Question : A differential equation \(\frac{di}{dt} - 0.2i = 0\) is applicable over \(-10 < t < 10\). If \(i(4) = 10\), then \(i(-5)\) is (Round off to 3 decimal places).
(Numerical Answer Type)
Solution:
1. General Solution: The differential equation \(\frac{di}{dt} = 0.2i\) is solved by separation of variables: \(i(t) = A e^{0.2t}\).
2. Find Constant \(A\): Using the condition \(i(4) = 10\): \[10 = A e^{0.2(4)} \implies A = 10 e^{-0.8}\]
3. Calculate \(i(-5)\): \[i(-5) = A e^{0.2(-5)} = (10 e^{-0.8}) e^{-1}\] \[i(-5) = 10 e^{-1.8}\]
\(i(-5) \approx 1.65298...\) Rounding off to 3 decimal places, the answer is \(\mathbf{1.653}\).
Question : Let a causal LTI system be characterised by the following differential equation, with initial rest condition \[\frac{d^2y}{dt^2} + 7\frac{dy}{dt} + 10y(t) = 4x(t) + 5\frac{dx(t)}{dt}\] where, \(x(t)\) and \(y(t)\) are the input and output respectively. The impulse response of the system is (\(u(t)\) is the unit step function)
Solution:
1. Transfer Function \(H(s)\): Applying the Laplace transform with zero initial conditions: \[s^2 Y(s) + 7s Y(s) + 10 Y(s) = 4 X(s) + 5s X(s)\] \[H(s) = \frac{Y(s)}{X(s)} = \frac{4 + 5s}{s^2 + 7s + 10} = \frac{4 + 5s}{(s+2)(s+5)}\]
2. Partial Fraction Expansion: \[H(s) = \frac{A}{s+2} + \frac{B}{s+5}\] Solving for \(A\) and \(B\): \(A = -2\), \(B = 7\). \[H(s) = \frac{7}{s+5} - \frac{2}{s+2}\]
3. Impulse Response \(h(t)\): \[h(t) = \mathcal{L}^{-1}\{H(s)\} = 7 e^{-5t}u(t) - 2 e^{-2t}u(t)\]
The correct option is B. \(-2e^{-2t}u(t) + 7e^{-5t}u(t)\).
Question : Consider the line integral \(I = \int_{c} (x^2 + iy^2)dz\), where \(z = x + iy\). The line \(c\) is the straight line path from \((0,0)\) to \((1,1)\) in the \(xy\)-plane. The value of \(I\) is
\(\frac{1}{2}i\)
\(\frac{2}{3}i\)
\(\frac{3}{4}i\)
\(\frac{4}{5}i\)
Solution:
1. Parametrize the Path \(c\): The path \(c\) is the straight line from \((0,0)\) to \((1,1)\), so \(y = x\), with \(x\) ranging from \(0\) to \(1\). The differential is \(dz = dx + idy = dx + idx = (1+i)dx\). The integrand is \(x^2 + iy^2 = x^2 + i(x)^2 = x^2(1+i)\).
2. Evaluate the Integral: \[I = \int_{0}^{1} x^2(1+i) (1+i) dx = \int_{0}^{1} x^2(1+2i+i^2) dx\] \[I = \int_{0}^{1} x^2(2i) dx = 2i \int_{0}^{1} x^2 dx\] \[I = 2i \left[ \frac{x^3}{3} \right]_0^{1} = \frac{2i}{3}\]
The correct option is B. \(\frac{2}{3}i\).
Question : For a complex number \(z\), \(\lim_{z \to i} \frac{z^2 + 1}{z^3 + 2z - i(z^2 + 2)}\) is
\(-2i\)
\(-i\)
\(i\)
\(2i\)
Solution:
1. Check Indeterminate Form: Substituting \(z=i\) into the expression yields the indeterminate form \(\frac{0}{0}\).
2. Factorization: \[\lim_{z \to i} \frac{z^2 + 1}{(z^3 + 2z) - i(z^2 + 2)} = \lim_{z \to i} \frac{z^2 - i^2}{z(z^2 + 2) - i(z^2 + 2)}\] \[\lim_{z \to i} \frac{(z-i)(z+i)}{(z-i)(z^2 + 2)}\]
3. Evaluate the Limit: Canceling the \((z-i)\) term (since \(z \ne i\) as \(z \to i\)): \[\lim_{z \to i} \frac{z+i}{z^2 + 2}\] Substituting \(z=i\): \[\frac{i+i}{i^2 + 2} = \frac{2i}{-1 + 2} = \frac{2i}{1} = 2i\]
The correct option is D. \(2i\).
Question : A function \(f(x)\) is defined as \[f(x) = \begin{cases} e^x, & x < 1 \\ \ln x + ax^2 + bx, & x \ge 1 \end{cases}\] where \(x \in \mathbb{R}\). Which one of the following statement is TRUE?
\(f(x)\) is NOT differentiable at \(x = 1\) for any values of \(a\) and \(b\).
\(f(x)\) is differentiable at \(x = 1\) for the unique values of \(a\) and \(b\).
\(f(x)\) is differentiable at \(x = 1\) for all values of \(a\) and \(b\) such that \(a + b = e\).
\(f(x)\) is differentiable at \(x = 1\) for all values of \(a\) and \(b\).
Solution:
1. Condition for Continuity at \(x=1\): For \(f(x)\) to be differentiable at \(x=1\), it must first be continuous at \(x=1\). This requires \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\). \[\lim_{x \to 1^-} e^x = e^1 = e\] \[\lim_{x \to 1^+} (\ln x + ax^2 + bx) = \ln 1 + a(1)^2 + b(1) = 0 + a + b = a + b\] Setting the limits equal gives the first equation: \[\mathbf{a + b = e} \quad \text{(Eq. 1)}\]
2. Condition for Differentiability at \(x=1\): For differentiability, the left-hand derivative (\(f'(1^-)\)) must equal the right-hand derivative (\(f'(1^+)\)). First, find the derivatives of the two parts: \[f'(x) = \begin{cases} \frac{d}{dx}(e^x) = e^x, & x < 1 \\ \frac{d}{dx}(\ln x + ax^2 + bx) = \frac{1}{x} + 2ax + b, & x > 1 \end{cases}\] The left-hand derivative at \(x=1\): \[f'(1^-) = \lim_{x \to 1^-} e^x = e^1 = e\] The right-hand derivative at \(x=1\): \[f'(1^+) = \lim_{x \to 1^+} \left( \frac{1}{x} + 2ax + b \right) = \frac{1}{1} + 2a(1) + b = 1 + 2a + b\] Setting them equal gives the second equation: \[\mathbf{1 + 2a + b = e} \quad \text{(Eq. 2)}\]
3. Solve the System of Equations: Subtract (Eq. 1) from (Eq. 2): \[(1 + 2a + b) - (a + b) = e - e\] \[1 + a = 0 \implies \mathbf{a = -1}\] Substitute \(a=-1\) into (Eq. 1): \[-1 + b = e \implies \mathbf{b = e + 1}\] Since there are unique values for \(a\) and \(b\) (\(a=-1\) and \(b=e+1\)) for which \(f(x)\) is both continuous and differentiable at \(x=1\), Option B is the correct statement.
The correct option is B. \(f(x)\) is differentiable at \(x = 1\) for the unique values of \(a\) and \(b\).
Question : A person decides to toss a fair coin repeatedly until he gets a head. He will make at most 3 tosses. Let the random variable \(Y\) denote the number of heads. The value of \(\text{var}\{Y\}\), where \(\text{var}\{\cdot\}\) denotes the variance, equals
\(\frac{7}{8}\)
\(\frac{49}{64}\)
\(\frac{7}{64}\)
\(\frac{105}{64}\)
Solution:
The possible outcomes and their probabilities for the number of heads \(Y\):
H (\(Y=1\)): \(P = 1/2\)
T H (\(Y=1\)): \(P = 1/4\)
T T H (\(Y=1\)): \(P = 1/8\)
T T T (\(Y=0\)): \(P = 1/8\)
The probability distribution for \(Y\): \(P(Y=1) = 7/8\) and \(P(Y=0) = 1/8\).
1. Expectation \(E[Y]\): \[E[Y] = \left( 1 \times \frac{7}{8} \right) + \left( 0 \times \frac{1}{8} \right) = \frac{7}{8}\]
2. Expectation \(E[Y^2]\): \[E[Y^2] = \left( 1^2 \times \frac{7}{8} \right) + \left( 0^2 \times \frac{1}{8} \right) = \frac{7}{8}\]
3. Variance \(\text{Var}\{Y\}\): \[\text{Var}\{Y\} = E[Y^2] - (E[Y])^2 = \frac{7}{8} - \left( \frac{7}{8} \right)^2 = \frac{56 - 49}{64} = \mathbf{\frac{7}{64}}\]
The correct option is C. \(\frac{7}{64}\).
Question : The eigenvalues of the matrix given below are \[A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -3 & -4 \end{pmatrix}\]
\((0, -1, -3)\)
\((0, -2, -3)\)
\((0, 2, 3)\)
\((0, 1, 3)\)
Solution:
The eigenvalues \(\lambda\) satisfy \(\det(A - \lambda I) = 0\): \[\begin{vmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 0 & -3 & -4 - \lambda \end{vmatrix} = 0\] Expanding along the first column: \[-\lambda \left( (-\lambda)(-4 - \lambda) - (-3) \right) = 0\] \[-\lambda \left( \lambda^2 + 4\lambda + 3 \right) = 0\] This yields \(\mathbf{\lambda = 0}\) and the quadratic \(\lambda^2 + 4\lambda + 3 = 0\). Factoring the quadratic gives \((\lambda + 1)(\lambda + 3) = 0\), so \(\mathbf{\lambda = -1}\) and \(\mathbf{\lambda = -3}\).
The eigenvalues are \(\mathbf{0, -1, \text{ and } -3}\).
The correct option is A. \((0, -1, -3)\).
Question : The value of the contour integral in the complex plane \(\oint_{|z|=3} \frac{z^3 - 2z + 3}{z-2} dz\) along the contour \(|z| = 3\), taken counter-clockwise is
\(-18\pi i\)
\(0\)
\(14\pi i\)
\(48\pi i\)
Solution:
We use the Cauchy Integral Formula: \(\oint_{C} \frac{f(z)}{z-a} dz = 2\pi i \cdot f(a)\). The integrand is \(\frac{f(z)}{z-a}\) where \(f(z) = z^3 - 2z + 3\) and the pole is at \(a=2\). The contour is \(|z|=3\), which encloses the pole \(z=2\).
Applying the Formula: \[f(2) = (2)^3 - 2(2) + 3 = 8 - 4 + 3 = \mathbf{7}\] The integral value is: \(2\pi i \cdot f(2) = 2\pi i (7) = \mathbf{14\pi i}\).
The correct option is C. \(14\pi i\).
Question : Let \(g(x) = \begin{cases} -x & x \le 1 \\ x+1 & x > 1 \end{cases}\) and \(f(x) = \begin{cases} 1-x & x \le 0 \\ x^2 & x > 0 \end{cases}\). Consider the composition of \(f\) and \(g\), i.e., \((f \circ g)(x) = f(g(x))\). The number of discontinuities in \((f \circ g)(x)\) present in the interval \((-\infty, 0)\) is:
0
1
2
4
Solution:
We analyze \((f \circ g)(x) = f(g(x))\) in the interval \(x \in (-\infty, 0)\).
For \(x \in (-\infty, 0)\), we have \(x \le 1\), so \(g(x) = -x\).
The range of \(g(x)\) for \(x < 0\) is \((0, \infty)\), so \(g(x) > 0\).
Since \(g(x) > 0\), we use the definition \(f(t) = t^2\) where \(t = g(x)\).
\[(f \circ g)(x) = f(-x) = (-x)^2 = x^2 \quad \text{for } x \in (-\infty, 0)\]
The resulting function \(x^2\) is a polynomial, which is **continuous** everywhere.
Therefore, the number of discontinuities in the interval \((-\infty, 0)\) is \(\mathbf{0}\).
The correct option is A. 0.
Question : Assume that in traffic junction, the cycle of the traffic signal lights is 2 minutes of green (Vehicle does not stop) and 3 minutes of red (Vehicle stops). Consider that the arrival time of vehicles at the junction is uniformly distributed over 5 minute cycle. The expected waiting time (in minutes) for the vehicle at the junction is
Solution:
The arrival time \(X \sim U(0, 5)\), with \(f(x) = 1/5\). The waiting time \(Y\) is \(Y=5-X\) for \(2 \le X \le 5\) (red light), and \(Y=0\) otherwise. \[E[Y] = \int_{2}^{5} (5-x) \cdot \frac{1}{5} dx = \frac{1}{5} \left[ 5x - \frac{x^2}{2} \right]_{2}^{5} = \frac{1}{5} \left[ \left( 25 - 12.5 \right) - \left( 10 - 2 \right) \right] = \frac{1}{5} (4.5) = \mathbf{0.9}\]
Question : Let \(y^2 - 2y + 1 = x\) and \(\sqrt{x} + y = 5\). The value of \(x + \sqrt{y}\) equals
(Give the answer up to three decimal places).
Solution:
The equations are \((y-1)^2 = x\) and \(\sqrt{x} = 5 - y\). Solving \((y-1)^2 = (5-y)^2\) gives \(y=3\). Then \(x = (3-1)^2 = 4\). The final value is \(x + \sqrt{y} = 4 + \sqrt{3} \approx 4 + 1.73205... \approx \mathbf{5.732}\).
Question : Let \(x\) and \(y\) be integers satisfying the following equations \[2x^2 + y^2 = 34\] \[x + 2y = 11\] The value of \((x + y)\) is
Solution:
Substitute \(x = 11 - 2y\) into the first equation to get \(9y^2 - 88y + 208 = 0\). The integer solution for \(y\) is \(y = 4\). Then \(x = 11 - 2(4) = 3\). The value of \(x + y\) is \(3 + 4 = \mathbf{7}\).
Question : An urn contains 5 red balls and 5 black balls. In the first draw, one ball is picked at random and discarded without noticing its colour. The probability to get a Red ball in the second draw is
\(\frac{1}{2}\)
\(\frac{4}{9}\)
\(\frac{5}{9}\)
\(\frac{6}{9}\)
Solution:
Using the Total Probability Theorem, the probability of drawing a red ball in the second draw (\(R_2\)) is: \[P(R_2) = P(R_2 | R_1) P(R_1) + P(R_2 | B_1) P(B_1)\] \[P(R_2) = \left( \frac{4}{9} \right) \left( \frac{5}{10} \right) + \left( \frac{5}{9} \right) \left( \frac{5}{10} \right) = \frac{4}{18} + \frac{5}{18} = \frac{9}{18} = \mathbf{1/2}\]
The correct option is A. \(\frac{1}{2}\).
Question : The figures show diagrammatic representations of vector fields \(\vec{X}\), \(\vec{Y}\), and \(\vec{Z}\) respectively. Which one of the following choices is true?
\(\nabla \cdot \vec{X} = 0, \nabla \times \vec{Y} \neq 0, \nabla \times \vec{Z} = 0\)
\(\nabla \cdot \vec{X} \neq 0, \nabla \times \vec{Y} = 0, \nabla \times \vec{Z} \neq 0\)
\(\nabla \cdot \vec{X} \neq 0, \nabla \times \vec{Y} \neq 0, \nabla \times \vec{Z} \neq 0\)
\(\nabla \cdot \vec{X} = 0, \nabla \times \vec{Y} = 0, \nabla \times \vec{Z} = 0\)
Solution:
We analyze the divergence (\(\nabla \cdot \vec{V}\)) and curl (\(\nabla \times \vec{V}\)) for each vector field based on the diagram:
1. Divergence (\(\nabla \cdot \vec{V}\)): Measures the outward flow (source) or inward flow (sink). It is zero if total outward flow equals total inward flow.
\(\vec{X}\) (Radial Outward): The vectors point away from the center. This indicates a **net outward flow** (source). \[\nabla \cdot \vec{X} \neq 0\]
\(\vec{Y}\) (Concentric Circles): The flow is neither inward nor outward; it’s purely rotational. The flow passing through any small surface is zero. \[\nabla \cdot \vec{Y} = 0\]
\(\vec{Z}\) (Spiral Outward): The vectors spiral outward, indicating a **net outward flow** combined with rotation. \[\nabla \cdot \vec{Z} \neq 0\]
2. Curl (\(\nabla \times \vec{V}\)): Measures the rotation of the field. It is zero if there is no rotation in space.
\(\vec{X}\) (Radial Outward): The field has no rotation. \[\nabla \times \vec{X} = 0\]
\(\vec{Y}\) (Concentric Circles): The field is purely rotational. \[\nabla \times \vec{Y} \neq 0\]
\(\vec{Z}\) (Spiral Outward): The field has clear rotation (whirlpool motion). \[\nabla \times \vec{Z} \neq 0\]
Summary of properties: \[\vec{X}: \nabla \cdot \vec{X} \neq 0, \nabla \times \vec{X} = 0\] \[\vec{Y}: \nabla \cdot \vec{Y} = 0, \nabla \times \vec{Y} \neq 0\] \[\vec{Z}: \nabla \cdot \vec{Z} \neq 0, \nabla \times \vec{Z} \neq 0\]
Comparing this to the options, the correct combination that matches the analysis for \(\vec{Y}\) and \(\vec{Z}\) is found in option C, where the fields are listed as \(\vec{X}\), \(\vec{Y}\), and \(\vec{Z}\) respectively. The full correct choice is: \[\nabla \cdot \vec{X} \neq 0, \nabla \times \vec{Y} \neq 0, \nabla \times \vec{Z} \neq 0\]
The correct option is C.
\(\nabla \cdot \vec{X} \neq 0, \nabla \times \vec{Y} \neq 0, \nabla \times \vec{Z} \neq 0\). (Note: The solution image identifies Option C as the correct choice, which corresponds to the \(\nabla \cdot \vec{X} \neq 0\), \(\nabla \times \vec{Y} \neq 0\), and \(\nabla \times \vec{Z} \neq 0\) properties of the fields).