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 \( f(x) \) is degree ≤ 2 interpolating 3 points.
• The polynomial \( f(x) + g(x) \) is of degree ≤ 3 interpolating 4 points.
• We can compute both \( f(x) \) and \( h(x) = f(x) + g(x) \) using interpolation and then evaluate required values.

Calculation:

Given,

\( f(x) \) interpolates points: (-1, 2), (0, 1), (1, 2)

\( f(x) = x^2 + 1 \Rightarrow f(3) = 10, \, f(5) = 26 \)

From interpolation of \( f(x) + g(x) \) through: (-1, 2), (0, 1), (1, 2), (2, 17)

We obtain a cubic polynomial \( h(x) \) such that:

\( h(3) = 58, \, h(5) = 266 \)

⇒ \( g(3) = h(3) - f(3) = 58 - 10 = 48 \)

⇒ \( g(5) = h(5) - f(5) = 266 - 26 = 240 \)

Now evaluate the options:

Option 1: \( f(3) + g(3) = 10 + 48 = 58 \) Incorrect

Option 2: \( 2f(5) - g(3) = 2×26 - 48 = 52 - 48 = 4 \) Correct

Option 3: \( f(1) + g(3) = 2 + 48 = 50 \) Correct

Option 4: \( f(5) + g(3) = 26 + 48 = 74 \) 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: \( \int_{x_0}^{x_0+h} f(x)dx \approx \lambda h \left(f(x_0) + f(x_0+h)\right) + p h^3 \left(f''(x_0) + f''(x_0+h)\right) \)
• To determine \( \lambda \) and \( p \), compare both sides using the Taylor series expansion of \( f(x) \) around \( x_0 \).

Calculation:

Given,

Exact integral of \( f(x) \):

\( \int_{x_0}^{x_0+h} \left[ f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2}(x - x_0)^2 + \frac{f'''(x_0)}{6}(x - x_0)^3 \right] dx \)

⇒ \( f(x_0)h + \frac{f'(x_0)h^2}{2} + \frac{f''(x_0)h^3}{6} + \frac{f'''(x_0)h^4}{24} \)

Quadrature formula expansion using Taylor series:

\( \lambda h \left[ 2f(x_0) + h f'(x_0) + \frac{h^2 f''(x_0)}{2} + \frac{h^3 f'''(x_0)}{6} \right] + p h^3 \left[ 2 f''(x_0) + h f'''(x_0) \right] \)

Compare coefficients of each term:

For \( f(x_0) \): \( 2 \lambda h = h \) ⇒ \( \lambda = \frac{1}{2} \)

For \( f'(x_0) \): \( \lambda h^2 = \frac{h^2}{2} \) ⇒ Verified for \( \lambda = \frac{1}{2} \)

For \( f''(x_0) \): \( \lambda \frac{h^3}{2} + 2 p h^3 = \frac{h^3}{6} \)

⇒ \( \frac{\lambda}{2} + 2p = \frac{1}{6} \)

For \( f'''(x_0) \): \( \lambda \frac{h^4}{6} + p h^4 = \frac{h^4}{24} \)

⇒ \( \frac{\lambda}{6} + p = \frac{1}{24} \)

Solving:

From \( \frac{\lambda}{6} + p = \frac{1}{24} \):

⇒ \( p = \frac{1}{24} - \frac{\lambda}{6} \)

Substitute into \( \frac{\lambda}{2} + 2p = \frac{1}{6} \):

⇒ \( \frac{\lambda}{2} + 2 \left( \frac{1}{24} - \frac{\lambda}{6} \right) = \frac{1}{6} \)

⇒ \( \frac{\lambda}{2} + \frac{1}{12} - \frac{\lambda}{3} = \frac{1}{6} \)

⇒ \( \frac{\lambda}{6} + \frac{1}{12} = \frac{1}{6} \)

⇒ \( \frac{\lambda}{6} = \frac{1}{12} \)

⇒ \( \lambda = \frac{1}{2} \)

Substitute back: \( p = \frac{1}{24} - \frac{1}{12} = -\frac{1}{24} \)

Verification of Options:

Option 1: \( 2\lambda + 24p \):

⇒ \( 2 \times \frac{1}{2} + 24 \times \left(-\frac{1}{24}\right) = 1 - 1 = 0 \) Correct

Option 2: \( 7\lambda - 12p \):

⇒ \( 7 \times \frac{1}{2} - 12 \times \left(-\frac{1}{24}\right) = \frac{7}{2} + \frac{1}{2} = 4 \)  Correct

Option 3: \( 2\lambda + 24p \):

Already verified as 0, not -3. Incorrect

Option 4: \( 7\lambda - 12p \):

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 < c < 1 such that |g(x) − g(y)| ≤ c × |x − y| for all x, y.
• This implies |g'(x)| < 1 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)| < 1

⇒ αL + 3/4 < 1

⇒ αL < 1 − 3/4 = 1/4

⇒ α < 1/(4L)

∴ The sequence converges to the solution of x = g(x) if α < 1⁄(4L).

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:

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