y cos xy 1 and y 2 sin xy respectively n n
A. y i a.n b xi
C. y cos xy 1 and y 2 sin xy respectively i 1 i 1
n n n
D. y cos xy 1 and y 2 sin xy respectively B. xi yi a xi2 b xi
i 1 i 1 i 1
Numerical Method & Computer Programming MCQ on Unit 3 Page 5
Matoshri College of Engineering & Research Center, Nashik
n n
B. y 7.5 x 7.3
C. yi a xi b
i 1 i 1 y 3x 4
C.
D. All of these
D. None of these
40. For fitting a parabola with the given data
37. In curve fitting by method of least square, below, the normalized equations are?
second order curve fitting equation are as
follows except x -5 -3 0 8 5
y -11 -8 -2 0 -4
n n n x 5 y 25 xy 59
A. y
i 1
i a.n b xi c xi2
i 1 i 1
x y 447 x 30 x 485 x
2 2 3 4
5427
n n n n
B. xi yi a xi b xi2 c xi3 5427a 485b 123c 447
i 1 i 1 i 1 i 1
A. 123a 485b 5c 59
n n n n 123a 5b 5c 25
C. x y
i 1
3
i i a xi2 b xi0 c xi2
i 1 i 1 i 1
485a 5427b 123c 447
n n n n
D. x
i 1
2
i yi a x b x c x
i 1
2
i
i 1
3
i
i 1
4
i
B. 485a 123b 5c 59
123a 5b 5c 25
5427a 485b 123c 447
38. Find the missing value from the following C. 485a 123b 5c 59
data 123a b c 25
x 0 1 2 3 4 5427a 485b 123c 447
y -4 -1 4 11 20 D. 485a 123b 5c 59
123a 5b 5c 25
x 10 y 30 xy 120
x y 434 x 100 x 30 x
2 3 2 4
354
A. y x 2 6x 4 41. Find the missing value from the following
data
B. y x2 2x 4
C. y x 2 3x 4 x 1 1.5 2 2.5 3
y 17 19 14 10 7
D. None of these
x 10 y 67 xy 119.5
39. Find the missing value from the following x 22.5
2
data
A. y 5.8 x 25
x 1 2 3 4 5
y 1 5 10 38 22 B. y 5.8 x 25
y 5.8 x 25
x 15 y 76 xy 303 x 2
55 C.
D. y 25 x 5.8
A. y 7.5 x 7.3
Numerical Method & Computer Programming MCQ on Unit 3 Page 6
Matoshri College of Engineering & Research Center, Nashik
NMCP
Test No. 1
Name: __________________________ Div: _______ Roll No.: ______
Write Answers here
Q.no Option Q.no Option Q.no Option Q.no Option
1 11 21 31
2 12 22 32
3 13 23 33
4 14 24 34
5 15 25 35
6 16 26 36
7 17 27 37
8 18 28 38
9 19 29 39
10 20 30 40
41
Numerical Method & Computer Programming MCQ on Unit 3 Page 7
