Numerical Analysis MCQ Quiz in मराठी - Objective Question with Answer for Numerical Analysis - मोफत PDF डाउनलोड करा

Concept:

Newton's Forward Difference formula:

y(x) = y₀ + pΔy0 + \(\frac{p(p-1)}{2!}\) Δ²y0 + \(\frac{p(p-1)(p-2)}{3!}\)Δ3y0 + ....

where p = \(\frac{x-x_0}{h}\) 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 + \(\frac{(-2)(-3)}{2!}\) × 2.5 + \(\frac{(-2)(-3)(-4)}{3!}\)× 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

\(x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\)

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

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

(3) is correct

Concept:

Taylor series expansion of f(x + h) is

f(x + h) = f(x) + h f'(x) + \(\frac{h^2}{2!}\)f''(x) + \(\frac{h^3}{3!}\)f'''(x) + ... 

Explanation:

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

f'(x) = Af(x) + B{f(x) + h f'(x) + \(\frac{h^2}{2!}\)f''(x) + \(\frac{h^3}{3!}\)f'''(x) + ...} + C{f(x) + 2h f'(x) + \(\frac{(2h)^2}{2!}\)f''(x) + \(\frac{(2h)^3}{3!}\)f'''(x) + ...} 

f'(x) = (A + B + C)f(x) + (Bh + 2Ch)f'(x) + (\(\frac{Bh^2}{2}+\frac{4Ch^2}{2}\))f''(x) + (\(\frac{Bh^3}{6}\) + \(\frac{8Ch^3}{6}\))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 = -\(\frac5{6h}\), B = \(\frac3{2h}\), C = -\(\frac2{3h}\) 

A + B + C = \(\frac{-5+9-4}{6h}\) = 0

Bh + 2Ch = \(\frac32\) - \(\frac43\)  = 1/6 ≠ 1

Option (3) is false

(4):  A = \(\frac5{6h}\), B = \(\frac3{2h}\), C = \(\frac2{3h}\)

A + B + C = \(\frac{5+9+4}{6h}\) ≠ 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

\(\frac{dy}{y}\) = λdx 

Integrating

y = c₁e^λx 

so y_n approaches to eλx_n 

 yn approaches to eλnh 

(1), (3) correct

Explanation:

\(\int^1_{-1}(1 - x)f(x)dx = Af(-α) + Bf(0) + Cf(α)\)

for f(x) = 1 we get

\(\int^1_{-1}(1 - x)dx = A + B + C\\ [x-\frac{x^2}{2}]_{-1}^1=A+B+C\)

A + B + C = 2...(i)

for f(x) = x we get

\(\int^1_{-1}(x - \frac{x^2}{2})dx = -α A + 0 + Cα\\ [\frac{x^2}{2}-\frac{x^3}{6}]_{-1}^1=(-A+C)α\)

(A-C)α = \(\frac23\)....(ii)

Option (1): 

A + B + C = \(\frac{5}{9}+\frac{\sqrt5}{3\sqrt3}+\frac{8}{9}+\frac{5}{9}-\frac{\sqrt5}{3\sqrt3}\) = 2

and (A-C)α = \((\frac{5}{9}+\frac{\sqrt5}{3\sqrt3}-\frac{5}{9}+\frac{\sqrt5}{3\sqrt3})\sqrt{\frac{3}{5}}=\frac23\)

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

Option (1) is correct.

Option (2): 

A + B + C = \(\frac{5}{9}-\frac{\sqrt5}{3\sqrt3}+\frac{8}{9}+\frac{5}{9}+\frac{\sqrt5}{3\sqrt3}\) = 2

and (A-C)α = \((\frac{5}{9}-\frac{\sqrt5}{3\sqrt3}-\frac{5}{9}-\frac{\sqrt5}{3\sqrt3})\sqrt{\frac{3}{5}}=-\frac23\)

It is not satisfying  (ii).

Option (2) is not correct.

Option (3): 

A + B + C = \(\frac{5}{9}\left(1-\frac{\sqrt3}{\sqrt5}\right)+\frac{8}{9}+\frac{5}{9}\left(1+\frac{\sqrt3}{\sqrt5}\right)\) = 2

and (A-C)α = \(\left(\frac{5}{9}\left(1-\frac{\sqrt3}{\sqrt5}\right)-\frac{5}{9}\left(1+\frac{\sqrt3}{\sqrt5}\right)\right)\sqrt{\frac{3}{5}}=-\frac23\)

It is not satisfying  (ii).

Option (3) is not correct.

Option (4): 

A + B + C = \(\frac{5}{9}\left(1+\frac{\sqrt3}{\sqrt5}\right)+\frac{8}{9}+\frac{5}{9}\left(1-\frac{\sqrt3}{\sqrt5}\right)\) = 2

and (A-C)α = \(\left(\frac{5}{9}\left(1+\frac{\sqrt3}{\sqrt5}\right)-\frac{5}{9}\left(1-\frac{\sqrt3}{\sqrt5}\right)\right)\sqrt{\frac{3}{5}}=\frac23\)

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

Option (4) is correct.

Explanation:

Given \(\int_{-1}^1g(x)dx\) ≈  g(α) + g(-α), α = (0.2)¼ - (i)

(a) ∵  \(\rm \int_{-1}^1(a+bx)dx=\left[ax+\frac{bx^2}{2}\right]^1_{-1}\)

           = =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) ∵ \(\rm \int_{-1}^1(a+bx+cx^2)dx=\left[ax+\frac{bx^2}{2}+\frac{cx^3}{3}\right]^1_{-1}\)

= 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 ∈ \(\mathbb{R}\) any thing

\(\int_{-1}^1\) g(x) = 2a + 2c/3 = 2/3,  g(α) + g(-α) = 2(0.2)\(^{\frac{1}{4}}\)

which is NOT equal ⇒ opt (2) False

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

\(∵ \int_{-1}^1\left(a+b x+c x^2+d x^3\right) d x\)

\(=\left[a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}-(-a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4})\right]\)

\(=2 a+\frac{2 c}{3}\)

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): ∵ \(\int_{-1}^1\) (a + bx + cx3 + dx4)dx = \(\left[a x+\frac{b x^2}{2}+\frac{c x^3}{4}+\frac{d x^5}{5}\right]_{-1}^1\) 

\(=a+\frac{b}{2}+\frac{c}{4}+\frac{d}{5}-(-a+\frac{b}{2}+\frac{c}{4}+\frac{d}{5})\)

= 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 +\( \frac{2 d}{5}\) ⇒ op(4) true 

Correct statements are

\({\lim _{n \to \infty }}\left( {x_n^n,y_n^n} \right) = \left( {\cos \left( 1 \right),\sin \left( 1 \right)} \right)\)

\({\lim _{n \to \infty }}\left( {x_2^n,y_2^n} \right) = \left( {1,0} \right)\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) \({\left( {x_i^n} \right)^2} + {\left( {y_i^n} \right)^2} > 1\) for i ≥ 1

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

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:

f(x) = x2|x| at the points x = −1, 0, 1

f(x) = \(\begin{cases}x^3, x≥0\\-x^3, x\le0\end{cases}\)

So, 

Then Pf(x) = x²

Now let g(x) = f(x) - Pf(x) = x2|x| - x2 

So, g(x) = \(\begin{cases}x^3-x^2, x\geq0\\-x^3-x^2, x\leq0\end{cases}\)

g'(x) = \(\begin{cases}3x^2-2x, x\geq0\\-3x^2-2x, x\leq0\end{cases}\)

For critical point

g'(x) = 0 ⇒ x = \(\begin{cases}0, \frac23, x\geq0\\0, -\frac23, x\leq0\end{cases}\)

Now, g(0) = 0 

g(2/3) = \(\frac8{27}-\frac49=-\frac{4}{27}\)

and g(-2/3) = \(\frac8{27}-\frac49=-\frac{4}{27}\)

max |g(x)| = \(\frac{4}{27}\) ≈ 0.15

Hence max g(x) = \(\rm \displaystyle \sup _{x \in[-1,1]}\left|f(x)-P_f(x)\right|=0.15\) 

Answer is 0.15

Explanation:

Let a = 0 and h = 1 then the points are −1, 0, 1

So, 

Then Pf(x) = x2

So, g''(0) = P_g''(0) = 2

So, g''(x) = Pg''(x) = 2

then g(x) = x² + c₁x + c₂

So, g(x) is a polynomial of degree 2.

The value of d is equal to 2.