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 x^{2}-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(x

_{0})=1 and f(x

_{1})=0.25

Hence x

_{2}=2.33.

8. A quadratic equation x^{2}-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(x

_{0})=0 and f(x

_{1})=0.25

Hence x

_{2}=2. Hence for x

_{2}=2 f(x) converges to exact root.

9. A quadratic equation x^{2}-4x+4=0 is defined with an initial guess of 10 and 20. Find the approximated value of x_{2} 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(x

_{0})=0 and f(x

_{1})=0.25

Hence x

_{2}= 7.538.

10. A quadratic equation x^{4}-x-8=0 is defined with an initial guess of 1 and 2. Find the approximated value of x_{2} 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(x

_{0})=-8 and f(x

_{1})=6

Hence x

_{2}= 1.571.

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