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 : (Based on \(f(x) = 2x - x^2 - x^3 + 3\)):
Find Critical Points: Calculate the first derivative, \(f'(x)\), and set it to zero: \[f(x) = -x^3 - x^2 + 2x + 3\] \[f'(x) = -3x^2 - 2x + 2\] Set \(f'(x) = 0\): \[-3x^2 - 2x + 2 = 0 \implies 3x^2 + 2x - 2 = 0\] Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=3, b=2, c=-2\): \[x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-2)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 24}}{6} = \frac{-2 \pm \sqrt{28}}{6}\] \[x = \frac{-2 \pm 2\sqrt{7}}{6} = \frac{-1 \pm \sqrt{7}}{3}\] The critical points are \(x_1 \approx 0.548\) and \(x_2 \approx -1.215\). Note: \(x=1\) is NOT a critical point.
Second Derivative Test: Calculate the second derivative: \[f''(x) = -6x - 2\] Evaluate at \(x_1\):
\[\begin{aligned} f''\left(\frac{-1 + \sqrt{7}}{3}\right) &= -6\left(\frac{-1 + \sqrt{7}}{3}\right) - 2 \\ &= -2(-1 + \sqrt{7}) - 2 = 2 - 2\sqrt{7} - 2 = -2\sqrt{7} \end{aligned}\]
Since \(f''(x_1) < 0\), \(f(x)\) has a local maxima at \(x \approx 0.548\).
Evaluate at \(x_2\):
\[\begin{aligned} f''\left(\frac{-1 - \sqrt{7}}{3}\right) &= -6\left(\frac{-1 - \sqrt{7}}{3}\right) - 2 \\ &= -2(-1 - \sqrt{7}) - 2 = 2 + 2\sqrt{7} - 2 = 2\sqrt{7} \end{aligned}\]
Since \(f''(x_2) > 0\), \(f(x)\) has a local minima at \(x \approx -1.215\). Based on the function in the image, the options provided are likely incorrect or the function itself is misstated in the question.
Solution Assume the function was intended to be \(f(x) = 2x - x^2 + 3\). \[f(x) = -x^2 + 2x + 3\] \[f'(x) = -2x + 2\] \[f'(x) = 0 \implies -2x + 2 = 0 \implies x = 1\] \[f''(x) = -2\] Since \(f''(1) = -2 < 0\), the function has a **maxima** at \(x=1\).
The function \(f(x) = -x^2 + 2x + 3\) is a parabola opening downwards, so it has only one maximum and no minimum. The answer follows the derivation in the EXP: The function has only maxima at \(x=1\).
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.