Question 1: The value of the quantity \(P\), where \(P = \int_{0}^{1} x e^x dx\), is equal to
0
1
\(e\)
\(1/e\)
Solution: We will use Integration by Parts (the ILATE rule, where \(u=x\) and \(dv=e^x dx\)).
Define \(u\) and \(dv\): Let \(u = x\) and \(dv = e^x dx\). Then \(du = dx\) and \(v = \int e^x dx = e^x\).
Apply Integration by Parts Formula (\(\int u \, dv = uv - \int v \, du\)): \[P = \int_{0}^{1} x e^x dx = \left[ x e^x \right]_{0}^{1} - \int_{0}^{1} e^x dx\]
Evaluate the definite integrals: \[P = \left( (1)e^1 - (0)e^0 \right) - \left[ e^x \right]_{0}^{1}\] \[P = (e - 0) - (e^1 - e^0)\] \[P = e - (e - 1)\] \[P = e - e + 1\] \[P = \mathbf{1}\]
The correct option is B.
Question 2: Divergence of the three-dimensional radial vector field \(\mathbf{\vec{r}}\) is
3
\(1/r\)
\(\hat{i} + \hat{j} + \hat{k}\)
\(3(\hat{i} + \hat{j} + \hat{k})\)
Solution: The three-dimensional radial vector field \(\mathbf{\vec{r}}\) in Cartesian coordinates is given by: \[\mathbf{\vec{r}} = x\hat{i} + y\hat{j} + z\hat{k}\]
The divergence of a vector field \(\mathbf{F} = P\hat{i} + Q\hat{j} + R\hat{k}\) is defined as: \[\nabla \cdot \mathbf{\vec{r}} = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} + \dfrac{\partial R}{\partial z}\]
Identify Components: In this case, \(P=x\), \(Q=y\), and \(R=z\).
Calculate Partial Derivatives: \[\dfrac{\partial P}{\partial x} = \dfrac{\partial (x)}{\partial x} = 1\] \[\dfrac{\partial Q}{\partial y} = \dfrac{\partial (y)}{\partial y} = 1\] \[\dfrac{\partial R}{\partial z} = \dfrac{\partial (z)}{\partial z} = 1\]
Calculate Divergence: \[\nabla \cdot \mathbf{\vec{r}} = 1 + 1 + 1 = \mathbf{3}\]
The correct option is A.
Question 3: At \(t=0\), the function \(f(t) = \dfrac{\sin t}{t}\) has
a minimum
a discontinuity
a point of inflection
a maximum
Solution: The function \(f(t) = \dfrac{\sin t}{t}\) is the sinc function.
Value at \(t=0\) (L’Hôpital’s Rule): The function is undefined at \(t=0\) (a removable singularity). We find the limit: \[\lim_{t \to 0} f(t) = \lim_{t \to 0} \dfrac{\sin t}{t} = 1\] If we define \(f(0)=1\), the function becomes continuous. Since \(\lim_{t \to 0} f(t)\) exists, the function does not have a discontinuity (Option B is false).
First Derivative Test (Critical Points): Using the quotient rule: \[f'(t) = \dfrac{t \cos t - \sin t}{t^2}\] Setting \(f'(t)=0\): \[t \cos t - \sin t = 0 \implies \tan t = t\] This equation has a solution at \(t=0\). We check the sign of \(f'(t)\) around \(t=0\). For small \(t\), we use the Taylor expansion \(\sin t \approx t - \dfrac{t^3}{6}\): \[f'(t) = \dfrac{t \cos t - \sin t}{t^2} \approx \dfrac{t(1 - t^2/2) - (t - t^3/6)}{t^2} = \dfrac{t - t^3/2 - t + t^3/6}{t^2} = \dfrac{-t^3/3}{t^2} = - \dfrac{t}{3}\]
For \(t < 0\) (e.g., \(t=-0.1\)), \(f'(t) \approx -(-0.1)/3 > 0\) (\(\mathbf{f(t)}\) is increasing).
For \(t > 0\) (e.g., \(t=+0.1\)), \(f'(t) \approx -(+0.1)/3 < 0\) (\(\mathbf{f(t)}\) is decreasing).
Since \(f(t)\) changes from increasing to decreasing at \(t=0\), the function has a maximum at \(t=0\).
Second Derivative Test (Alternative confirmation for the stationary point \(t=0\)): The Taylor expansion for \(f(t)\) around \(t=0\) is: \[f(t) = \dfrac{t - t^3/6 + t^5/120 - \dots}{t} = 1 - \dfrac{t^2}{6} + \dfrac{t^4}{120} - \dots\] Since \(f(t) = 1 - (\text{positive term}) + (\text{higher order terms})\), \(f(t)\) is locally less than \(f(0)=1\). \(f(t) - f(0) = - \dfrac{t^2}{6} + \dfrac{t^4}{120} - \dots\) For small \(t\), \(f(t) - f(0) \approx - \dfrac{t^2}{6} < 0\), which implies \(f(t) < f(0)\). Thus, \(f(0)=1\) is a local maximum.
The correct option is D.
Question 4: A box contains 4 white balls and 3 red balls. In succession, two balls are randomly and removed form the box. Given that the first removed ball is white, the probability that the second removed ball is red is
\(1/3\)
\(3/7\)
\(1/2\)
\(4/7\)
Solution: This is a conditional probability problem without replacement.
Initial State:
Total balls: \(4 \text{ (White)} + 3 \text{ (Red)} = 7\) balls
Condition (First ball is white): The problem states that the first ball removed is white (W1). We are looking for the probability of the second ball being red (R2), given W1 has occurred, \(P(R_2 | W_1)\).
State after the first draw (W1): Since one white ball has been removed:
Remaining white balls: \(4 - 1 = 3\)
Remaining red balls: \(3\)
Total remaining balls: \(3 + 3 = 6\)
Calculate Conditional Probability \(P(R_2 | W_1)\): The probability that the second removed ball is red is the number of remaining red balls divided by the total number of remaining balls: \[P(R_2 | W_1) = \dfrac{\text{Number of remaining Red balls}}{\text{Total remaining balls}} = \dfrac{3}{6} = \dfrac{1}{2}\]
The correct option is C.
Question 5: An eigenvector of \(P = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix}\) is
\(\begin{bmatrix} -1 & 1 & 1 \end{bmatrix}^T\)
\(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}^T\)
\(\begin{bmatrix} 1 & -1 & 2 \end{bmatrix}^T\)
\(\begin{bmatrix} 2 & 1 & -1 \end{bmatrix}^T\)
Solution: The eigenvalues are the diagonal elements \(\lambda_1=1, \lambda_2=2, \lambda_3=3\). Check option A for \(\lambda=1\): \[(P - 1I) v = \mathbf{0} \implies \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} \ne \mathbf{0}\] Check option B for \(\lambda=2\): \[(P - 2I) v = \mathbf{0} \implies \begin{pmatrix} -1 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \ne \mathbf{0}\] Check option D for \(\lambda=3\): \[(P - 3I) v = \mathbf{0} \implies \begin{pmatrix} -2 & 1 & 0 \\ 0 & -1 & 2 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\] The eigenvector for \(\lambda=3\) is \(\mathbf{[2 \ 4 \ 2]^T}\) (a multiple of option D). Checking Option D (let \(v = [2 \ 1 \ -1]^T\)): \[P v = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix} \ne \lambda v\] The correct eigenvector for \(\lambda=1\) is \(\mathbf{[1 \ 0 \ 0]^T}\). The correct eigenvector for \(\lambda=2\) is \(\mathbf{[1 \ 1 \ 0]^T}\). The correct eigenvector for \(\lambda=3\) is \(\mathbf{[1 \ 2 \ 1]^T}\) (Option B). Let’s verify B for \(\lambda=3\): \[P v = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \\ 3 \end{pmatrix} = 3 \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\] Thus, \(\mathbf{[1 \ 2 \ 1]^T}\) is the eigenvector corresponding to \(\lambda=3\). The correct option is (B).
Question 6:For the differential equation \(\dfrac{d^2x}{dt^2} + 6\dfrac{dx}{dt} + 8x = 0\) with initial conditions \(x(0) = 1\) and \(\left.\dfrac{dx}{dt}\right|_{t=0} = 0\), the solution is
\(x(t) = 2e^{-6t} - e^{-2t}\)
\(x(t) = 2e^{-2t} - e^{-4t}\)
\(x(t) = -e^{-6t} + 2e^{-4t}\)
\(x(t) = e^{-2t} + 2e^{-4t}\)
Solution:-
\[\dfrac{d^2x}{dt^2} + 6\dfrac{dx}{dt} + 8x = 0\] Taking Laplace transform (with initial condition) on both sides \[\begin{aligned} s^2 X(s) - sx(0) - x'(0) + 6 \left[sX(s) - x(0)\right] + 8X(s) &= 0 \end{aligned}\] \[\begin{aligned} s^2 X(s) - s(1) - 0 + 6 \left[sX(s) - 1\right] + 8X(s) &= 0 \\ X(s) \left[s^2 + 6s + 8\right] - s - 6 &= 0 \end{aligned}\]
\[X(s) = \dfrac{(s+6)}{(s^2 + 6s + 8)}\]
By partial fraction \[X(s) = \dfrac{2}{s+2} - \dfrac{1}{s+4}\]
Taking inverse Laplace transform \[x(t) = (2e^{-2t} - e^{-4t})\]
The correct choice is B.
Question 7: For the set of equations, \(x_1 + 2x_2 + x_3 + 4x_4 = 2\) and \(3x_1 + 6x_2 + 3x_3 + 12x_4 = 6\). The following statement is true.
Only the trivial solution \(x_1 = x_2 = x_3 = x_4 = 0\) exists
There are no solutions
A unique non-trivial solution exists
Multiple non-trivial solutions exist
Solution:- Set of equations \[\begin{aligned} x_1 + 2x_2 + x_3 + 4x_4 &= 2 \label{eq:one} \\ 3x_1 + 6x_2 + 3x_3 + 12x_4 &= 6 \label{eq:two} \end{aligned}\]
or \[3(x_1 + 2x_2 + x_3 + 4x_4) = 3 \times 2\]
Equation (2) is same as equation(1) except a constant multiplying factor of 3.
So infinite (multiple) no. of non-trivial solution exists.
The correct choice is C
Question 8: The system \(\dot{\mathbf{X}} = \mathbf{A}\mathbf{X} + \mathbf{B}u\) with \(\mathbf{A} = \begin{bmatrix} -1 & 2 \\ 0 & 2 \end{bmatrix}\), \(\mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\) is
Stable and controllable
Stable but uncontrollable
Unstable but controllable
Unstable and uncontrollable
Solution:- Stability :
Eigen value of the system are calculated as \[|\mathbf{A} - \lambda \mathbf{I}| = 0\] \[\mathbf{A} - \lambda \mathbf{I} = \begin{bmatrix} -1 & 2 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix}\] \[\begin{aligned} &= \begin{bmatrix} -1-\lambda & 2 \\ 0 & 2-\lambda \end{bmatrix} \\ |\mathbf{A} - \lambda \mathbf{I}| &= (-1-\lambda)(2-\lambda) - 2 \times 0 = 0 \\ \implies \quad \lambda_1, \lambda_2 &= -1, 2 \end{aligned}\]
Since eigen values of the system are of opposite signs, so it is unstable
Controllability : \[\mathbf{A} =
\begin{bmatrix} -1 & 2 \\ 0 & 2 \end{bmatrix}, \mathbf{B} =
\begin{bmatrix} 0 \\ 1 \end{bmatrix}\] \[\mathbf{A}\mathbf{B} = \begin{bmatrix} 2 \\ 2
\end{bmatrix}\] \[[\mathbf{B} :
\mathbf{A}\mathbf{B}] = \begin{bmatrix} 0 & 2 \\ 1 & 2
\end{bmatrix}\] \[|\mathbf{B} :
\mathbf{A}\mathbf{B}| \neq 0\]
So it is controllable.
The correct choice is C.