Mathematical Methods of Physics MCQ Quiz in मराठी - Objective Question with Answer for Mathematical Methods of Physics - मोफत PDF डाउनलोड करा

Explanation:

⇒ The probability of drawing a white ball on the first draw is \(\frac{6}{10}\), or \(\frac{3}{5}\),

since there are 6 white balls out of a total of 10 balls in the bag.

⇒ Assuming a white ball is drawn first, the probability of drawing a red ball on the second draw is \(\frac{4}{9}\), since there are 4 red balls left out of a total of 9 balls in the bag (one white ball has already been drawn and removed).

⇒ Assuming a white ball is drawn first, and a red ball is drawn second, the probability of drawing a white ball on the third draw is \(\frac{5}{8}\), since there are 5 white balls left out of a total of 8 balls in the bag (one white ball and one red ball have already been drawn and removed).

⇒ Therefore, the overall probability of drawing a white ball on the first draw, a red ball on the second draw, and a white ball on the third draw is:

⇒ \(\frac{3}{5}\) \(\times\) \(\frac{4}{9}\) \(\times\) \(\frac{5}{8}\) = .\(\frac{1}{6}\)

So, the correct option is \(\frac{1}{6}\) .

Concept:

The linear algebra is the study of linear equations and their representation in the vector space. 

Explanation:

The divergence of a vector field \(\vec{F} = \langle F_1, F_2, F_3\rangle\) in \(\mathbb{R}^3 \) is given by the formula: \(\text{div}(\vec{F}) = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}.\)

In this case, we have \(F_1(x,y) = x^3 \), \(F_2(x,y) = 3x^2y \), and  \(F_3(x,y) = 3xy^2 \).

Taking partial derivatives with respect to x, y and z we get: \(\frac{\partial F_1}{\partial x} = 3x^2, \quad \frac{\partial F_2}{\partial y} = 3x^2, \quad \frac{\partial F_3}{\partial z} = 0.\)  Now putting these in the expression of  \(\text{div}(\vec{F}) \) given above we have.

the divergence of \(\vec{F} \) at the point \((1,1)\) is given as: \(\text{div}(\vec{F})(1,1) = \frac{\partial F_1}{\partial x}(1,1) + \frac{\partial F_2}{\partial y}(1,1) + \frac{\partial F_3}{\partial z}(1,1) = 3 + 3 = 6.\)

The correct option is option (1) 6.

Explanation:

f, g entire function such that \(\rm\displaystyle\lim_{z \rightarrow \infty} \frac{f(z)}{z^n}=\lim_{z \rightarrow \infty} \frac{g(z)}{z^{n}}\) = 1 (for some n ∈ N)

Now

(1) Let n = 1, f = z, and g(z) = z + 1

\(\rm\displaystyle\lim_{z \rightarrow \infty} \frac{f(z)}{z} = \frac{z}{z} = 1 \) 

 \(\lim_{z \rightarrow \infty} \frac{g(z)}{z} = \frac{z+1}{z} = 1 + \frac{1}{z}\)= 1

But f(z) ≠ g(z)

Option (1) is false.

option (3) and option (4):

Replacing z by 1/z we get

\(\rm\displaystyle\lim_{z \rightarrow \infty} \frac{f(z)}{z^n} = \lim_{z \rightarrow 0} z^n f\left(\frac{1}{z}\right)\) = 1 ≠ 0

⇒ f(1/z) has a pole of order n at 0

⇒ f(z) has pole of order n at z = ∞

As we know that entire fn has a pole if and only if it is a non-constant polynomial and order of pole is degree of polynomial.

⇒ f(z) = a0 + a1z + a2z2 + … + an−1zn−1 + zn

similarly, g(z) = b0 + b1z + b2z2 + … + bn−1zn−1 + zn

⇒ f − g = (a0 − b0) + (a1 − b1)z + … + (an−1 − bn−1)zn−1 + 0

So f − g is a polynomial of deg. n−1

⇒ option (3) and option (4) are false

Hence option (2) is true

Concept:

Trapezoidal Rule is a rule that evaluates the area under the curves by dividing the total area into smaller trapezoids rather than using rectangles. This integration works by approximating the region under the graph of a function as a trapezoid, and it calculates the area. This rule takes the average of the left and the right sum.

Calculation:

Given:

h = 1/2

The Trapezoidal rule states

I = \({h\over 2}\)[y₀+y_n+2(y₁+y₂+....)]

\({5\over16}\)= \({1 \over 4}\)[1-\({5\over 4}\)+2(0+a)]

\({5\over 4}\) = [\(1- {5 \over 4} +2a\)]

2a = \({10 \over 4} - 1\)

a = 3/4

The correct answer is option (1).

Explanation:

Here, we have:

\( f(z) = \cosh z \)

The derivatives of f(z) are:

• \( f'(z) = \sinh z \)
• \( f''(z) = \cosh z \)
• \( f'''(z) = \sinh z \)

At z = πi  :

• \( f(π i) = \cosh (π i) = \frac{e^{π i} + e^{-π i}}{2} \)
• \( f'(π i) = \sinh (π i) = \frac{e^{π i} - e^{-π i}}{2} \)
• \( f''(π i) = \cosh (π i) \)
• \( f'''(π i) = \sinh (π i) \)

Using Taylor's series expansion around z = πi :

\( f(z) = f(π i) + (z - π i) f'(π i) + \frac{(z - π i)^2}{2!} f''(π i) + \frac{(z - π i)^3}{3!} f'''(π i) + \ldots \)

Substituting the values:

\( f(z) = \cosh (π i) + (z - π i) \sinh (π i) + \frac{(z - π i)^2}{2!} \cosh (π i) + \frac{(z - π i)^3}{3!} \sinh (π i) + \ldots \)

The third term is \(\frac{(z - π i)^2}{2!} \cosh (π i)\)

The correct option is 1).

Concept:

To determine the number of zeros of a polynomial f(z) inside a given contour, we can use various tools from complex analysis, such as Rouche's theorem. Rouche's theorem is particularly useful for comparing two functions to establish the number of zeros inside a contour.

Rouche's Theorem: If two holomorphic functions f and g satisfy \( |f(z) - g(z)| < |g(z)| \ \) on a simple closed contour C, then f and g have the same number of zeros inside C.

Solution:

We are given the function \(f(z) = z^3 - 6z + 2 \ \) and the contour \(|z| = 2\ \). To apply Rouche's theorem, we choose a simpler function g(z) to compare with f(z).

Step 1: Choose g(z) = z³. Now we check the inequality \( |f(z) - g(z)| < |g(z)| \ \) on |z| = 2:

\(f(z) = z^3 - 6z + 2 \\ g(z) = z^3 \\ f(z) - g(z) = (z^3 - 6z + 2) - z^3 = -6z + 2 \ \ \)

So, we need to check \( |-6z + 2| < |z^3| \ \) on |z| = 2:\( |-6z + 2| = |6z - 2| = 6|z| - 2 = 6 \cdot 2 + 2 = 12 - 2 = 10 \ \ \text{and}\ \ |z^3| = |2^3| = 8 \)Clearly, this inequality does not hold since 10 is not less than 8.

Step 2: Try another choice. Let's choose \(g(z) = -6z \ \).

\( f(z) = z^3 - 6z + 2 \\ g(z) = -6z \\ f(z) - g(z) = (z^3 - 6z + 2) + 6z = z^3 + 2 \ \ \)

So, we need to verify \( |z^3 + 2| < |-6z| \ \ \) on |z| = 2: \( |z^3 + 2| = |2^3 + 2| = |8 + 2| = 10 \ \ \text{and} \ \ |-6z| = 6|z| = 6 \cdot 2 = 12 \ \ \) Since 10 < 12, the inequality \(|z^3 + 2| < 6|z| \ \) holds on |z| = 2.

Hence, By Rouche's theorem, the function \(f(z) = z^3 - 6z + 2 \ \) and the function \(g(z) = -6z \ \) have the same number of zeros inside |z| = 2. Since, g(z) = -6z has exactly one zero, f(z) must also have one zero inside the contour |z| = 2.

So, the correct answer is option 2.

Concept:

• The Legendre polynomial of degree (n), denoted as \((P_n(x)),\) is defined by the formula: \( [P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}[(x^2 - 1)^n] ]\)
• This formula is called Rodrigues's formula for Legendre polynomials.
• The first few Legendre polynomials are given by:

1. \(P_0(x) = 1\)
2. \(P_1(x) = x\)
3. \(P_2(x) = \frac{1}{2}(3x^2 - 1)\)
4. \(P_3(x) = \frac{1}{2}(5x^3 - 3x)\)
5. \(P_4(x) = \frac{1}{8}(35x^4 - 30x^2 + 3)\)

Explanation:

Substituting x = -1 in the polynomial P₃: 

\(P_3(x) = \frac{1}{2}(5(-1)^3 - 3(-1))\)

\(P_3(x) = \frac{1}{2}(-5 + 3)\)

\(P_3(x) = \frac{1}{2}(-2)\)

\(P_3(x) = -1\)

Explanation:

The trapezoidal rule is a method in numerical integration used to approximate definite integrals. To evaluate the integral \(∫\frac{1}{(1+x)}dx\) from 0 to 2 using the trapezoidal rule with four intervals, we first determine the width of the intervals.

The interval [0,2] is divided into 4 subintervals, so the width of each interval, \(h= \frac{(2 - 0)}{ 4} = 0.5\)

Our x points are: \(x₀ = 0, x₁ = 0.5, x₂ = 1.0, x₃ = 1.5, x₄ = 2.0.\)

Now, applying the trapezoidal rule:

\(∫\frac{1}{(1+x)}dx ≈ \frac h2[f(x₀) + 2f(x₁) + 2f(x₂) + 2f(x₃) + f(x₄)]\)

Substitute the values:

\(≈ \frac{0.5} 2 [f(0) + 2f(0.5) + 2f(1.0) + 2f(1.5) + f(2.0)]\)

In this case, \(f(x) = \frac1 { (1 + x)}\), so calculate the values for f(x) at each point:

\(≈ 0.25 [\frac1{(1+0)} + 2\frac{1}{(1+0.5)} + 2\frac{1}{(1+1.0)} + 2\frac{1}{(1+1.5)} + \frac{1}{(1+2)}]\)

\(≈0.25[1+1.33+1+0.8+0.33]\)

\(≈0.25\times4.46\)

\(≈1.115\)

Explanation:

Given:

• n (Number of trials): 5
• k (Number of successes): 3
• p (Probability of success): \(\frac12\)
• The formula for binomial probability is: \( P(X=k) = C(n, k) \times (p^k) \times ((1-p)^{(n-k)})\)
• We can substitute the values into this formula: \(P(X=3) = C(5, 3) \times (\frac12)^3 \times (\frac12)^2\)
• Calculating the binomial coefficient : \(C(5, 3): C(n, k) = \frac{n!}{k! \times (n-k)!} = \frac{5!}{3! \times (5-3)!} = 10\)
• Substitute this into the equation: \(P(X=3) = 10 \times (\frac12)^3 \times (\frac12)^2 = 10 \times (\frac18) \times (\frac14) = 10 \times (\frac1{32}) = \frac{10}{32}=\frac{5}{16}\)
• So, the probability that the captain will win 3 times and lose 2 times in a series of 5 matches is : \(\frac{5}{16} \)

Concept:

(i) A square matrix is said to be a triangular matrix if it is similar to a triangular matrix

(ii): Let A be a square matrix whose characteristic polynomial factors into linear polynomials, then A is similar to a triangular matrix i.e., there exists an invertible matrix P such that P^-1AP is triangular.

Explanation:

 A ∈ M3(ℝ) and X = {C ∈ GL3(ℝ) | CAC-1 is triangular}

So CAC-1 is similar to A

then CAC-1 is triangular if and only if A is triangularizable

Thus if A is not triangularizable then A = ϕ

(1) is false

The characteristic polynomial of A is of degree 3

So it has at least one real root

(4) is false

If X = Ø then the characteristic polynomial of A has 3 distinct roots on ℂ 

So A is diagonalizable over ℂ 

(3) is correct, (2) is false