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

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:

h = x1 - x0 = 

Then 

f(x0, x1, x2, x3) = 

                       =  = 

Option (1) is true.

Concept:

The equation f(x) = 0 is convergent at x = a in iteration method if |ϕ'(x)|

Explanation:

x2 - x - 1 = 0....(i)

(1): 

⇒ x² = x - 1

⇒ x² - x + 1 = 0 not same as (i)

Option (1) is false.

(2): x = 2x - x2 + 1

⇒ - x2 - x + 1 = 0 not same as (i)

Option (2) is false.

(3): 

⇒ x2 = x + 1

⇒ x2 - x - 1 = 0 same as (i)

So, ϕ(x) = 

Then ϕ'(x) = 

Hence |ϕ'(x)|

Option (3) is true.

(1): x = x2 - 1

⇒ x2 - x - 1 = 0, same as (i)

ϕ(x) = x2 - 1

Then ϕ'(x) = 2x

Hence |ϕ'(-0.62)| = 1.24 not less than 1.

Option (4) is false.

Concept:

By Simpson's 3/8 rule

where h = 

Explanation:

From given data

a = 0, a + h = π / 6, a + 2h = π / 3, a + 4h = b = π / 2, 

h = π / 6 - 0 = π / 6

 = 

                = h[f(0) + 3f(π / 6) + 3f(π / 3) + f(π / 2)] 

              = [1+3 × 1.64872 + 3 × 2.3632 + 2.71828]

             ≈ 3.09329

Option (1) is true.

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