Numerical Analysis Questions and Answers – Muller’s Method

This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Muller’s Method”.

1. What will be the root of the equation f(x)=2x-5 using Muller method?
a) 2.5
b) 1.8
c) 4.2
d) 3.6
View Answer

Answer: a
Explanation: Given,
Let f(x)=2x-5
Here

x 0 1 2 3
f(x) -5 -3 -1 1

x0=2
x1=3
x2=2.5
1st iteration :
f(x0)=f(2)=(2⋅2)-5=-1
f(x1)=f(3)=(2⋅3)-5=1
f(x2)=f(2.5)=(2⋅2.5)-5=0
h1=x1-x0=3-2=1
h2=x2-x1=2.5-3=-0.5
δ1=f(x1)-f(x0)/h1=(1-(-1))/1=2
δ2=f(x2)-f(x1)/h2=(0-1)/-0.5=2
a=(δ2-δ1)/(h2+h1)=(2-2)/(-0.5+1)=0
b=a×h2+d2=(0×(-0.5))+2=2

c=f(x2)=0
x3 = x2 + \(\frac{-2c}{b ± \sqrt{b^2-4ac}}\)
x3 = x2 + \(\frac{-2c}{b+(b)\sqrt{b^2-4ac}}\)
= 2.5 + \(\frac{-2×0}{2+\sqrt{2^2-4×0×0}}\)
= 2.5+(0/(2+√4))
= 2.5+(0/(2+2))
= 2.5
Approximate root of the equation 2x-5=0 using Muller method is 2.5

advertisement
advertisement

2. What will be the root of an equation f(x)=x2-x-1 using Muller method?
a) -0.619
b) -0.611
c) -0.614
d) -0.618
View Answer

Answer: d
Explanation: Given,
Here x2-x-1=0
Let f(x)=x2-x-1
Here

x 0 -1 -2
f(x) -1 1 5

x0=-1
x1=0
x2=-0.5

n x0 x1 x2 f(x0) f(x1) f(x2) a b c x3 ɛan
1 -1 0 -0.5 1 -1 -0.25 1 -2 -0.25 -0.618 19.0983
2 0 -0.5 -0.618 -1 -0.25 0 1 -2.2361 0 -0.618 0
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

Approximate root of the equation x2-x-1=0 using Muller method is -0.618.

3. What will be the root of the equation f(x)=x3+2x2+x-9 using Muller method?
a) 1.4722
b) 1.4724
c) 1.4721
d) 1.4727
View Answer

Answer: b
Explanation: Given,
Here x3+2x2+x-9=0
Let f(x)=x3+2x2+x-9
Here

x 0 1 2
f(x) -9 -5 9
advertisement

x0=1
x1=2
x2=1.5

n x0 x1 x2 f(x0) f(x1) f(x2) a b c x3 ɛan
1 1 2 1.5 -5 9 0.375 6.5 14 0.375 1.4729 1.8418
2 2 1.5 1.4729 9 0.375 0.0068 6.9729 13.3853 0.0068 1.4724 0.0343
3 1.5 1.4729 1.4724 0.375 0.0068 0 6.4452 13.3931 0 1.4724 0

Approximate root of the equation x3+2x2+x-9=0 using Muller method is 1.4724.

advertisement

4. What will be the root of the equation f(x)=tan(x) using Muller method?
a) 4
b) 2
c) 0
d) 1
View Answer

Answer: c
Explanation: Given,
Here tan(x)=0
Let f(x)=tan(x)
Here

x 0 1
f(x) 0 1.5574

x0=0
x1=1
x2=0.5
1st iteration :
f(x0)=f(0)=tan(0)=0

f(x1)=f(1)=tan(1)=1.5574
f(x2)=f(0.5)=tan(0.5)=0.5463
h1=x1-x0=(1-0)=1
h2=x2-x1=(0.5-1)=-0.5
δ1=f(x1)-f(x0)/h1=(1.5574-0)/1=1.5574
δ2=f(x2)-f(x1)/h2=(0.5463-1.5574)/(-0.5)=2.0222
a=(δ2-δ1)/(h2+h1)=(2.0222-1.5574)/(-0.5+1)=0.9296
b=a×h2+d2=(0.9296×(-0.5))+2.0222=1.5574
c=f(x2)=0.5463
x3 = x2 + \(\frac{-2c}{b±\sqrt{b^2-4ac}}\)
x3 = x2 + \(\frac{-2c}{(b+(b)\sqrt{b^2-4ac}}\)
= 0.5 + \(\frac{-2×0.5463}{1.5574+\sqrt{(1.5574)^2-4×0.9296×0.5463}}\)
= 0.5 + \(\frac{-1.0926}{1.5574+\sqrt{0.3941}}\)
= 0.5 + \(\frac{-1.0926}{1.5574+0.6278}\)
= 0

Now,
x0=x1=1
x1=x2=0.5
x2=x3=0

2nd iteration:

f(x0)=f(1)=tan(1)=1.5574
f(x1)=f(0.5)=tan(0.5)=0.5463
f(x2)=f(0)=tan(0)=0
h1=x1-x0=0.5-1=-0.5

h2=x2-x1=0-0.5=-0.5
δ1=f(x1)-f(x0)/h1=(0.5463-1.5574)/-0.5=2.0222
δ2=f(x2)-f(x1)/h2=(0-0.5463)/-0.5=1.0926
 a=(δ2-δ1)/(h2+h1)=(1.0926-2.0222)/(-0.5±0.5)=0.9296
b=a×h2+d2=(0.9296×(-0.5))+1.0926=0.6278
c=f(x2)=0
x3=x2+\(\frac{-2c}{b±\sqrt{b^2-4ac}}\)
x3=x2+\(\frac{-2c}{b+ (b)\sqrt{b^2-4ac}}\)
=0+\(\frac{-2×0}{0.6278+\sqrt{0.6278^2-4×0.9296×0}}\)
=0+\(\frac{0}{0.6278+\sqrt{0.3941}}\)
=0+\(\frac{0}{0.6278+0.6278}\)
=0

Approximate root of the equation tan(x)=0 using Muller method is 0.

5. What will be the root of the equation f(x)=2x-x-1.5 using Muller method?
a) 1
b) -1
c) 2
d) -2
View Answer

Answer: b
Explanation: Given,
Here -x+2x-1.5=0
Let f(x)=-x+2x-1.5
Here

x 0 -1 -2
f(x) -0.5 0 0.75

x0=-1
x1=0
x2=-0.5

n x0 x1 x2 f(x0) f(x1) f(x2) a b c x3 ɛan
1 -1 0 -0.5 0 -0.5 -0.2929 0.1716 -0.5 -0.2929 -1 50
2 0 -0.5 -1 -0.5 -0.2929 0 0.1716 -0.6716 0 -1 0

Approximate root of the equation -x+2x-1.5=0 using Muller method is -1.

6. What will be the root of the equation f(x)=2⋅cos(x)-x+1 using Muller method?
a) 1.3799
b) 1.3791
c) 1.3794
d) 1.3798
View Answer

Answer: d
Explanation: Given,
Here 2cos(x)-x+1=0
Let f(x)=2cos(x)-x+1
Here

x 0 1 2
f(x) 3 1.0806 -1.8323

x0=1
x1=2
x2=1.5

n x0 x1 x2 f(x0) f(x1) f(x2) a b c x3 ɛan
1 1 2 1.5 1.0806 -1.8323 -0.3585 -0.0693 -2.9129 -0.3585 1.3766 8.9676
2 2 1.5 1.3766 -1.8323 -0.3585 0.0095 0.054 -2.9879 0.0095 1.3797 0.2302
3 1.5 1.3766 1.3797 -0.3585 0.0095 0.0001 -0.1514 -2.9635 0.0001 1.3798 0.0019

Approximate root of the equation 2cos(x)-x+1=0 using Muller method is 1.3798.

7. What will be the root of the equation f(x)=log(x) using Muller method?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: Given,
Here log(x)=0
Let f(x)=log(x)
Here

x 0 1 2
f(x) "Undefined" 0 0.301

x0=1

x1=2
x2=1.5

1st iteration :
f(x0)=f(1)=log(1)=0
f(x1)=f(2)=log(2)=0.301
f(x2)=f(1.5)=log(1.5)=0.1761
h1=x1-x0=2-1=1
h2=x2-x1=1.5-2=-0.5
δ1=f(x1)-f(x0)/h1=(0.301-0)/1=0.301
δ2=f(x2)-f(x1)/h2=(0.1761-0.301)/-0.5=0.2499
a=(δ2-δ1)/(h2+h1)=(0.2499-0.301)/(-0.5+1)=-0.1023
b=a×h2+d2=((-0.1023)×(-0.5))+0.2499=0.301
c=f(x2)=0.1761
x3=x2+\(\frac{-2c}{b±\sqrt{b^2-4ac}}\)
x3=x2+\(\frac{-2c}{b+(b)\sqrt{b^2-4ac}}\)
=1.5+\(\frac{-2×0.1761}{0.301+\sqrt{(0.301)^2-4×(-0.1023)×0.1761}}\)
=1.5+\(\frac{-0.3522}{0.301+\sqrt{0.1627}}\)
=1.5+\(\frac{-0.3522}{0.301+0.4033}\)

=1

Now,
x0=x1=2
x1=x2=1.5
x2=x3=1

2nd iteration:

f(x0)=f(2)=log(2)=0.301
f(x1)=f(1.5)=log(1.5)=0.1761
f(x2)=f(1)=log(1)=0
h1=x1-x0=1.5-2=-0.5
h2=x2-x1=1-1.5=-0.5
δ1=f(x1)-f(x0)/h1=(0.1761-0.301)/-0.5=0.2499
δ2=f(x2)-f(x1)/h2=(0-0.1761)/-0.5=0.3522
 a=(δ2-δ1)/(h2+h1)=(0.3522-0.2499)/(-0.5±0.5)=-0.1023
b=a×h2+d2=(-0.1023×(-0.5))+0.3522=0.4033
c=f(x2)=0
x3=x2+\(\frac{-2c}{b±\sqrt{b^2-4ac}}\)
x3=x2+\(\frac{-2c}{b+(b)\sqrt{b^2-4ac}}\)
=1+(-2)x\(\frac{0}{0.4033+\sqrt{(0.4033)^2-4×(-0.1023)×0}}\)
=1+\(\frac{0}{0.4033+\sqrt{0.1627}}\)
=1+\(\frac{0}{0.4033+0.4033}\)
=1
Approximate root of the equation log(x)=0 using Muller method is 1.

8. Muller’s method takes a similar approach as secant method, but projects a parabola through the
points.
a) True
b) False
View Answer

Answer: a
Explanation: Muller’s method takes a similar approach as secant method, but projects a parabola through the points, just as in secant method where we obtain a root estimate by projecting a straight line to the x-axis through two function values.

9. Which of the following is correct about Muller’s method?
a) Muller’s method is a straight-ward approach
b) It gives an exact root of the function
c) It projects a hyperbola through the points
d) It solves equation of the form f(x)=0
View Answer

Answer: d
Explanation: Muller’s method solves equation of the form f(x)=0, where it follows an iterative approach. It takes a similar approach as Secant’s method, but projects a parabola through the points. Furthermore, instead of giving an exact root of the function, it generates the approximate value for the root.

10. What is the order of convergence for Muller’s method?
a) 1.84
b) 1.62
c) 2
d) 2.56
View Answer

Answer: a
Explanation: The order of convergence for Muller’s method is 1.84. It is a quantity that represent how quickly the sequence approaches the limit. As similar to Muller’s method order of convergence for Secant’s method is 1.62 and order of convergence for Newton’s method is 2.

Sanfoundry Global Education & Learning Series – Numerical Analysis.

To practice all areas of Numerical Analysis, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.