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

Explanation:

We will check the constraints and graph them to identify the feasible region:

1. First constraint:  \(3x - 7y \leq 21 \)

Rewriting:  \(y \leq \frac{3x - 21}{7} \)

2. Second constraint: \( y - 2x \leq 10 \)

Rewriting:  \(y \leq 2x + 10 \)

3. Non-negativity constraints:  \(x \geq 0 , y \geq 0 \)

To check if the system is feasible or unbounded, let’s solve the system of inequalities.

Intersection of 3x - 7y = 21 and y - 2x = 10 :

these two lines do not intersect in the first quadrant, and we need to check further whether the feasible region is unbounded.

If the feasible region is unbounded, this means that the objective function z = 2x + y can continue to increase without any limit

To determine this, we need to check if there are directions in which the feasible region extends infinitely

The constraint  \(y - 2x \leq 10 \)  defines an upper boundary for y relative to x ,

but since the region is constrained by non-negativity for x and y , the feasible region may stretch infinitely along the axes

Since the region is not bounded (i.e., it extends infinitely in some direction, as indicated by the inequalities),

and there is no specific point where the region ends, the problem is unbounded.

Correct Answer is The LPP is unbounded

Concept:

The Cauchy-Riemann (C-R) equations are the necessary conditions for a function \( f(z) \) to be differentiable (analytic) at a point in the complex plane.

If \( f(z) = u(x, y) + iv(x, y) \), then the C-R equations are:

\( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \) and \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \)

Given Options:

1. \( f(z) = \text{Re}(z) = x \)
2. \( f(z) = \text{Im}(z) = y \)
3. \( f(z) = z = x + iy \)
4. \( f(z) = \sin z \)

Calculation:

• Options 1 and 2 are purely real and imaginary parts — not analytic → do not satisfy C-R equations.
• Option 3: \( f(z) = z \) → clearly analytic everywhere. It satisfies C-R equations.
• Option 4: \( f(z) = \sin z \) is also analytic, but the question asks for the **only** function — this may be a trick; among options, z is simplest and surely satisfies C-R equations.

Correct Answer: 3) f(z) = z

Concept:

Greatest Integer Function:

• The greatest integer function, denoted by [x], returns the largest integer less than or equal to x.
• The function is also known as the floor function. Mathematically, [x] is defined as the greatest integer less than or equal to x.
• The greatest integer function is continuous everywhere except at integer points, where it is not differentiable.
• For differentiability, the function must have no "sharp corners" at the points of discontinuity.

Calculation:

Let's analyze each function in the options to match with the correct descriptions.

• (B)  x − |x|: This is a combination of absolute value functions. These are continuous and differentiable everywhere except at the points where the absolute values change, which are x = 1 and x = 2. Therefore, this function is differentiable everywhere except at x = 0.
• (A) |x – 1| + |x – 2| : This function involves the absolute value function. The greatest integer function has a discontinuity at integer points, and this function involves absolute values, which means it is continuous everywhere but not differentiable at x = 0. Hence, it is continuous everywhere.
• (C) x − [x]: This function involves the greatest integer function (floor function), which is continuous but not differentiable at integer points. Therefore, this function is not differentiable at x = 1 because there is a discontinuity at integer points.
• (D) |x|: This function is continuous and differentiable at all points, including x = 0. Therefore, it is differentiable at x = 1.

∴ Correct Matching: (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

Concept:

(i) Trace of a square matrix = sum of the eigenvalues

(ii) determinant of a square matrix = product of eigenvalues

(ii) If sum of the elements of each row oa a matrix is λ then λ is an eigenvalue of that matrix.   

Explanation:

\(A=\left[\begin{array}{rrr}1 & 1 & 2 \\ 1 & -2 & 5 \\ 2 & 5 & -3\end{array}\right]\)

the sum of elements of each row is 4 so 4 is an eigenvalue

trace of A  = 1 - 2 - 3 = -4

det(A) = \(\begin{vmatrix}1 & 1 & 2 \\ 1 & -2 & 5 \\ 2 & 5 & -3\end{vmatrix}\) = 1(6 - 25)  + 1(10 + 3) + 2(5 + 4) = -19 + 13 + 18 = 12

(1): 4 is not an eigenvalue

Hence option (1) is false

(2): trace of matrix = sum of eigenvalues = 4 + 3 + 1 = 9 ≠ -4

Hence option (2) is false

(3): trace of matrix = 4 - 4 + √13 - 4 - √13 = -4

and determinant of A = 4(-4 + √13)(-4 - √13) = 4(16 - 13) = 12

Hence option (3) is true

(4): determinant of matrix = 4 (-2 + 2√7)(-2 - 2√7) = 4(4 - 28) = -96

Hence option (4) is false

Explanation:

Augmented matrix will be

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 3&1&{ - 3}&{13}\\ 2&{19}&{ - 47}&{32} \end{array}} \right] \)

R₃ → R₃ – R₁

R₂ → 2R₂ – 3R₁

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&{22}&{ - 54}&{27} \end{array}} \right]\)

R₃ → R₃ – 2R₂

\(\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&7&5\\ 0&{11}&{ - 27}&{11}\\ 0&0&0&5 \end{array}} \right]\)

Rank of A ≠ Rank of Augmented matrix.

Hence given system of equations has no solution.

Hence Option(5) is the correct answer.

Concept:

Explanation:

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

Singularities of \(\rm\displaystyle\int_C \frac{d z}{z^2\left(e^z−e^{−z}\right)}\) is given by

z² = 0 ⇒ z = 0 and

e^z - e^-z = 0 ⇒ ez = e-z  ⇒ z = 0

Now,

 \(\frac{1}{z^2\left(e^z−e^{−z}\right)}\) = \(\frac{1}{z^2}(e^z-e^{-z})^{-1}\)

          = \(\frac{1}{z^2}(2z+\frac{2z^3}{3!}+\frac{2z^5}{5!}+...)^{-1}\) (Expansion of ez - e-z)

         = \(\frac{1}{z^2}.\frac{1}{2z}(1+(\frac{z^2}{3!}+\frac{z^4}{5!}+...))^{-1}\)

        = \(\frac{1}{2z^3}(1-(\frac{z^2}{3!}+\frac{z^5}{5!}+...)+(\frac{z^2}{3!}+\frac{z^5}{5!}+...)^2+...)\) (expansion of (1 + x)^-1)

So Residue of \(\frac{1}{z^2\left(e^z−e^{−z}\right)}\) = coefficient of 1/z =      \(\frac12.(-\frac{1}{3!})=-\frac1{12}\)

Hence \(\rm\displaystyle\int_C \frac{d z}{z^2\left(e^z−e^{−z}\right)}\) = 2πi(sum of residues) = \(-\frac{2\pi i}{12}\) = −iπ/6

Option (4) is correct

Explanation:

Given 

utt − uxx = 0, 0 < x < 2, t > 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) = \(\sum D_n \sin({n\pi x\over l})\cos({n\pi ct\over l})\) where

\(D_n=\frac2l\int_0^lf(x)\sin({n\pi x\over l})dx\)

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

So, \(D_n=\frac22\int_0^2(\sin(\pi x)+2\sin (2\pi x))\sin({n\pi x\over 2})dx\)

and u(x, t) = \(\sum D_n \sin({n\pi x\over 2})\cos({n\pi ct\over 2})\)

u(x, 0) = \(\sum D_n \sin({n\pi x\over 2})\) = 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)

\(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = f\left( {a,b} \right)\)

Note:

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

Calculation:

Given:

\(f\left( {x,y} \right) = \;\left\{ {\begin{array}{*{20}{c}} {\frac{{xy}}{{\sqrt {{x^2} + {y^2}} }}\;\;\;{x^2} + {y^2} \ne 0}\\ {0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = y = 0} \end{array}} \right.\)

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

\(\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} f\left( {x,y} \right) = f\left( {a,b} \right)\) and finite.

f(a,b) = f(0,0) ⇒ 0 (given)

\(\mathop {\lim }\limits_{\left( {r,\theta } \right) \to \left( {0,0} \right)} f\left( {r,\theta} \right) =\frac{ r^2cos\theta rsin\theta }{r} \) = 0 

f_x(0, 0) = \(\mathop {\lim }\limits_{\left( {h,0 } \right) \to \left( {0,0} \right)}\){f(h, 0) - f(0, 0)} / h = 0 

fy(0, 0) = \(\mathop {\lim }\limits_{\left( {0,k } \right) \to \left( {0,0} \right)}\){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:

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

Explanation:

g'(z) =\(f(z)=\frac{\sin z}{z^p}\) z ∈ \(\mathbb{C}\) \{0}

⇒ g'(z) = \(\frac{1}{z^p}(z-\frac{z^3}{3!}+\frac{z^5}{5!}-...)\)

⇒ g'(z) = \((\frac{z^{1-p}}{1!}-\frac{z^{3-p}}{3!}+\frac{z^{5-p}}{5!}-...)\)

Integrating both sides we get

g(z) = \((\frac{z^{2-p}}{1!(2-p)}-\frac{z^{4-p}}{3!(4-p)}+\frac{z^{6-p}}{5!(6-p)}-...)\)

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

Concept:

Leibniz's test: A series of the form \(\rm\displaystyle\sum_{n=1}^{\infty}\)(-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) \(\lim_{n\to\infty}b_n=0\)

Explanation:

an = (−1)n+1\(\rm (\sqrt{n+1}−\sqrt{n})\)

    = (−1)n+1  \(\rm \frac{(\sqrt{n+1}−\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})}\) 

   = (−1)n+1\(\rm \frac{1}{(\sqrt{n+1}+\sqrt{n})}\)

So series is \(\rm\displaystyle\sum_{n=1}^{\infty}\)\(\rm \frac{(-1)^{n+1}}{(\sqrt{n+1}+\sqrt{n})}\) 

So here b_n = \(\rm \frac{1}{(\sqrt{n+1}+\sqrt{n})}\), bn+1 = \(\rm \frac{1}{(\sqrt{n+2}+\sqrt{n+1})}\)

\(\frac{b_{n+1}}{b_n}<1\) so bn+1 < bn  

Also \(\lim_{n\to\infty}b_n\) = \(\lim_{n\to\infty}\) \(\rm \frac{1}{(\sqrt{n+1}+\sqrt{n})}\) = 0

Hence by Leibnitz's test \(\rm\displaystyle\sum_{n=1}^{\infty}\) an is convergent.

Now the series is \(\rm\displaystyle\sum_{n=1}^{\infty}\)\(\rm |\frac{(-1)^{n+1}}{(\sqrt{n+1}+\sqrt{n})}|\) =  \(\rm\displaystyle\sum_{n=1}^{\infty}\)\(\rm \frac{1}{(\sqrt{n+1}+\sqrt{n})}\) = \(\rm\displaystyle\sum_{n=1}^{\infty}\) \(\rm \frac{1}{\sqrt n(\sqrt{1+\frac{1}{n}}+1)}\)

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:

Residue of f(z)  at z = 0 is the coefficient of \(\frac1z\) in the Maclaurin series expansion of f(z)

Explanation:

f(z) = exp\(\rm\left(z+\frac{1}{z}\right)\)

     = \(e^z.e^{\frac1z}\)

    = \((1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+...)\)\(.(1+\frac1zz+\frac{1}{z^22!}+\frac{1}{z^33!}+...)\)

Hence the coefficient of \(\frac1z\) in the above expression

 = \(\frac11+\frac1{2!.1!}+\frac1{3!.2!}+\frac1{4!.3!}+...\)

 = \(\sum_{l=0}^{\infty} \frac{1}{l !(l+1) !}\)

Hence option (3) is correct 

Concept: 

Explanation:

A = \(\left[\begin{array}{ll}5 & 4 \\1 & 2\end{array}\right]\)

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

\(\left[\begin{array}{ll}4 & 4 \\1 & 1\end{array}\right]\)\(\left[\begin{array}{ll}u_1 \\u_2\end{array}\right]\) = 0  

u₁ + u₂ = 0 ⇒ u1 = - u2 

Eigenvector is u = \(\begin{bmatrix}-1 \\1\end{bmatrix}\)

Eigenvector corresponding to eigenvalue λ = 6 is given by

\(\begin{bmatrix}-1 & 4 \\1 & -4\end{bmatrix}\)\(\left[\begin{array}{ll}v_1 \\v_2\end{array}\right]\) = 0  

v1 - 4v2 = 0 ⇒ v1 = 4v2 

Eigenvector is v = \(\left[\begin{array}{ll}4 \\1\end{array}\right]\)

Hence solution is

x(t) = c1\(\begin{bmatrix}-1 \\1\end{bmatrix}\)et + c2\(\left[\begin{array}{ll}4 \\1\end{array}\right]\)e6t

x(t) = \(\begin{bmatrix}-c_1e^t+4c_2e^{6t} \\c_1e^t+c_2e^{6t}\end{bmatrix}\)

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)| = \(\begin{bmatrix}-c_1e^{-5t}+4c_2 \\c_1e^{-5t}+c_2\end{bmatrix}\) → \(\begin{bmatrix}4c_2 \\c_2\end{bmatrix}\) does not tends to 0

So (2) is false

e−t|x(t)| = \(\begin{bmatrix}-c_1+4c_2e^{5t} \\c_1+c_2e^{5t}\end{bmatrix}\)

Let x(0) = \(\begin{bmatrix}1 \\-1\end{bmatrix}\) then

-c₁ + 4c₂ = 1 and c1 + c2 = -1

Solving them we get c1 = -1, c2 = 0

Hence x(t) = \(\begin{bmatrix}-1 \\1\end{bmatrix}\) does not tends to ∞ as t → ∞   

(3) is false

e−10t|x(t)| = \(\begin{bmatrix}-c_1e^{-9t}+4c_2e^{-4t} \\c_1e^{-9t}+c_2e^{-4t}\end{bmatrix}\) → 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 - \(\rm M=\begin{bmatrix}4&-3\\\ 1&0\end{bmatrix}\)

Now characteristic equation for the given matrix is -

⇒ | M - λ I | = 0

⇒ \(\rm det\begin{bmatrix}4-λ&-3\\\ 1&-λ\end{bmatrix}=0\)

⇒ -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  \((1)^8 + (1)^{12} = 2 \ \ and \ \ (3)^8 +(3)^{12} =82(3)^8\)

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  \((1)^7 + (1)^9 = 2 \ \ and \ \ (3)^7 +(3)^9 =10(3)^7\)

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

Hence the option (4) is true.

Explanation:

Characteristic polynomial p(T) is divisible by T2.

p(x)/x2

So p(x) = x2(x + a) where a can be zero also

Option (1): Let A = \(\begin{bmatrix}0&2&3\\0&0&1\\0&0&a\end{bmatrix}\) Here 0 and a are the eigenvalues and eigenspace of A for 0 is 1

So option (1) is false

Here A is 3 × 3 matrix and two eigenvalues are 0, 0. Since complex eigenvalue are always complex conjugate and they are in pairwise. So here third eigenvalue must be real.

Option (2) is correct

 For A = \(\begin{bmatrix}0&2&3\\0&0&1\\0&0&a\end{bmatrix}\), A3 ≠ 0

Option (3) is false

also AM of eigenvalue 0 is 2 and GM of eigenvalue 0 is 1 

Since AM ≠ GM so not diagonalizable.

Option (4) is false

Concept:

If xⁿ = 1 then o(x) divides n

Explanation:

ℤ/105ℤ ≅ ℤ₁₀₅

105 = 3 × 5 × 7

So \(U_{ℤ_{105}}\) ≅ U(3) × U(5) × U(7) ≅ ℤ₂ × ℤ₄ × ℤ₆

Given x2 = 1 so o(x) divides 2 Hence o(x) = 1 or 2

Element of ℤ2 of order 1 and 2 is 2

Element of ℤ4 of order 1 and 2 is 2

Element of ℤ6 of order 1 and 2 is 2

Hence total such elements = 2 × 2 × 2 = 8 

Option (4) is correct