# Numerical Analysis Questions and Answers – Secant Method

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

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

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

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)=$$\frac{x(n)f[x(n)]-x(n-1)f[x(n-1)]}{[f[x(n)]-f[x(n-1)]]}$$

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

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.
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6. The number of iterations in Secant Method are less as compared to Newton Raphson.
a) True
b) False

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

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

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

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

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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