ENGINEERING MATHEMATICS QUESTION PAPER GATE 2015 SET 1 & SET 2 (COMBINED)

Solution:

Let \(p\) be the probability of getting a six (success) in a single roll, and \(q\) be the probability of not getting a six (failure). \[p = P(\text{getting a six}) = \frac{1}{6}\] \[q = P(\text{not getting a six}) = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}\] Player \(A\) wins on rolls 1, 3, 5, ... The total probability that \(A\) wins is the sum of the infinite geometric series: \[P(A \text{ wins}) = p + q^2 p + q^4 p + \cdots = \frac{p}{1 - q^2}\] Substituting the values \(p=1/6\) and \(q=5/6\): \[P(A \text{ wins}) = \frac{1/6}{1 - (5/6)^2} = \frac{1/6}{1 - 25/36} = \frac{1/6}{11/36} = \frac{6}{11}\] The correct option is D. \(6/11\).

Solution:

The given ordinary differential equation is: \[\frac{d^2y}{dt^2} + 5\frac{dy}{dt} + 6y = 0\] The characteristic equation is \(\lambda^2 + 5\lambda + 6 = 0\), which yields roots \(\lambda_1 = -3\) and \(\lambda_2 = -2\). The general solution is: \[y(t) = C_1e^{-3t} + C_2e^{-2t}\] Apply \(y(0) = 2\): \[y(0) = C_1 + C_2 = 2 \quad \mathbf{(i)}\] Apply \(y(1) = -\frac{1 - 3e}{e^3}\): The given boundary condition simplifies to \(y(1) = -e^{-3} + 3e^{-2}\). Substituting \(t=1\): \[y(1) = C_1e^{-3} + C_2e^{-2}\] Comparing coefficients gives \(C_1 = -1\) and \(C_2 = 3\). Find \(\frac{dy}{dt}(0)\): The derivative of the solution is: \[\frac{dy}{dt} = -3C_1e^{-3t} - 2C_2e^{-2t}\] Substituting \(C_1 = -1\) and \(C_2 = 3\) and evaluating at \(t=0\): \[\frac{dy}{dt}(0) = -3(-1)e^{0} - 2(3)e^{0} = 3 - 6 = -3\] The value of \(\frac{dy}{dt}(0)\) is \(\boxed{-3}\).

Solution:

A matrix has three linearly independent real eigenvectors if and only if its characteristic equation, \(f(\lambda) = \det(A - \lambda I) = 0\), has three real roots.

1. Characteristic Equation: \[f(\lambda) = \lambda^3 + 6\lambda^2 + 11\lambda + (6 + 2a) = 0\] The equation has three real roots if the local minimum \(f(\lambda_{\text{min}}) \leq 0\).

2. Find Extrema: \[f'(\lambda) = 3\lambda^2 + 12\lambda + 11\] The critical points are: \(\lambda = -2 \pm \frac{\sqrt{3}}{3}\). The local minimum occurs at \(\lambda_{\text{min}} = -2 + \frac{\sqrt{3}}{3}\).

3. Apply the Condition \(f(\lambda_{\text{min}}) \leq 0\): By substituting \(\lambda_{\text{min}}\) into \(f(\lambda)\), we find: \[f(\lambda_{\text{min}}) = -\frac{2\sqrt{3}}{9} + 2a\] The condition \(f(\lambda_{\text{min}}) \leq 0\) yields: \[-\frac{2\sqrt{3}}{9} + 2a \leq 0 \implies a \leq \frac{\sqrt{3}}{9}\]

4. Maximum Value: The maximum value of \(a\) is \(\frac{\sqrt{3}}{9}\). This is equivalent to option B: \[\frac{\sqrt{3}}{9} = \frac{1}{3\sqrt{3}}\] The correct option is B. \(\frac{1}{3\sqrt{3}}\).

Solution:

Let \(A\) be a \(2 \times 2\) matrix. The sum of the diagonal elements is the Trace of the matrix, which is equal to the sum of its eigenvalues (\(\lambda_1 + \lambda_2\)). The determinant (\(\det(A)\)) is the product of its eigenvalues (\(\lambda_1 \lambda_2\)). We are given the sum of eigenvalues: \[\lambda_1 + \lambda_2 = -6\] We want to maximize the product: \[\det(A) = \lambda_1 \lambda_2\] To maximize the product of two numbers whose sum is constant, the numbers must be equal (by the AM-GM inequality, or calculus). Set \(\lambda_1 = \lambda_2\): \[\lambda_1 + \lambda_1 = -6 \implies 2\lambda_1 = -6 \implies \lambda_1 = -3\] Thus, the maximum determinant occurs when \(\lambda_1 = -3\) and \(\lambda_2 = -3\). \[\text{Maximum } \det(A) = (-3) \times (-3) = 9\] The maximum possible value of the determinant is \(\boxed{9}\).

Solution:

This problem relates to the Intermediate Value Theorem (IVT). The IVT states that for a continuous function \(f(x)\) on the interval \([a, b]\):

• If \(f(a)\) and \(f(b)\) have opposite signs, i.e., \(f(a) \cdot f(b) < 0\), then there must be at least one root (a point \(c\) where \(f(c)=0\)) in the interval \((a, b)\).

• If the function does not have a root in \([a, b]\), this implies that the function cannot cross the x-axis, and the endpoints \(f(a)\) and \(f(b)\) must have the same sign and cannot be zero.

If \(f(a)\) and \(f(b)\) have the same sign, their product must be positive: \[\text{If } f(a) > 0 \text{ and } f(b) > 0, \text{ then } f(a) \cdot f(b) > 0\] \[\text{If } f(a) < 0 \text{ and } f(b) < 0, \text{ then } f(a) \cdot f(b) > 0\] Therefore, the statement that must be TRUE is \(\mathbf{f(a) \cdot f(b) > 0}\). The correct option is C. \(f(a) \cdot f(b) > 0\).

Solution:

We use the two defining properties of a Probability Density Function (PDF) to find the constants \(a\) and \(b\).

1. Normalization Condition (\(\int_{-\infty}^{\infty} f(x)\,dx = 1\)): \[\int_{0}^{1} (a+bx)\,dx = 1\] \[\left[ ax + \frac{bx^2}{2} \right]_{0}^{1} = 1\] \[a + \frac{b}{2} = 1 \implies \mathbf{2a + b = 2} \quad \mathbf{(i)}\]

2. Expected Value Condition (\(E[X] = 2/3\)): \[E[X] = \int_{-\infty}^{\infty} x f(x)\,dx = \frac{2}{3}\] \[\int_{0}^{1} x(a+bx)\,dx = \int_{0}^{1} (ax+bx^2)\,dx = \frac{2}{3}\] \[\left[ \frac{ax^2}{2} + \frac{bx^3}{3} \right]_{0}^{1} = \frac{2}{3}\] \[\frac{a}{2} + \frac{b}{3} = \frac{2}{3} \implies 3a + 2b = 4 \quad \mathbf{(ii)}\]

3. Solve for \(a\) and \(b\): Multiply (i) by 2: \(\quad 4a + 2b = 4\) Subtract (ii) from the new equation: \[(4a + 2b) - (3a + 2b) = 4 - 4\] \[a = 0\] Substitute \(a=0\) into (i): \[2(0) + b = 2 \implies b = 2\] Thus, the PDF is \(f(x) = 2x\) for \(0 < x < 1\).

4. Calculate \(\Pr[X < 0.5]\): \[\Pr[X < 0.5] = \int_{0}^{0.5} f(x)\,dx = \int_{0}^{0.5} 2x\,dx\] \[\left[ x^2 \right]_{0}^{0.5} = (0.5)^2 - 0^2 = 0.25\] The value of \(\Pr[X < 0.5]\) is \(\boxed{0.25}\).

Solution:

The discrete-time update equations can be expressed in matrix form as: \[\begin{bmatrix} x_{k+1} \\ y_{k+1} \end{bmatrix} = \begin{bmatrix} 1 & h \\ -h & 1 \end{bmatrix} \begin{bmatrix} x_k \\ y_k \end{bmatrix}\] The eigenvalues of this matrix determine the system’s stability. The characteristic equation is: \[|A - \lambda I| = (1 - \lambda)^2 + h^2 = 0\] Hence, the eigenvalues are: \[\lambda = 1 \pm jh\] The magnitude of each eigenvalue is: \[|\lambda| = \sqrt{1 + h^2} > 1 \text{ for all } h > 0\] Therefore, the eigenvalues lie outside the unit circle for any nonzero \(h\), meaning the system is unstable for all \(h > 0\).

Correct Option: A. The system is not stable for \(h > 0\).

Solution:

For independence, we must have \[P(H_RH_S)=P(H_R)P(H_S)\] Compute marginal probabilities: \[P(H_R)=0.28+0.30 = 0.58,\qquad P(H_S)=0.28+0.24 = 0.52\] If independent, \[P(H_RH_S)=0.58 \times 0.52 = 0.3016\] Given: \[P(H_RH_S)=0.28 \ne 0.3016\] Thus, the probabilities do not factor as the product of marginals.

Correct Option: D. The coin tosses are dependent.

Solution:

The triangular region is bounded by \(x=0\), \(y=1\), and \(x=y\). So the limits are: \[0 \le x \le 1, \qquad x \le y \le 1\] Required volume: \[\iint_{R} e^x\, dy\, dx = \int_{0}^{1} \int_{x}^{1} e^x\, dy\, dx\] Integrate w.r.t \(y\) first: \[= \int_{0}^{1} e^x (1-x)\, dx\] \[= \int_{0}^{1} (e^x - xe^x)\, dx\] \[= \left[e^x\right]_{0}^{1} - \left[(xe^x - e^x)\right]_{0}^{1}\] \[= (e - 1) - \left[(e - e) - (0 - 1)\right]\] \[= (e - 1) - (0 + 1) = e - 2\] Correct Option: B. \(e - 2\)

Solution:

Given: \[f(t)=2\sqrt{\frac{t}{\pi}} \quad \Rightarrow \quad \mathcal{L}\{f(t)\}=s^{-3/2}\] Now, \[g(t)=\sqrt{\frac{1}{\pi t}} = \frac{1}{\sqrt{\pi}}t^{-1/2}\] We know: \[\frac{d}{dt}t^{1/2} = \frac{1}{2}t^{-1/2} \quad \Rightarrow \quad t^{-1/2} = 2\frac{d}{dt}(t^{1/2})\] Thus, \[g(t)=\frac{1}{\sqrt{\pi}}t^{-1/2} = \frac{2}{\sqrt{\pi}}\frac{d}{dt}(t^{1/2})\] Using Laplace property \[\mathcal{L}\left\{\frac{df}{dt}\right\} = sF(s)\] We already know that \[\mathcal{L}\{2\sqrt{\tfrac{t}{\pi}}\}=s^{-3/2}\] So, \[\mathcal{L}\{g(t)\} = s \cdot s^{-3/2} = s^{-1/2}\] Correct Option: B. \(s^{-1/2}\)

Solution:

Stokes’s theorem: \[\oint A \cdot dl = \iint (\nabla \times A)\cdot ds \quad \Rightarrow \quad P \rightarrow 4\] Gauss’s theorem (Flux theorem): \[\iint D \cdot ds = Q \quad \Rightarrow \quad Q \rightarrow 1\] Divergence theorem: \[\iiint (\nabla \cdot A)\,dv = \iint A \cdot ds \quad \Rightarrow \quad R \rightarrow 3\] Cauchy’s integral theorem: \[\oint f(z)\,dz = 0 \quad \Rightarrow \quad S \rightarrow 2\] \[\boxed{P-4,\; Q-1,\; R-3,\; S-2}\] Correct Option: B

Solution:

For a system of 3 linear equations in 3 variables: - \(P\): A unique solution exists \(\iff\) coefficient matrix is invertible. - \(Q\): Equations are linearly independent \(\iff\) rows of coefficient matrix are linearly independent \(\iff\) determinant \(\neq 0\). - \(R\): All eigenvalues of the coefficient matrix are nonzero \(\iff\) determinant (product of eigenvalues) \(\neq 0\). - \(S\): Determinant \(\neq 0\) means invertible matrix. Thus: \[P \iff Q \iff R \iff S\] All four statements are equivalent. Correct Option: A

Solution:

We have \(f(z) = g(z) + h(z)\). If \(g(z)\) and \(h(z)\) are differentiable at \(z_0\), then both satisfy the Cauchy–Riemann equations. Since the sum of two differentiable complex functions is also differentiable, \[f(z) = g(z) + h(z)\] must be differentiable at \(z_0\). Thus, the correct statement is: \[\boxed{\text{If } g(z) \text{ and } h(z) \text{ are differentiable at } z_0, \text{ then } f(z) \text{ is also differentiable at } z_0.}\] Correct Option: B