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$f(x_0+\frac{h}{2},y_0+\frac{k_1}{2})$ 

Explanation:

Given 

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

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

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

   = 0.5 $f(0+0.25,1+0.25)$ 

  = 0.5 × f(0.25, 1.25)

 = 0.5 × $\frac{1}{1.50}$ = 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

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

$x_{n+1}=x_n-\frac{x_n}{3}+\frac{\alpha }{3x_n^2}$

$x_{n+1}=\frac{2}{3}{x}_n+\frac{\alpha }{3x_n^2}$

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

hence we get

$e_{n+1}+{\alpha }^{1/3}=\frac{2}{3}(e_n+{\alpha }^{1/3})+\frac{\alpha }{3{(e_n+{\alpha }^{1/3})}^2}$

$e_{n+1}\approx \frac{e_n^2}{e_n+2{\alpha }^{1/3}}$

(2) is correct

Explanation:

h = x1 - x0 = $\frac{1}{2}=({\mathrm{x}}_2-{\mathrm{x}}_1)=({\mathrm{x}}_3-{\mathrm{x}}_2)$

Then 

f(x0, x1, x2, x3) = $\frac{{\mathrm{△}}^{3}f(x_0)}{3!h^3}$

                       = $\frac{{\mathrm{△}}^{3}f(x_0)}{6\times \frac{1}{8}}$ = $\frac{4}{3}{\mathrm{\Delta }}^3\mathrm{f}({\mathrm{x}}_0)$

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): $\mathrm{x}=1-\frac{1}{\mathrm{x}}$

⇒ 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): $\mathrm{x}=\sqrt{\mathrm{x}+1}$

⇒ x2 = x + 1

⇒ x2 - x - 1 = 0 same as (i)

So, ϕ(x) = $\sqrt{x+1}$

Then ϕ'(x) = $\frac{1}{2\sqrt{x+1}}$

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

${\int }_a^{b}f(x)[f(a)+3f(a+h)+3f(a+2h)+f(b)]$

where h = $\frac{b-a}{3}$

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

${\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{\sin \mathrm{x}}\mathrm{d}\mathrm{x}$ = ${\int }_0^{\frac{\pi }{2}}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}$

                = $\frac{3}{8}$h[f(0) + 3f(π / 6) + 3f(π / 3) + f(π / 2)] 

              = $\frac{3}{8}\times \frac{\pi }{6}$[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

$\frac{2}{3}$ = 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 = $\frac{2}{3}$

2c - 2d = $\frac{4}{3}$ ⇒ c - d = $\frac{2}{3}$ ...(ix)

adding (viii) and (ix) we get

2c = $\frac{2}{3}$ so c = $\frac{1}{3}$

Hence d = - $\frac{1}{3}$    

Therefore we get a = 1, b = 1, c = $\frac{1}{3}$, d = $-\frac{1}{3}$

Option (1) is correct

The quantity

$\underset{x\in 1}{sup}|p_{2n-1}(x)-q_{2n-1}(x)|$ converges to 0 

Option (3) is correct

Concept:

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

Explanation:

For P,

M = $\left[\begin{array}{cc}2& -1\\ -4& 3\end{array}\right]$

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 = $\left[\begin{array}{cc}1& 0\\ a& 1\end{array}\right]$ and U = $\left[\begin{array}{cc}b& c\\ 0& d\end{array}\right]$ 

Then

$\left[\begin{array}{cc}2& -1\\ -4& 3\end{array}\right]$ = $\left[\begin{array}{cc}1& 0\\ a& 1\end{array}\right]$$\left[\begin{array}{cc}b& c\\ 0& d\end{array}\right]$

⇒ $\left[\begin{array}{cc}2& -1\\ -4& 3\end{array}\right]$ = $\left[\begin{array}{cc}b& c\\ ab& ac+d\end{array}\right]$

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 = $\left[\begin{array}{cc}2& -1\\ 0& 1\end{array}\right]$

Hence trace(U) = 1 + 2 = 3

P is TRUE

For Q,

M = $\left[\begin{array}{cc}2& -1\\ -4& 3\end{array}\right]$

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

Then H_Jacobi = D^-1(L + U)  where

D is diagonal matric ie..e, $\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]$ and L + U = $\left[\begin{array}{cc}0& -1\\ -4& 0\end{array}\right]$

So, D^-1 = $\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{3}\end{array}\right]$

So  HJacobi = $\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{3}\end{array}\right]$$\left[\begin{array}{cc}0& -1\\ -4& 0\end{array}\right]$ = $\left[\begin{array}{cc}0& -\frac{1}{2}\\ -\frac{4}{3}& 0\end{array}\right]$

Hence eigenvalues re given by

λ² - 0λ - 2/3 = 0 

⇒ λ = $\pm \sqrt{\frac{2}{3}}$

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:

$\frac{dy}{dx}=\sqrt{3x+2y+1},y(1)=1\text{, }$ 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)

$e_{n+1}+\surd \alpha =(e_{n+1}+\surd \alpha )-\frac{{(e_n+\surd \alpha )}^2-\alpha }{2(e_n+\surd \alpha )}$

$\Rightarrow e_{n+1}+\surd \alpha =\frac{2[e_n^2+\alpha +2e_n\surd \alpha ]-{(e_n+\surd \alpha )}^2+\alpha }{2(e_{n+}+\surd \alpha )}$

$e_{n+1}=\frac{e_n^2+\alpha +2e_n\sqrt{\alpha }+\alpha }{2(e_n+\surd \alpha )}-\surd \alpha$

$=\frac{e_n^2+\alpha +2e_n\sqrt{\alpha }+\alpha -2e_n\sqrt{\alpha }-2\alpha }{2(e_n+\sqrt{\alpha })}$

$=e_n^2/2(e_n+\sqrt{\alpha })$

(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 

${\int }_{-1}^{1}1dx$ = 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$f(x_0+\frac{h}{2},y_0+\frac{k_1}{2})$ 

Explanation:

Given 

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

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

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

   = 0.5 $f(0+0.25,1+0.25)$ 

  = 0.5 × f(0.25, 1.25)

 = 0.5 × $\frac{1}{1.50}$ = 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, $\frac{d^{4}q}{d{x}^4}$ at x = 0, will be 4! × coefficient of x⁴ 

So, $\frac{d^{4}q}{d{x}^4}$ = 24($\frac{-2}{30}+\frac{-3}{8}+\frac{5}{12}+\frac{1}{-12}+\frac{23}{120}$) 

             = - $\frac{8}{5}$ - 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

$\frac{2}{3}$ = 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 = $\frac{2}{3}$

2c - 2d = $\frac{4}{3}$ ⇒ c - d = $\frac{2}{3}$ ...(ix)

adding (viii) and (ix) we get

2c = $\frac{2}{3}$ so c = $\frac{1}{3}$

Hence d = - $\frac{1}{3}$    

Therefore we get a = 1, b = 1, c = $\frac{1}{3}$, d = $-\frac{1}{3}$

Option (1) is correct