Mathematical Science MCQ Quiz - Objective Question with Answer for Mathematical Science - Download Free PDF

Concept:

Understanding Sequences and Their Limits:

• Sequence: An ordered list of numbers defined by a function on natural numbers.
• Convergence: A sequence converges if it approaches a unique real number as .
• lim sup: The greatest accumulation point (limit of suprema of tails).
• lim inf: The smallest accumulation point (limit of infima of tails).
• Max sequence: is non-decreasing.
• Min sequence: is non-increasing.

Calculation:

Given,

⇒ For even :

⇒ For odd :

⇒ So the sequence oscillates between numbers close to +1 and -1, hence:

does not exist 

Option 1: " does not converge"

⇒ Correct.

Since the sequence oscillates between two values without settling on one, it diverges.

Option 2: ""

→ False.

Since 1 \), and this is the max forever,

⇒ not equal to 1

Option 3: ""

→ False.

which is minimum forever ⇒ for all large

So

Option 4: ""

→ False.

As shown,

Hence, not equal

∴ Only correct statement is Option 1.

Concept:

Explanation:

C be the positively oriented circle in the complex plane of radius 3 centered at the origin.

Singularities of  is given by

z² = 0 ⇒ z = 0 and

e^z - e^-z = 0 ⇒ ez = e-z  ⇒ z = 0

Now,

  = 

          =  (Expansion of ez - e-z)

         = 

        =  (expansion of (1 + x)^-1)

So Residue of  = coefficient of 1/z =      

Hence  = 2πi(sum of residues) =  = −iπ/6

Option (4) is correct

Concept:

A function f(z) is said to be holomorphic in a domain D if f(z) has no singularities in D.

Explanation:

g'(z) = z ∈  \{0}

⇒ g'(z) = 

⇒ g'(z) = 

Integrating both sides we get

g(z) = 

So g(z) can not be holomorphic if p is a multiple of 2, 3 and 4.

∴ Options (1), (3) and (4) are not correct.

Hence option (2) is correct

Explanation:

Given 

utt − uxx = 0, 0 0

u(0, t) = 0 = u(2, t), ∀ t > 0, 

u(x, 0) = sin(πx) + 2sin(2πx), 0 ≤ x ≤ 2,

ut(x, 0) = 0, 0 ≤ x ≤ 2.

which is a wave equation of finite length. So solution is

u(x, t) =  where

Here c = 1, l = 2, f(x) = sin(πx) + 2sin(2πx)

So, 

and u(x, t) = 

u(x, 0) =  = sin(πx) + 2sin(2πx)

Comparing we get

D₂ = 1, D₄ = 2, D_n = 0 for other natural number n

Hence we get

u(x, t) = sin(πx) cos(πt) + 2sin(2πx)cos(2πt)

Then u(1, 1) = 0

u(1/2, 1) = -1

u(1/2, 2) = 1

u(1/2, 1/2) = 0

Option (3) is correct, other are false.

Concept:

A function f(x,y) is defined for (x,y) = (a,b) is said to be continuous at (x,y) = (a,b) if:

i) f(a,b) = value of f(x,y) at (x,y) = (a,b) is finite.

ii) The limit of the function f(x,y) as (x,y) → (a,b) exists and equal to the value of f(x,y) at (x,y) = (a,b)

Note:

For a function to be differentiable at a point, it should be continuous at that point too.

Calculation:

Given:

For function f(x,y) to be continuous:

 and finite.

f(a,b) = f(0,0) ⇒ 0 (given)

 = 0 

f_x(0, 0) = {f(h, 0) - f(0, 0)} / h = 0 

fy(0, 0) = {f(0, k) - f(0, 0)} / k = 0 

∵ the limit value is defined and function value is 0 at (x,y) = (0,0), ∴ the function f(x,y) is continuous.

Hence, Option 2, 3 & 4 all are correct 

Hence, Option 1 is not correct 

Hence, The Correct Answer is option 1.

Concept:

Residue of f(z)  at z = 0 is the coefficient of  in the Maclaurin series expansion of f(z)

Explanation:

f(z) = exp

     =

    = 

Hence the coefficient of  in the above expression

 = 

 = 

Hence option (3) is correct 

Concept:

Leibniz's test: A series of the form (-1)ⁿbn, where either all bn are positive or all bn are negative is convergent if

(i) |bn| decreases monotonically i.e., |bn+1| ≤ |bn|

(ii) 

Explanation:

an = (−1)n+1

    = (−1)n+1   

   = (−1)n+1

So series is  

So here b_n = , bn+1 = 

 so bn+1 n  

Also  =   = 0

Hence by Leibnitz's test  an is convergent.

Now the series is  =   =  

Hence by Limit comparison Test, it is divergent series by P - Test.

Hence the given series is conditionally convergent.

Option (3) is correct.

In Official answer key - Options (2) & (3) both are correct.

Concept: Solution of ODE x'(t) = Ax is x(t) = c₁ue^λ₁t + c2veλ2t, where u, v are the eigenvectors corresponding to the eigenvalues λ1 and λ2 respectively and c1 and c2 are constants.

Explanation:

A = 

tr(A) = 5 + 2 = 7 and det(A) = 10 - 4 = 6

Eigenvalues are given by

λ² - tr(A)λ + det(A) = 0

λ2 - 7λ + 6 = 0

(λ - 1)(λ - 6) = 0

λ = 1, 6

Eigenvector corresponding to eigenvalue λ = 1 is given by

 = 0  

u₁ + u₂ = 0 ⇒ u1 = - u2 

Eigenvector is u = 

Eigenvector corresponding to eigenvalue λ = 6 is given by

 = 0  

v1 - 4v2 = 0 ⇒ v1 = 4v2 

Eigenvector is v = 

Hence solution is

x(t) = c1et + c2e6t

x(t) = 

e^t → ∞ as t → ∞ also e6t → ∞ as t → ∞

So x(t) is not bounded solution for any x0 ≠ 0

(1) is false

e−6t|x(t)| =  →  does not tends to 0

So (2) is false

e−t|x(t)| = 

Let x(0) =  then

-c₁ + 4c₂ = 1 and c1 + c2 = -1

Solving them we get c1 = -1, c2 = 0

Hence x(t) =  does not tends to ∞ as t → ∞   

(3) is false

e−10t|x(t)| =  → 0 as t → ∞, for all x0 ≠ 0.

Option (4) is correct

Concept -

(1) If a is the eigen value of M then the eigen value of Mⁿ is aⁿ.

(2) If all the eigen value of the matrix is different then the matrix is diagonalizable.

Explanation -

Given - 

Now characteristic equation for the given matrix is -

⇒ | M - λ I | = 0

⇒ 

⇒ -λ (4 - λ ) + 3 = 0 ⇒ λ² - 4λ + 3 = 0

Now solve this equation we get the eigen values of the matrix -

 ⇒ λ2 - 3λ - λ  + 3 = 0  ⇒ (λ -3)(λ -1) = 0  ⇒ λ = 1, 3 

So the eigen value of M are 1 and 3.

Now solve the given statements -

(P) 𝑀8 + 𝑀12 is diagonalizable.

Now the eigen value of 𝑀8 + 𝑀12  are  

Hence both the eigen value of 𝑀8 + 𝑀12  are different So 𝑀8 + 𝑀12  is diagonalizable.

(𝑄) 𝑀7 + 𝑀9 is diagonalizable.

Now the eigen value of 𝑀7 + 𝑀9  are  

Hence both the eigen value of 𝑀7 + 𝑀9 are different So 𝑀7 + 𝑀9 is diagonalizable.

Hence the option (4) is true.

Concept:

Odd degree polynomial must have at least one real root

Explanation:

A is a a 3 × 3 matrix with real entries.

So characteristic polynomial of A will be of degree 3.

(1): Since we know that, odd degree polynomial must have at least one real root so A must have a real eigenvalue.

(1) is true

(2): As we know that determinant of a matrix is equal to the product of eigenvalues. So if the determinant of A is 0, then 0 is an eigenvalue of A.

(2) is true

(3): The determinant of A is negative and 3 is an eigenvalue of A.

If possible let the other two eigenvalues of A are not real and they are α + iβ, α - iβ

So determinant = 3(α + iβ)(α - iβ) = 3(α² + β²) > 0 for all α, β which is a contradiction.

So A must have three real eigenvalues.

(3) is true and (4) is false statement