Question 1. Two independent random variables \(X\) and \(Y\) are uniformly distributed in the interval \([-1, 1]\). The probability that \(\max[X, Y]\) is less than \(1/2\) is

1. \(3/4\)

2. \(9/16\)

3. \(1/4\)

4. \(2/3\)

Solution: Uniform distribution \(X\), \(Y\) on \([-1, 1]\); \(f(x) = f(y) = \frac{1}{2}\)

\[P\left(\max(X, Y) \le \frac{1}{2}\right) = P\left(X \le \frac{1}{2}, -1 \le Y \le \frac{1}{2}\right) \cdot P\left(-1 \le X \le \frac{1}{2}, Y \le \frac{1}{2}\right)\]

\[= \int_{-1}^{1/2} \frac{1}{2} \, dx \int_{-1}^{1/2} \frac{1}{2} \, dy = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}\]

The correct option is B

Question 2. With initial condition \(x(1) = 0.5\), the solution of the differential equation \(t\frac{dx}{dt} + x = t\) is

1. \(x = t - \frac{1}{2}\)

2. \(x = t^2 - \frac{1}{2}\)

3. \(x = \frac{t^2}{2}\)

4. \(x = \frac{t}{2}\)

Solution: The given differential equation is a first-order linear differential equation.

Standard Form: The given equation is \(t\frac{dx}{dt} + x = t\). Dividing by \(t\) (for \(t \ne 0\)) gives the standard linear form \(\frac{dx}{dt} + P(t)x = Q(t)\): \[\frac{dx}{dt} + \frac{x}{t} = 1\] Here, \(P(t) = \frac{1}{t}\) and \(Q(t) = 1\).

Integrating Factor (IF): \[IF = e^{\int P(t) dt} = e^{\int \frac{1}{t} dt}\] \[IF = e^{\log|t|} = t \quad (\text{assuming } t > 0)\]

General Solution: The solution is given by \(x \cdot (IF) = \int Q(t) \cdot (IF) \, dt + C\): \[x \cdot t = \int 1 \cdot t \, dt + C\] \[xt = \int t \, dt + C\] \[xt = \frac{t^2}{2} + C\]

Apply Initial Condition: Given the initial condition \(x(1) = 0.5\), substitute \(t=1\) and \(x=0.5 = \frac{1}{2}\): \[(1) \left(\frac{1}{2}\right) = \frac{(1)^2}{2} + C\] \[\frac{1}{2} = \frac{1}{2} + C\] \[C = 0\]

Final Solution: Substitute \(C=0\) back into the general solution \(xt = \frac{t^2}{2} + C\): \[xt = \frac{t^2}{2}\] \[x = \frac{t^2}{2t} = \frac{t}{2}\]

The required solution is \(x = \frac{t}{2}\). The correct option is D.

Question 3. The maximum value of \(f(x) = x^3 - 9x^2 + 24x + 5\) in the interval \([1, 6]\) is

1. 21

2. 25

3. 41

4. 46

Solution: To find the maximum value of \(f(x)\) on the closed interval \([1, 6]\), we check the values at the critical points inside the interval and at the endpoints.

Find Critical Points: Calculate the first derivative, \(f'(x)\), and set it to zero for stationary values: \[f(x) = x^3 - 9x^2 + 24x + 5\] \[f'(x) = 3x^2 - 18x + 24\] Set \(f'(x) = 0\): \[3x^2 - 18x + 24 = 0\] \[3(x^2 - 6x + 8) = 0\] Factor the quadratic: \[(x-2)(x-4) = 0\] The critical points are \(x=2\) and \(x=4\). Both \(x=2\) and \(x=4\) are within the interval \([1, 6]\).

Evaluate \(f(x)\) at Critical Points and Endpoints:

•  Endpoint (\(x=1\)): \[f(1) = (1)^3 - 9(1)^2 + 24(1) + 5 = 1 - 9 + 24 + 5 = 21\]

•  Critical Point (\(x=2\)): \[f(2) = (2)^3 - 9(2)^2 + 24(2) + 5 = 8 - 36 + 48 + 5 = 25\]

•  Critical Point (\(x=4\)): \[f(4) = (4)^3 - 9(4)^2 + 24(4) + 5 = 64 - 144 + 96 + 5 = 21\]

•  Endpoint (\(x=6\)): \[f(6) = (6)^3 - 9(6)^2 + 24(6) + 5 = 216 - 324 + 144 + 5 = 41\]

Determine Maximum Value: Comparing the values: \(\{21, 25, 21, 41\}\). The maximum value of the function in the interval \([1, 6]\) is \(\mathbf{41}\). The correct option is C.

Question 4. Given that \(A = \begin{bmatrix} -5 & -3 \\ 2 & 0 \end{bmatrix}\) and \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), the value of \(A^3\) is

1. \(15A + 12I\)

2. \(19A + 30I\)

3. \(17A + 15I\)

4. \(17A + 21I\)

Solution: We use the Cayley-Hamilton Theorem, which states that every square matrix satisfies its own characteristic equation.

Find the Characteristic Equation: The characteristic equation is given by \(\det(A - \lambda I) = 0\): \[\det\left(\begin{bmatrix} -5-\lambda & -3 \\ 2 & 0-\lambda \end{bmatrix}\right) = 0\] \[(-5-\lambda)(-\lambda) - (-3)(2) = 0\] \[5\lambda + \lambda^2 + 6 = 0\] \[\lambda^2 + 5\lambda + 6 = 0\]

Apply Cayley-Hamilton Theorem to \(A\): Replace \(\lambda\) with the matrix \(A\) and the constant term with \(6I\): \[A^2 + 5A + 6I = 0\] \[\mathbf{A^2 = -5A - 6I}\]

Calculate \(A^3\): Multiply the equation for \(A^2\) by \(A\): \[A^3 = A \cdot A^2 = A(-5A - 6I)\] \[A^3 = -5A^2 - 6AI\] \[A^3 = -5A^2 - 6A\]

Substitute \(A^2\) again: Substitute the expression for \(A^2\) found in Step 2: \[A^3 = -5(-5A - 6I) - 6A\] \[A^3 = 25A + 30I - 6A\] \[\mathbf{A^3 = 19A + 30I}\]

The correct option is B.

Question 5. Consider the differential equation \[\frac{d^2y(t)}{dt^2} + 2\frac{dy(t)}{dt} + y(t) = \delta(t)\] with \(y(t)|_{t=0^-} = -2\) and \(\frac{dy}{dt}\bigg|_{t=0^-} = 0\).

The numerical value of \(\frac{dy}{dt}\bigg|_{t=0^+}\) is

1. -2

2. -1

3. 0

4. 1

Solution:

\[\frac{d^2y(t)}{dt^2} + 2 \frac{dy(t)}{dt} + y(t) = \delta(t)\] Converting to s-domain, with \(y(0^-)=-2\) and \(y'(0^-)=0\): \[s^2Y(s) - sy(0) - y'(0) + 2[sY(s) - y(0)] + Y(s) = 1\] Substituting I.C.s: \[[s^2 + 2s + 1]Y(s) + 2s + 4 = 1\] \[Y(s) = \frac{1 - 2s - 4}{(s^2 + 2s + 1)} = \frac{-3 - 2s}{(s^2 + 2s + 1)}\] Find inverse lapalce transform (using Partial Fraction Decomposition): \[y(t) = [-2e^{-t} - te^{-t}]u(t)\] The derivative for \(t>0\) is: \[\frac{dy(t)}{dt} = 2e^{-t} + te^{-t} - e^{-t}\] Evaluating at \(t=0^+\): \[\frac{dy(t)}{dt}\bigg|_{t=0^+} = 2e^{-0} + 0 \cdot e^{-0} - e^{-0}\] \[\frac{dy(t)}{dt}\bigg|_{t=0^+} = 2 - 1 = \mathbf{1}\]

The correct option is D.

Question 6 A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

1. \(1/3\)

2. \(1/2\)

3. \(2/3\)

4. \(3/4\)

Solution:

Exp:- \(P(\text{odd tosses}) = P(H) + P(TTH) + P(TTTTH) + \dots\dots\) \[= \frac{1}{2} + \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^5 + \dots\dots\] \[= \frac{1}{2} \left( 1 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^4 + \dots\dots \right)\] \[= \frac{1}{2} \left[ 1 + \left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)^2 + \dots\dots \right]\] \[= \frac{1}{2} \left[ \frac{1}{1 - \frac{1}{4}} \right] = \frac{1}{2} \cdot \frac{1}{\frac{3}{4}} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}\] The correct option is C.