Numerical Analysis MCQ Quiz - Objective Question with Answer for Numerical Analysis - Download Free PDF

Concept:

Polynomial Interpolation:

• A unique polynomial of degree ≤ n passes through (n+1) points.
• We are given that is degree ≤ 2 interpolating 3 points.
• The polynomial is of degree ≤ 3 interpolating 4 points.
• We can compute both and using interpolation and then evaluate required values.

Calculation:

Given,

interpolates points: (-1, 2), (0, 1), (1, 2)

From interpolation of through: (-1, 2), (0, 1), (1, 2), (2, 17)

We obtain a cubic polynomial such that:

⇒

⇒

Now evaluate the options:

Option 1: Incorrect

Option 2: Correct

Option 3: Correct

Option 4: Correct

∴ The correct options are: Option 2 , Option 3 and Option 4. 

Concept:

Quadrature Formula and Degree of Exactness:

• A quadrature formula is a numerical method to approximate the definite integral of a function over an interval.
• The degree of exactness of a quadrature formula is the highest degree of polynomial for which the formula integrates exactly.
• Given formula is:
• To determine and , compare both sides using the Taylor series expansion of around .

Calculation:

Given,

Exact integral of :

⇒

Quadrature formula expansion using Taylor series:

Compare coefficients of each term:

For : ⇒

For : ⇒ Verified for

For :

⇒

For :

⇒

Solving:

From :

⇒

Substitute into :

⇒

⇒

⇒

⇒

⇒

Substitute back:

Verification of Options:

Option 1: :

⇒ Correct

Option 2: :

⇒  Correct

Option 3: :

Already verified as 0, not -3. Incorrect

Option 4: :

Already verified as 4, not 11. Incorrect

∴ The correct answers are 1 and 2.

Concept:

Fixed Point Iteration Method:

• In numerical analysis, the fixed point iteration method is used to find a solution of the equation x = g(x).
• The method defines a sequence using the recursive formula: x_k+1 = g(x_k).
• The sequence converges to a fixed point if g(x) is a contraction mapping.
• Contraction Mapping: A function g(x) is a contraction if there exists a constant 0 such that |g(x) − g(y)| ≤ c × |x − y| for all x, y.
• This implies |g'(x)| for convergence.
• Given Functions:
  • g(x) = αf(x) + h(x)
  • With f(x) being Lipschitz with constant L i.e., |f'(x)| ≤ L
  • And h(x) having derivative bound |h'(x)| ≤ 3/4
• Objective: Find the condition on α such that the iteration converges to the solution of x = g(x)

Calculation:

Given,

Lipschitz constant of f(x) = L

Bound on |h'(x)| = 3/4

Function: g(x) = αf(x) + h(x)

Derivative: g'(x) = αf'(x) + h'(x)

⇒ |g'(x)| ≤ |αf'(x)| + |h'(x)|

⇒ |g'(x)| ≤ αL + 3/4

⇒ For convergence, we need |g'(x)|

⇒ αL + 3/4

⇒ αL

⇒ α

∴ The sequence converges to the solution of x = g(x) if α

Concept:

Let y' = f(x, y) then using Runge-Kutta method

k₁ = hf(x₀, y₀) and

k₂ = h 

Explanation:

Given 

, x0 = 0, y(x0) = 1, h = 0.5. 

k₁ = hf(x0, y0) = 0.5 × f(0, 1) = 0.5 × 1 = 0.5

k2 = h 

   = 0.5  

  = 0.5 × f(0.25, 1.25)

 = 0.5 ×  = 0.333

Option (2) and (3) are true. 

Concept:

Iterative formula for NR-method

Explanation

Iterative formula for NR-method

 --- (i)

Let f(x) = x³ - α 

⇒ f'(x) = 3x²

So, putting in (i) we get

and given en = xn -  ⇒ xn = en + .

hence we get

(2) is correct

Explanation:

Given 

 f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)....(i)

Let f(x) = 1 then from (i)

dx = a + b 

⇒ 2 = a + b...(ii)

For f(x) = x ⇒ f'(x) = 1 so from (i)

x dx = -a + b + c + d

0 = -a + b + c + d....(iii)

For f(x) = x² ⇒ f'(x) = 2x so from (i)

x² dx = a + b - 2c + 2d

 = a + b - 2c + 2d....(iv)

and for f(x) = x³ ⇒ f'(x) = 3x² so from (i)

x3 dx = -a + b + 3c + 3d

0 = -a + b + 3c + 3d....(v)

Multiply (iii) by 3 we get

-3a  +3b + 3c + 3d = 0...(vi)

Subtract (v) and (vi) we get

2a - 2b = 0 ⇒ a = b...(vii)

Putting a = b in (ii)

2a = 2 ⇒ a = 1

Hence a = b = 1

Putting these values of a and b in (iii) and (iv) we get

c + d = 0 ...(viii) and 

2 - 2c + 2d = 

2c - 2d =  ⇒ c - d =  ...(ix)

adding (viii) and (ix) we get

2c =  so c = 

Hence d = -     

Therefore we get a = 1, b = 1, c = , d = 

Option (1) is correct

The quantity

 converges to 0 

Option (3) is correct

Concept:

A square matrix (a_ij) is called a diagonally dominant matrix if |a_ii| ≥  for all i

Explanation:

For P,

M = 

Let M = LU, where 𝐿 and U are lower triangular and upper triangular square matrices

Each element of the main diagonal of 𝐿 is 1

Let L =  and U =  

Then

 = 

⇒  = 

Comparing both sides

b = 2, c = -1, ab = -4 and ac + d = 3

Substituting b = 2 in ab = -4 we get a = -2

Again substituting a = -2 and c = -1 in ac + d = 3 we get

(-2)(-1) + d = 3 ⇒  2 + d = 3 ⇒ d = 1

So U = 

Hence trace(U) = 1 + 2 = 3

P is TRUE

For Q,

M = 

M is not a diagonally dominant matrix as 3  |-4|

Then H_Jacobi = D^-1(L + U)  where

D is diagonal matric ie..e,  and L + U = 

So, D^-1 = 

So  HJacobi =  = 

Hence eigenvalues re given by

λ² - 0λ - 2/3 = 0 

⇒ λ = 

Since |λ|

Hence for any choice of the initial vector 𝑥(0) , the Jacobi iterates 𝑥(𝑘) , 𝑘 = 1,2,3 … converge to the unique solution of the linear system 𝑀𝑥 = 𝑏.

Q is TRUE

both 𝑃 and 𝑄 are TRUE

(1) is correct

Concept:

Euler method to solve y' = f(x, y), y(x₀) = y₀, with step size h is

y_n+1 = y_n + h.[f(x_n, y_n)]

Explanation:

 h = 0.05, 

x₀ = 1, y₀ = 1, h = 0.05

x₁ = x₀ + h = 1 + 0.05 = 1.05

x₂ = x₀ + 2h = 1 + 2(0.05) = 1 + 0.1 = 1.1

So we need to find y(x₂) = y₂

y₁ = y₀ + h. f(x₀, y₀) = 1 + 0.05 × f(1, 1) = 1.12247

y₂ = y₁ + h. f(x₁, y₁) = 1.12247 + 0.05 × f(1.05, 1.12247) = 1.248 ≈ 1.25 rounded off to two decimal places

Hence option (3) is correct

Explanation

Recall: Iterative formula for NR-method

 --- (i)

Nos let x = √α ⇒ x2 - α = 0

so assume that f(x) = x2 - α

⇒ f'(x) = 2x

and given en = xn - √α ⇒xn = en + √α

so, by (i)

(3) is correct

Concept:

Basic ideas of polynomial .

Calculation:

Given 

  = af(-1) + bf'(0) + cf'(1)  . . . . . . (i)

is exact for all polynomials of degree less than or equal to 2

let f(x) = 1

So from equation (i), we get 

 = a + 0 + 0

2 = a . . . . . . . (ii)

let f(x) = x 

now from equation (i) , we get 

 = -a + b + c

b + c = 2 . . . . . . (iii)

now adding equation (ii) and (iii) , we get 

a + b + c = 2 + 2

a + b + c = 4

Hence option (1) correct

Concept:

Let y' = f(x, y) then using Runge-Kutta method

k₁ = hf(x₀, y₀) and

k₂ = h 

Explanation:

Given 

, x0 = 0, y(x0) = 1, h = 0.5. 

k₁ = hf(x0, y0) = 0.5 × f(0, 1) = 0.5 × 1 = 0.5

k2 = h 

   = 0.5  

  = 0.5 × f(0.25, 1.25)

 = 0.5 ×  = 0.333

Option (2) and (3) are true. 

Concept:

Explanation:

𝑝(𝑥) = 𝑥3 − 2𝑥 + 2

Given 

Now, p(-2) = -8 + 4 + 2 = -2; p(-1) = -1 + 2 + 2 = 3

p(1) = 1 - 2 + 2 = 1, p(3) = 27 - 6 + 2 = 23

Then the given data becomes

Then using Lagrange interpolating polynomial

q(x) =  × (-2) +  × 3 +  × 2.5 +  × 1 +  × 23

Now,  at x = 0, will be 4! × coefficient of x⁴ 

So,  = 24() 

             = -  - 9 + 10 - 2 +  = 2

Hence answer is 2

Explanation:

Given 

 f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)....(i)

Let f(x) = 1 then from (i)

dx = a + b 

⇒ 2 = a + b...(ii)

For f(x) = x ⇒ f'(x) = 1 so from (i)

x dx = -a + b + c + d

0 = -a + b + c + d....(iii)

For f(x) = x² ⇒ f'(x) = 2x so from (i)

x² dx = a + b - 2c + 2d

 = a + b - 2c + 2d....(iv)

and for f(x) = x³ ⇒ f'(x) = 3x² so from (i)

x3 dx = -a + b + 3c + 3d

0 = -a + b + 3c + 3d....(v)

Multiply (iii) by 3 we get

-3a  +3b + 3c + 3d = 0...(vi)

Subtract (v) and (vi) we get

2a - 2b = 0 ⇒ a = b...(vii)

Putting a = b in (ii)

2a = 2 ⇒ a = 1

Hence a = b = 1

Putting these values of a and b in (iii) and (iv) we get

c + d = 0 ...(viii) and 

2 - 2c + 2d = 

2c - 2d =  ⇒ c - d =  ...(ix)

adding (viii) and (ix) we get

2c =  so c = 

Hence d = -     

Therefore we get a = 1, b = 1, c = , d = 

Option (1) is correct