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\(f\left(x_0+\frac h2, y_0+\frac {k_1}2\right)\) 

Explanation:

Given 

\({dy\over dx}={1\over x+y}\), x0 = 0, y(x0) = 1, h = 0.5. 

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

k2 = h\(f\left(x_0+\frac h2, y_0+\frac {k_1}2\right)\) 

   = 0.5 \(f\left(0+0.25, 1+0.25\right)\) 

  = 0.5 × f(0.25, 1.25)

 = 0.5 × \(1\over 1.50\) = 0.333

Option (2) and (3) are true. 

Concept:

Iterative formula for NR-method

\(x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime}\left(x_n\right)}\) 

Explanation

Iterative formula for NR-method

\(x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime}\left(x_n\right)}\) --- (i)

Let f(x) = x³ - α 

⇒ f'(x) = 3x²

So, putting in (i) we get

\(x_{n+1}=x_n-\frac{\left(x_n^3-\alpha\right)}{3x_n^2}\)

\(x_{n+1}=x_n-\frac{x_n}{3}+{\alpha\over 3x_n^2}\)

\(x_{n+1}=\frac23x_n+{\alpha\over 3x_n^2}\)

and given en = xn - \(\alpha^{1/3}\) ⇒ xn = en + \(\alpha^{1/3}\).

hence we get

\(e_{n+1}+\alpha^{1/3} =\frac23\left(e_{n}+\alpha^{1/3}\right)+\frac{α}{3\left(e_n+\alpha^{1/3}\right)^2}\)

\(e_{n+1}\approx {e_n^2\over e_n+2\alpha^{1/3}}\)

(2) is correct

Explanation:

h = x1 - x0 = \(\rm \frac{1}{2}=(x_2-x_1)=(x_3-x_2)\)

Then 

f(x0, x1, x2, x3) = \(\triangle^3f(x_0)\over 3!h^3\)

                       = \(\triangle^3f(x_0)\over 6\times\frac18\) = \(\rm \frac{4}{3}\Delta^3f(x_0)\)

Option (1) is true.

Concept:

The equation f(x) = 0 is convergent at x = a in iteration method if |ϕ'(x)| < 1 in the neighbourhood of x = a where x = ϕ(x) is obtained form f(x) = 0.

Explanation:

x2 - x - 1 = 0....(i)

(1): \(\rm x=1-\frac{1}{x}\)

⇒ 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): \(\rm x=\sqrt{x+1}\)

⇒ x2 = x + 1

⇒ x2 - x - 1 = 0 same as (i)

So, ϕ(x) = \(\sqrt{x+1}\)

Then ϕ'(x) = \(1\over2\sqrt{x+1}\)

Hence |ϕ'(x)| < 1 in the neighbourhood of x = -0.62

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

\(\int_a^bf(x)\left[f(a)+3f(a+h)+3f(a+2h)+f(b)\right]\)

where h = \(b-a\over3\)

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

\(\rm \int_0^{\frac{π}{2}}e^{\sin x}dx\) = \(\rm \int_0^{\frac{π}{2}}f(x)dx\)

                = \(\frac38\)h[f(0) + 3f(π / 6) + 3f(π / 3) + f(π / 2)] 

              = \(\frac38× {\pi\over 6}\)[1+3 × 1.64872 + 3 × 2.3632 + 2.71828]

             ≈ 3.09329

Option (1) is true.

Explanation:

Given 

\(\displaystyle\int_{−1}^1\) f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)....(i)

Let f(x) = 1 then from (i)

\(\displaystyle\int_{−1}^1\)dx = a + b 

⇒ 2 = a + b...(ii)

For f(x) = x ⇒ f'(x) = 1 so from (i)

\(\displaystyle\int_{−1}^1\)x dx = -a + b + c + d

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

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

\(\displaystyle\int_{−1}^1\)x² dx = a + b - 2c + 2d

\(\frac23\) = a + b - 2c + 2d....(iv)

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

\(\displaystyle\int_{−1}^1\)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 = \(\frac23\)

2c - 2d = \(\frac43\) ⇒ c - d = \(\frac23\) ...(ix)

adding (viii) and (ix) we get

2c = \(\frac23\) so c = \(\frac13\)

Hence d = - \(\frac13\)    

Therefore we get a = 1, b = 1, c = \(\frac{1}{3}\), d = \(−\frac{1}{3}\)

Option (1) is correct

The quantity

\(\sup _{x \in 1}\left|p_{2 n-1}(x)-q_{2 n-1}(x)\right|\) converges to 0 

Option (3) is correct

Concept:

A square matrix (a_ij) is called a diagonally dominant matrix if |a_ii| ≥ \(\sum_{j\neq i}|a_{ij}|\) for all i

Explanation:

For P,

M = \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\)

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 = \(\begin{bmatrix}1&0\\a&1\end{bmatrix}\) and U = \(\begin{bmatrix}b&c\\0&d\end{bmatrix}\) 

Then

\(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\) = \(\begin{bmatrix}1&0\\a&1\end{bmatrix}\)\(\begin{bmatrix}b&c\\0&d\end{bmatrix}\)

⇒ \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\) = \(\begin{bmatrix}b&c\\ab&ac+d\end{bmatrix}\)

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 = \(\begin{bmatrix}2&-1\\0&1\end{bmatrix}\)

Hence trace(U) = 1 + 2 = 3

P is TRUE

For Q,

M = \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\)

M is not a diagonally dominant matrix as 3 \(\ngeq\) |-4|

Then H_Jacobi = D^-1(L + U)  where

D is diagonal matric ie..e, \(\begin{bmatrix}2&0\\\ 0&3\end{bmatrix}\) and L + U = \(\begin{bmatrix}0&-1\\\ -4&0\end{bmatrix}\)

So, D^-1 = \(\begin{bmatrix}\frac12&0\\\ 0&\frac13\end{bmatrix}\)

So  HJacobi = \(\begin{bmatrix}\frac12&0\\\ 0&\frac13\end{bmatrix}\)\(\begin{bmatrix}0&-1\\\ -4&0\end{bmatrix}\) = \(\begin{bmatrix}0&-\frac12\\\ -\frac43&0\end{bmatrix}\)

Hence eigenvalues re given by

λ² - 0λ - 2/3 = 0 

⇒ λ = \(\pm\sqrt{\frac23}\)

Since |λ| < 1

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:

\(\frac{d y}{d x}=\sqrt{3 x+2 y+1}, \quad y(1)=1 \text {, }\) 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

\(x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime}\left(x_n\right)}\) --- (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)

\(e_{n+1}+√{α} =\left(e_{n+1}+√{α}\right)-\frac{\left(e_n+√{α}\right)^2-α}{2\left(e_n+√{α}\right)}\)

\(\Rightarrow e_{n+1}+√{α} =\frac{2\left[e_n^2+α+2 e_n √{α}\right]-\left(e_n+√{α}\right)^2+α}{2\left(e_{n+}+√{α}\right)}\)

\(e_{n+1} =\frac{e_n^2+α+2 e_n \sqrtα+α}{2\left(e_n+√{α}\right)}-√{α}\)

\(=\frac{e_n^2+α+2 e_n \sqrt{α}+α-2 e_n \sqrt{α}-2 α}{2\left(e_n+\sqrt{α}\right)}\)

\(=e_n^2 / 2\left(e_n+\sqrt{α}\right)\)

(3) is correct

Concept:

Basic ideas of polynomial .

Calculation:

Given 

\(\int_{ - 1}^1 f(x)dx\)  = 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 

\(\int_{ - 1}^1 1 dx \) = a + 0 + 0

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

let f(x) = x 

now from equation (i) , we get 

\(\int_{ - 1}^1 xdx\) = -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\(f\left(x_0+\frac h2, y_0+\frac {k_1}2\right)\) 

Explanation:

Given 

\({dy\over dx}={1\over x+y}\), x0 = 0, y(x0) = 1, h = 0.5. 

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

k2 = h\(f\left(x_0+\frac h2, y_0+\frac {k_1}2\right)\) 

   = 0.5 \(f\left(0+0.25, 1+0.25\right)\) 

  = 0.5 × f(0.25, 1.25)

 = 0.5 × \(1\over 1.50\) = 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) = \(\frac{(x+1)(x-0)(x-1)(x-3)}{(-2+1)(-2-0)(-2-1)(-2-3)}\) × (-2) + \(\frac{(x+2)(x-0)(x-1)(x-3)}{(-1+2)(-1-0)(-1-1)(-1-3)}\) × 3 + \(\frac{(x+2)(x+1)(x-1)(x-3)}{(0+2)(0+1)(0-1)(0-3)}\) × 2.5 + \(\frac{(x+2)(x+1)(x-0)(x-3)}{(1+2)(1+1)(1-0)(0-3)}\) × 1 + \(\frac{(x+2)(x+1)(x-0)(x-1)}{(3+2)(3+1)(3-0)(3-1)}\) × 23

Now, \(\frac{d^4q}{dx^4}\) at x = 0, will be 4! × coefficient of x⁴ 

So, \(\frac{d^4q}{dx^4}\) = 24(\({-2\over30}+{-3\over8}+{5\over12}+{1\over-12}+{23\over120}\)) 

             = - \(\frac85\) - 9 + 10 - 2 + \(\frac{23}5\) = 2

Hence answer is 2

Explanation:

Given 

\(\displaystyle\int_{−1}^1\) f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)....(i)

Let f(x) = 1 then from (i)

\(\displaystyle\int_{−1}^1\)dx = a + b 

⇒ 2 = a + b...(ii)

For f(x) = x ⇒ f'(x) = 1 so from (i)

\(\displaystyle\int_{−1}^1\)x dx = -a + b + c + d

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

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

\(\displaystyle\int_{−1}^1\)x² dx = a + b - 2c + 2d

\(\frac23\) = a + b - 2c + 2d....(iv)

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

\(\displaystyle\int_{−1}^1\)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 = \(\frac23\)

2c - 2d = \(\frac43\) ⇒ c - d = \(\frac23\) ...(ix)

adding (viii) and (ix) we get

2c = \(\frac23\) so c = \(\frac13\)

Hence d = - \(\frac13\)    

Therefore we get a = 1, b = 1, c = \(\frac{1}{3}\), d = \(−\frac{1}{3}\)

Option (1) is correct