Numerical Analysis MCQ Quiz in मल्याळम - Objective Question with Answer for Numerical Analysis - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Given data is (−1, f(−1)), (0, f(0)) and (2, f(2)). 

f is an interpolating polynomial of degree 2

Let f(x) = ax² + bx + c

⇒ f'(x) = 2ax + b ⇒ f'(0) = b which is the coefficient of x

Now, using Lagrange's interpolation formula

f(x) = f(-1) + f(0) + f(2)

So, f'(0) is the coefficient of x

i.e., f'(0) = − f(−1) + f(0) + f(2).

Comparing with f′(0) ≈ − f(−1) + αf(0) + βf(2) we get

α = , β = 

So,  = 12

Option (4) is true

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:

Newton's Forward Difference formula:

y(x) = y₀ + pΔy0 +  Δ²y0 + Δ3y0 + ....

where p =  and h is the step length

Explanation:

From the given data we can write the difference table as

Here x₀ = 3, x = 1, h = 4 - 3 = 1 and so p = (1 - 3)/1 = - 2

Then using Newton's Forward Difference formula,

y(1) = 4.8+ (-2) × 3 +  × 2.5 + × 0.5 = 3.1

(3) is correct

Explanation:

Newton Raphson's method: Let x₀ be the initial guess of f(x) then the nth iteration to finding the root of f(x) is is given by

The order of convergence of Newton Raphson method is 2 or the convergence is quadratic.

It converges if |f(x).f’’(x)| 2. Also, this method fails if f’(x) = 0.

(3) is correct

Concept:

Trapezoidal rule: 

 ≈ 

where , x_n = b, x₀ = a

Explanation:

Here f(x) = ||x + 1| + |x - 1||, a = -2, b = 2, n = 4 as there are five points so intervals = 4

Dividing [-2, 2] into 4 subintervals we get endpoints as

a = -2, -1, 0, 1, 2 = b

f(x₀) = f(-2) = 4, f(x1) = f(-1) = 2

f(x2) = f(0) = 2, f(x3) = f(1) = 2, f(x4) = f(2) = 4

Then using the Trapezoidal rule

I =  = 10

(2) is correct

Concept:

Taylor series expansion of f(x + h) is

f(x + h) = f(x) + h f'(x) + f''(x) + f'''(x) + ... 

Explanation:

f'(x) = Af(x) + Bf(x + h)  +Cf(x + 2h)

f'(x) = Af(x) + B{f(x) + h f'(x) + f''(x) + f'''(x) + ...} + C{f(x) + 2h f'(x) + f''(x) + f'''(x) + ...} 

f'(x) = (A + B + C)f(x) + (Bh + 2Ch)f'(x) + ()f''(x) + ( + )f'''(x) + .....

In this expression, we can see that h² is associated with f''(x)

So options (1) and (2) are false.

Now, the magnitude of the truncation error will be of the form f''(x) if

A + B + C = 0, Bh + 2Ch = 1

(3): A = -, B = , C = - 

A + B + C =  = 0

Bh + 2Ch =  -   = 1/6 ≠ 1

Option (3) is false

(4):  A = , B = , C = 

A + B + C =  ≠ 0

Option (4) is false

All options are wrong here

Explanation:

 y' = λy; y(0) = 1

x_n = x0 + nh = 0 +nh = nh

y' = λy

 = λdx 

Integrating

y = c₁e^λx 

so y_n approaches to eλx_n 

 yn approaches to eλnh 

(1), (3) correct

Explanation:

for f(x) = 1 we get

A + B + C = 2...(i)

for f(x) = x we get

(A-C)α = ....(ii)

Option (1): 

A + B + C =  = 2

and (A-C)α = 

Hence it satisfies both equations (i) and (ii).

Option (1) is correct.

Option (2): 

A + B + C =  = 2

and (A-C)α = 

It is not satisfying  (ii).

Option (2) is not correct.

Option (3): 

A + B + C =  = 2

and (A-C)α = 

It is not satisfying  (ii).

Option (3) is not correct.

Option (4): 

A + B + C =  = 2

and (A-C)α = 

Hence it satisfies both equations (i) and (ii).

Option (4) is correct.

Explanation:

Given  ≈  g(α) + g(-α), α = (0.2)¼ - (i)

(a) ∵  

           = =a + b/2 - (-a + b/2) = 2a

while g(α) + g(-α) = (a + bα) + (a - bα) = 2a 

⇒ for a + bx form, (i) gives exact value

⇒ Opt (1) True

(b) ∵ 

= a + b/2 + c/3 - (-a + b/2 - c/3)

= 2a + 2c/3

While g(α) + g(-α)  = a + bα + cα2 + [a - bα + cα2]

= 2a + 2cα2 ≠ 2a + 2c/3

so, Not exact in this case.

Eg. Let a = 0, c = 1, b ∈  any thing

 g(x) = 2a + 2c/3 = 2/3,  g(α) + g(-α) = 2(0.2)

which is NOT equal ⇒ opt (2) False

Op(c). (Note: similar argument as (a) & (b)

While g(α) + g(-α) = a + bα + cα2 + dα3 + a - bα + cα2 - dα3 

=2 (a + cα2)

Again, in this case, both are not equal.

so opt (3) False.

Op(d): ∵  (a + bx + cx3 + dx4)dx =  

= 2a + 2d/5

while g(α) + g(-α) = a + bα + cα3 + dα4 + (a - bα - cα3 + dα4)

= 2a + 2dα4 = 2a + 2d(0.2)4/4

= 2a + 2d(0.2) = 2a + ⇒ op(4) true 

Correct statements are

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)  1\) for i ≥ 1

Options (2), (3) (4) are correct