Question 1. Roots of the algebraic equation \(x^3 + x^2 + x + 1 = 0\) are
\((+1, +j, -j)\)
\((+1, -1, +1)\)
\((0, 0, 0)\)
\((-1, +j, -j)\)
Solution:
\[\begin{aligned} x^3 + x^2 + x + 1 &= 0 \\ x^2(x + 1) + 1(x + 1) &= 0 \\ (x^2 + 1)(x + 1) &= 0 \end{aligned}\] This yields two separate equations for the roots:
\(x + 1 = 0 \implies x = -1\)
\(x^2 + 1 = 0 \implies x^2 = -1 \implies x = \pm \sqrt{-1} \implies x = \pm j\)
The roots are \(\mathbf{-1, +j, -j}\) The correct option is D.
Question 2. With \(K\) as a constant, the possible solution for the first order differential equation \(\frac{dy}{dx} = e^{-3x}\) is
\(y = - \frac{1}{3}e^{-3x} + K\)
\(y = \frac{1}{3}e^{-3x} + K\)
\(y = - \frac{1}{3}e^{3x} + K\)
\(y = -3e^{-x} + K\)
Answer:- (A)
Exp: The given differential equation is separable: \[\frac{dy}{dx} = e^{-3x}\] Separate the variables: \[dy = e^{-3x} \, dx\] Integrate on both sides: \[\int dy = \int e^{-3x} \, dx\] \[y = \frac{e^{-3x}}{-3} + K\] \[y = - \frac{1}{3}e^{-3x} + K\] The correct option is A.
Question 3: The function \(f(x) = 2x - x^2 - x^3 + 3\) has
a maxima at \(x = 1\) and minimum at \(x = 5\)
a maxima at \(x = 1\) and minimum at \(x = -5\)
only maxima at \(x = 1\) and
only a minimum at \(x = 5\)
Solution: Taken exactly as printed, \(f(x) = -x^3 - x^2 + 2x + 3\) gives \[f'(x) = -3x^2 - 2x + 2 = 0 \implies x = \frac{-1 \pm \sqrt{7}}{3} \approx 0.548, \; -1.215\] which are not the points named in any option. The function of the paper is \(f(x) = 2x - x^2 + 3\).
For that function \[f'(x) = 2 - 2x = 0 \implies x = 1\] \[f''(x) = -2 < 0\] so \(x = 1\) is a maxima. Being a downward parabola, \(f(x)\) has this single stationary point and no minimum.
The correct option is C.
Question 4: The matrix \(A = \begin{bmatrix} 2 & 1 \\ 4 & -1 \end{bmatrix}\) is decomposed into a product of a lower triangular matrix \([L]\) and an upper triangular matrix \([U]\). The properly decomposed \([L]\) and \([U]\) matrices respectively are
\(\begin{bmatrix} 1 & 0 \\ 4 & -1 \end{bmatrix}\) and \(\begin{bmatrix} 1 & 1 \\ 0 & -2 \end{bmatrix}\)
\(\begin{bmatrix} 2 & 0 \\ 4 & -1 \end{bmatrix}\) and \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\) and \(\begin{bmatrix} 2 & 1 \\ 0 & -3 \end{bmatrix}\)
\(\begin{bmatrix} 2 & 0 \\ 4 & -3 \end{bmatrix}\) and \(\begin{bmatrix} 1 & 1.5 \\ 0 & 1 \end{bmatrix}\)
Solution: We need to find \(L\) (Lower Triangular) and \(U\) (Upper Triangular) such that \(A = LU\). We will use the common approach where \(L\) has unit diagonal elements ( \(L_{ii}=1\)).
Set up the decomposition \(A=LU\): \[\begin{bmatrix} 2 & 1 \\ 4 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ l_{21} & 1 \end{bmatrix} \begin{bmatrix} u_{11} & u_{12} \\ 0 & u_{22} \end{bmatrix}\]
Calculate the elements of \(L\) and \(U\): Perform the matrix multiplication: \[\begin{bmatrix} u_{11} & u_{12} \\ l_{21}u_{11} & l_{21}u_{12} + u_{22} \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & -1 \end{bmatrix}\]
First row (U): \(u_{11} = 2\) \(u_{12} = 1\)
Second row, first column (\(L\)): \(l_{21}u_{11} = 4 \implies l_{21}(2) = 4 \implies l_{21} = 2\)
Second row, second column (\(U\)): \(l_{21}u_{12} + u_{22} = -1 \implies (2)(1) + u_{22} = -1 \implies 2 + u_{22} = -1 \implies u_{22} = -3\)
Form the matrices \(L\) and \(U\): \[L = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \quad \text{and} \quad U = \begin{bmatrix} 2 & 1 \\ 0 & -3 \end{bmatrix}\]
Check the options: The calculated \(L\) and \(U\) match Option.
The correct option is C.
Question 5: The two vectors \([1, 1, 1]\) and \([1, a, a^2]\), where \(a = \left( - \frac{1}{2} + j \frac{\sqrt{3}}{2} \right)\), are
Orthonormal
Orthogonal
Parallel
Collinear
Solution:
The complex number \(a = - \frac{1}{2} + j \frac{\sqrt{3}}{2}\) is the complex cube root of unity, often denoted as \(\omega\) or \(e^{j2\pi/3}\). A key property of the cube roots of unity (\(1, a, a^2\)) is that their sum is zero: \[1 + a + a^2 = 0\]
To determine if the vectors \(V_1 = [1, 1, 1]\) and \(V_2 = [1, a, a^2]\) are orthogonal, we check their dot product. For complex vectors, the dot product is calculated as \(V_1 \cdot V_2 = V_1^T \overline{V_2}\) (or \(\overline{V_1}^T V_2\)), but since \(V_1\) is real, we can use \(V_1 \cdot V_2 = V_1^T V_2\) for simplicity here, as the options only distinguish between orthogonality and others.
Calculate the Dot Product \(V_1 \cdot V_2\): \[V_1 \cdot V_2 = (1)(1) + (1)(a) + (1)(a^2)\] \[V_1 \cdot V_2 = 1 + a + a^2\]
Apply the Cube Root of Unity Property: \[V_1 \cdot V_2 = 0\] Since the dot product is zero, the two vectors are Orthogonal.
Check for Orthonormal: A vector is orthonormal if it’s orthogonal and has a magnitude (norm) of 1. \[\|V_1\|^2 = 1^2 + 1^2 + 1^2 = 3 \implies \|V_1\| = \sqrt{3} \ne 1\] Thus, they are orthogonal but not orthonormal. The correct choice is B. Orthogonal.
Question 6: Given that \(f(y) = \frac{|y|}{y}\), and \(q\) is any non-zero real number, the value of \(|f(q) - f(-q)|\) is
0
-1
1
2
Solution:-
Given, \(f(y) = \frac{|y|}{y}\). This is the signum function, \(\text{sgn}(y)\).
Evaluate \(f(q)\): Since \(q\) is a non-zero real number, we consider two cases:
If \(q > 0\): \(f(q) = \frac{|q|}{q} = \frac{q}{q} = 1\)
If \(q < 0\): \(f(q) = \frac{|q|}{q} = \frac{-q}{q} = -1\)
In general, \(f(q) = \text{sgn}(q)\).
Evaluate \(f(-q)\): Since \(q\) is non-zero, \(-q\) is also non-zero. \[f(-q) = \frac{|-q|}{-q}\] We know that \(|-q| = |q|\). \[f(-q) = \frac{|q|}{-q} = - \frac{|q|}{q} = - f(q)\] In general, \(f(-q) = \text{sgn}(-q) = -\text{sgn}(q)\).
Calculate \(|f(q) - f(-q)|\): \[|f(q) - f(-q)| = |f(q) - (-f(q))| = |f(q) + f(q)| = |2f(q)| = 2|f(q)|\] Since \(f(q) = 1\) (if \(q>0\)) or \(f(q) = -1\) (if \(q<0\)), \(|f(q)| = 1\). \[|f(q) - f(-q)| = 2(1) = \mathbf{2}\]
Alternative Calculation (following image steps): \[|f(q) - f(-q)| = \left| \frac{|q|}{q} - \frac{|-q|}{-q} \right| = \left| \frac{|q|}{q} - \left( - \frac{|q|}{q} \right) \right|\] \[= \left| \frac{|q|}{q} + \frac{|q|}{q} \right| = \left| \frac{2|q|}{q} \right| = 2 \left| \frac{|q|}{q} \right| = 2(1) = \mathbf{2}\]
The correct choice is D.
Question 7: The sum of n terms of the series \(4+44+444+\dots\) is
\((4/81)[10^{n+1}-9n-1]\)
\((4/81)[10^{n}-9n-1]\)
\((4/81)[10^{n+1}-9n-10]\)
\((4/81)[10^{n}-9n-10]\)
Solution:
Let \(S\) be the sum of \(n\) terms. \[\begin{aligned} S &= 4 + 44 + 444 + \dots \text{ (n terms)} \\ S &= 4(1 + 11 + 111 + \dots) \\ S &= \frac{4}{9}(9 + 99 + 999 + \dots) \\ S &= \frac{4}{9} \{(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1) \} \\ S &= \frac{4}{9} \{(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{ n times}) \} \\ S &= \frac{4}{9} \left\{ 10 \left( \frac{10^n - 1}{10 - 1} \right) - n \right\} \quad \text{ (Sum of Geometric Progression)} \\ S &= \frac{4}{9} \left\{ \frac{10}{9} (10^n - 1) - n \right\} \\ S &= \frac{4}{81} \left\{ 10(10^n - 1) - 9n \right\} \\ S &= \frac{4}{81} \left\{ 10^{n+1} - 10 - 9n \right\} \\ S &= \frac{4}{81} [10^{n+1} - 9n - 10] \end{aligned}\]
The correct option is C.