This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Secant Method”.
1. What is the region of convergence of Secant Method?
a) 1.5
b) 1.26
c) 1.62
d) 1.66
View Answer
Explanation: The region of convergence of Secant Method is 1.62. It converges faster than Bisection method.
2. Secant Method is also called as?
a) 2-point method
b) 3-point method
c) 4-point method
d) 5-point method
View Answer
Explanation: Secant Method is also called as 2-point method. In Secant Method we require at least 2 values of approximation to obtain a more accurate value.
3. Secant method converges faster than Bisection method.
a) True
b) False
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Explanation: Secant method converges faster than Bisection method. Secant method has a convergence rate of 1.62 where as Bisection method almost converges linearly.
4. The iterative formula for a secant method is given as _________
a) x(n+1)=\(\frac{x(n-1)f[x(n)]-x(n)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]}\)
b) x(n+1)=\(\frac{x(n)f[x(n)]-x(n)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]} \)
c) x(n+1)=\(\frac{x(n-1)f[x(n)]-x(n-1)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]} \)
d) x(n+1)=\(\frac{x(n)f[x(n)]-x(n-1)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]} \)
View Answer
Explanation: The Iterative formula for a Secant Method is given by
x(n+1)=\(\frac{x(n-1)f[x(n)]-x(n)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]} \)
Since there are 2 points considered in the Secant Method, it is also called 2-point method.
5. Secant Method is slower than Newton Raphson Method.
a) True
b) False
View Answer
Explanation: Secant Method is faster as compares to Newton Raphson Method. Secant Method requires only 1 evaluation per iteration whereas Newton Raphson Method requires 2.
6. The number of iterations in Secant Method are less as compared to Newton Raphson.
a) True
b) False
View Answer
Explanation: The number of iterations required for Secant Method is less than Newton Raphson Method. Secant Method requires only 1 evaluation per iteration whereas Newton Raphson Method requires 2.
7. A quadratic equation x2-4x+4=0 is defined with an initial guess of 3 and 2.5. Find the approximated value of root using Secant Method.
a) 1.33
b) 2.33
c) 3.33
d) 4.33
View Answer
Explanation: By Secant Method the iterative formula is given as:
x(n+1)=\(\frac{x(n-1)f[x(n)]-x(n)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]} \)
For n=1
x(2)=\(\frac{x(0)f[x(1)]-x(1)f[x(0)]}{[f[x(1)]-f[x(0)]]} \)
x(0)=3 and x(1)=2.5, f(x0)=1 and f(x1)=0.25
Hence x2=2.33.
8. A quadratic equation x2-4x+4=0 is defined with an initial guess of 2 and 2.5. Find the approximated value of root using Secant Method.
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: By Secant Method the iterative formula is given as:
x(n+1)=\(\frac{x(n-1)f[x(n)]-x(n)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]} \)
For n=1
x(2)=\(\frac{x(0)f[x(1)]-x(1)f[x(0)]}{[f[x(1)]-f[x(0)]]} \)
x(0)=2 and x(1)=2.5, f(x0)=0 and f(x1)=0.25
Hence x2=2. Hence for x2=2 f(x) converges to exact root.
9. A quadratic equation x2-4x+4=0 is defined with an initial guess of 10 and 20. Find the approximated value of x2 using Secant Method.
a) 7.538
b) 7.853
c) 7.358
d) 7.835
View Answer
Explanation: By Secant Method the iterative formula is given as:
x(n+1)=\(\frac{x(n-1)f[x(n)]-x(n)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]} \)
For n=1
x(2)=\(\frac{x(0)f[x(1)]-x(1)f[x(0)]}{[f[x(1)]-f[x(0)]]} \)
x(0)=10 and x(1)=20, f(x0)=0 and f(x1)=0.25
Hence x2 = 7.538.
10. A quadratic equation x4-x-8=0 is defined with an initial guess of 1 and 2. Find the approximated value of x2 using Secant Method.
a) 7.538
b) 7.853
c) 7.358
d) 1.571
View Answer
Explanation: By Secant Method the iterative formula is given as:
x(n+1)=\(\frac{x(n-1)f[x(n)]-x(n)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]} \)
For n=1
x(2)=\(\frac{x(0)f[x(1)]-x(1)f[x(0)]}{[f[x(1)]-f[x(0)]]} \)
x(0)=1 and x(1)=2, f(x0)=-8 and f(x1)=6
Hence x2= 1.571.
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