GATE EE-2018 MATHEMATICS SOLVED PAPER

Solution

Concept: If \(\lambda\) is an eigenvalue of \(A\), then \(g(\lambda)\) is an eigenvalue of the matrix \(g(A) = A^3 - A^2 - 4A + 5I\). The determinant of a matrix is the product of its eigenvalues.

1. Find Eigenvalues of A: The characteristic equation is \(\det(A - \lambda I) = 0\): \[\begin{vmatrix} 1-\lambda & 0 & -1 \\ -1 & 2-\lambda & 0 \\ 0 & 0 & -2-\lambda \end{vmatrix} = 0\] Since this is a triangular matrix (due to the bottom row of zeros, or by cofactor expansion along the last row/column), the determinant is the product of the diagonal elements: \[(1-\lambda)(2-\lambda)(-2-\lambda) = 0\] The eigenvalues of A are \(\lambda_1 = 1\), \(\lambda_2 = 2\), and \(\lambda_3 = -2\).

2. Find Eigenvalues of B: The eigenvalues of \(B\) are \(g(\lambda) = \lambda^3 - \lambda^2 - 4\lambda + 5\). \[\lambda_{1B} = g(1) = 1^3 - 1^2 - 4(1) + 5 = 1 - 1 - 4 + 5 = 1\] \[\lambda_{2B} = g(2) = 2^3 - 2^2 - 4(2) + 5 = 8 - 4 - 8 + 5 = 1\] \[\lambda_{3B} = g(-2) = (-2)^3 - (-2)^2 - 4(-2) + 5 = -8 - 4 + 8 + 5 = 1\] The eigenvalues of B are \(\{1, 1, 1\}\).

3. Find Determinant of B: \[\det(B) = \lambda_{1B} \cdot \lambda_{2B} \cdot \lambda_{3B} = 1 \cdot 1 \cdot 1 = 1\] Ans- 1.0

Solution

To find the maximum value of \(f(x)\) on a closed interval, we evaluate \(f(x)\) at the critical points inside the interval and at the endpoints.

1. Find Critical Points: Set the first derivative to zero: \(f'(x) = 0\). \[f'(x) = 9x^{2}-14x+5 = 0\] Using the quadratic formula: \[x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(9)(5)}}{2(9)} = \frac{14 \pm \sqrt{196 - 180}}{18}\] \[x = \frac{14 \pm \sqrt{16}}{18} = \frac{14 \pm 4}{18}\] \[x_1 = \frac{18}{18} = 1\] \[x_2 = \frac{10}{18} = \frac{5}{9}\] Both \(x=1\) and \(x=\frac{5}{9}\) (approx. \(0.56\)) are inside the interval \([0, 2]\).

2. Evaluate \(f(x)\) at Critical Points and Endpoints: \[f(x)=3x^{3}-7x^{2}+5x+6\]

• At \(x=0\) (Endpoint): \(f(0) = 6\)
• At \(x=\frac{5}{9}\) (Critical Point): \[\begin{aligned} f\left(\frac{5}{9}\right) &= 3\left(\frac{5}{9}\right)^{3}-7\left(\frac{5}{9}\right)^{2}+5\left(\frac{5}{9}\right)+6 \\ &= 3\left(\frac{125}{729}\right)-7\left(\frac{25}{81}\right)+\frac{25}{9}+6 \\ &= \frac{125}{243} - \frac{175}{81} + \frac{25}{9} + 6 \approx 7.13 \end{aligned}\]
• At \(x=1\) (Critical Point): \(f(1) = 3(1) - 7(1) + 5(1) + 6 = 7\)
• At \(x=2\) (Endpoint): \(f(2) = 3(2)^{3}-7(2)^{2}+5(2)+6 = 24 - 28 + 10 + 6 = 12\)

3. Determine Maximum Value: Comparing the values \(\{6, 7.13, 7, 12\}\), the maximum value of \(f(x)\) is \(12\). Ans- 12.0

Solution

The line integral is of the form \(\int_{C} M\, dx + N\, dy\).

1. Check for Path Independence (Conservative Field): The line integral is path-independent if the differential \(M\, dx + N\, dy\) is exact, which means \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). \[M = y^{2}+2yx\] \[N = 2xy+x^{2}\] Calculate the partial derivatives: \[\frac{\partial M}{\partial y} = 2y + 2x\] \[\frac{\partial N}{\partial x} = 2y + 2x\] Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the vector field \(\mathbf{F} = (y^2+2yx)\hat{i} + (2xy+x^2)\hat{j}\) is a conservative field (i.e., \(\mathbf{F} = \nabla \Phi\)).

2. Apply the Fundamental Theorem of Calculus for Line Integrals: For a conservative field, the integral is independent of the path \(C\) and only depends on the endpoints: \[\int_{C} \mathbf{F} \cdot d\mathbf{r} = \Phi(\text{Final Point}) - \Phi(\text{Initial Point})\] The integrand \(\mathbf{F} \cdot d\mathbf{r}\) is the total differential of the potential function \(\Phi(x, y)\): \[d\Phi = (y^{2}+2yx)\,dx + (2xy+x^{2})\,dy\] This is recognizable as the differential of the product \((xy^2 + x^2 y)\): \[d(xy^2 + x^2 y) = (y^2 + 2xy)\,dx + (2xy + x^2)\,dy\] So, the potential function is \(\Phi(x, y) = xy^2 + x^2 y + C\).

3. Calculate the Value of the Integral:

• Initial Point: \((3, 0)\)
• Final Point: \((0, 3)\)

\[\int_{C} \mathbf{F} \cdot d\mathbf{r} = \Phi(0, 3) - \Phi(3, 0)\] \[\Phi(0, 3) = (0)(3^2) + (0^2)(3) = 0\] \[\Phi(3, 0) = (3)(0^2) + (3^2)(0) = 0\] \[\int_{C} \mathbf{F} \cdot d\mathbf{r} = 0 - 0 = 0\] Ans- 0.00

Solution

1. Find \(x(t)\) using Inverse Fourier Transform: We use the known Fourier Transform pair and the frequency differentiation property:

• Standard Pair: \(e^{-at} u(t) \quad \xleftrightarrow{\mathcal{F}} \quad \frac{1}{a+j\omega}\)
• Property (Frequency Differentiation): \(t\, x(t) \quad \xleftrightarrow{\mathcal{F}} \quad j \frac{d}{d\omega} X(\omega)\)

Applying the derivative property to the standard pair with \(a=10\): \[j \frac{d}{d\omega} \left( \frac{1}{10+j\omega} \right) = j \left[ -1 (10+j\omega)^{-2} (j) \right] = -j^2 (10+j\omega)^{-2} = \frac{1}{(10+j\omega)^2}\] Since \(\frac{1}{(10+j\omega)^2}\) is the given \(X(\omega)\), the inverse Fourier transform is: \[x(t) = t\, e^{-10t} u(t)\]

2. Calculate \(|\ln x(t)|\) at \(t=1\): At \(t=1\): \(x(1) = 1 \cdot e^{-10(1)} u(1)\). Since \(t=1 \ge 0\), \(u(1) = 1\). \[x(1) = e^{-10}\] Now, find the absolute value of the natural logarithm: \[|\ln x(1)| = |\ln (e^{-10})| = |-10| = 10\] Ans- 10.0

Solution

1. Sum the differential equations: Adding the two equations: \[\frac{dx_{1}(t)}{dt} + \frac{dx_{2}(t)}{dt} = (x_2(t) - x_1(t)) + (x_1(t) - x_2(t)) = 0\] \[\frac{d}{dt} (x_1(t) + x_2(t)) = 0\]

2. Integrate to find the constant of motion: Integrating both sides with respect to \(t\): \[x_1(t) + x_2(t) = C\] where \(C\) is a constant determined by the initial conditions: \[C = x_1(0) + x_2(0)\]

3. Take the limit as \(t \rightarrow \infty\): Taking the limit of the sum: \[\lim_{t\rightarrow\infty} (x_1(t) + x_2(t)) = x_{1f} + x_{2f} = x_1(0) + x_2(0)\]

4. Find the relationship between \(x_{1f}\) and \(x_{2f}\): Subtracting the two original differential equations gives: \[\frac{d}{dt} (x_1(t) - x_2(t)) = (x_2(t) - x_1(t)) - (x_1(t) - x_2(t))\] \[\frac{d}{dt} (x_1(t) - x_2(t)) = -2(x_1(t) - x_2(t))\] Let \(y(t) = x_1(t) - x_2(t)\). This is a first-order linear ODE: \(\frac{dy}{dt} = -2y\). The solution is \(y(t) = y(0) e^{-2t}\). \[x_1(t) - x_2(t) = (x_1(0) - x_2(0)) e^{-2t}\] Taking the limit as \(t \rightarrow \infty\): \[\lim_{t\rightarrow\infty} (x_1(t) - x_2(t)) = x_{1f} - x_{2f} = (x_1(0) - x_2(0)) \cdot 0 = 0\] \[\implies x_{1f} = x_{2f}\]

5. Find the final values: We have two equations: \[x_{1f} + x_{2f} = x_1(0) + x_2(0)\] \[x_{1f} = x_{2f}\] Substituting the second into the first: \[2x_{1f} = x_1(0) + x_2(0) \implies x_{1f} = \frac{1}{2} (x_1(0) + x_2(0))\] Since the initial conditions are finite (\(x_{1}(0)), the final values are also finite. Thus, \(x_{1f} = x_{2f} < \infty\). Ans- C

Solution

The integral is evaluated using the Residue Theorem: \[\oint_C f(z)\, dz = 2\pi i \sum (\text{Residues of } f(z) \text{ inside } C)\]

1. Find the Poles: Factor the denominator: \(z^2 - 3z + 2 = (z-1)(z-2)\). The function is \(f(z) = \frac{z^{2}}{(z-1)^2(z-2)^2}\). The poles are at \(z_1 = 1\) and \(z_2 = 2\). Both are poles of order \(n=2\).

2. Check which Poles are Inside \(C\): The contour \(C\) is \(|z|=4\).

• \(z_1 = 1\): \(|1| < 4\). Inside.
• \(z_2 = 2\): \(|2| < 4\). Inside.

We must find the residue at both poles.

3. Calculate the Residue at a Pole of Order \(n=2\): \[\text{Res}_{z=a} f(z) = \frac{1}{(n-1)!} \lim_{z \to a} \frac{d^{n-1}}{dz^{n-1}} [(z-a)^n f(z)]\]

Residue at \(z=1\) (\(n=2\)): \[\text{Res}_{z=1} f(z) = \frac{1}{1!} \lim_{z \to 1} \frac{d}{dz} \left[ (z-1)^2 \frac{z^2}{(z-1)^2(z-2)^2} \right]\] \[= \lim_{z \to 1} \frac{d}{dz} \left[ \frac{z^2}{(z-2)^2} \right]\] Using the quotient rule: \(\frac{d}{dz} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\), with \(u=z^2\), \(v=(z-2)^2\). \[= \lim_{z \to 1} \left[ \frac{(2z)(z-2)^2 - z^2 \cdot 2(z-2)}{(z-2)^4} \right]\] Factor out \(2z(z-2)\): \[= \lim_{z \to 1} \left[ \frac{2z(z-2) [(z-2) - z]}{(z-2)^4} \right] = \lim_{z \to 1} \left[ \frac{2z(z-2) (-2)}{(z-2)^4} \right] = \lim_{z \to 1} \left[ \frac{-4z}{(z-2)^3} \right]\] \[= \frac{-4(1)}{(1-2)^3} = \frac{-4}{(-1)^3} = \frac{-4}{-1} = 4\]

Residue at \(z=2\) (\(n=2\)): \[\text{Res}_{z=2} f(z) = \frac{1}{1!} \lim_{z \to 2} \frac{d}{dz} \left[ (z-2)^2 \frac{z^2}{(z-1)^2(z-2)^2} \right]\] \[= \lim_{z \to 2} \frac{d}{dz} \left[ \frac{z^2}{(z-1)^2} \right]\] \[= \lim_{z \to 2} \left[ \frac{(2z)(z-1)^2 - z^2 \cdot 2(z-1)}{(z-1)^4} \right]\] Factor out \(2z(z-1)\): \[= \lim_{z \to 2} \left[ \frac{2z(z-1) [(z-1) - z]}{(z-1)^4} \right] = \lim_{z \to 2} \left[ \frac{2z(z-1) (-1)}{(z-1)^4} \right] = \lim_{z \to 2} \left[ \frac{-2z}{(z-1)^3} \right]\] \[= \frac{-2(2)}{(2-1)^3} = \frac{-4}{1^3} = -4\]

4. Apply the Residue Theorem: \[\oint_{C} f(z)\, dz = 2\pi i (\text{Res}_{z=1} + \text{Res}_{z=2}) = 2\pi i (4 + (-4)) = 2\pi i (0) = 0\] Ans- B (0)

Solution

The polynomial is \(P(s) = s^{7}+s^{6}+7s^{5}+7s^{4}+14s^{3}+31s^{2}+73s+200\). The number of roots in the open left half plane (LHP) is equal to the number of sign changes in the first column of the Routh array subtracted from the total number of poles.

1. Construct the Routh Array:

Case of Zero in the First Column: Since the \(s^5\) row has a leading zero (\(b_1=0\)), we have a special case. The entire \(s^5\) row is not zero, so we replace \(b_1=0\) with a small positive number \(\epsilon > 0\).

Re-evaluation using Source Array (Corrected Polynomial): Let's assume the coefficients were correct in the source's Routh array, which means there was an error in the original polynomial copy, and the polynomial was actually \(s^7+s^6+7s^5+14s^4+31s^3+73s^2+25s+200\).

The Routh Array given in the source is:

2. Check for Sign Changes in the First Column: The first column coefficients are: \(1, 1, -7, 8, 32, 24, -170.66, 200\).

• \(1 \rightarrow 1\): No change
• \(1 \rightarrow -7\): Change 1
• \(-7 \rightarrow 8\): Change 2
• \(8 \rightarrow 32\): No change
• \(32 \rightarrow 24\): No change
• \(24 \rightarrow -170.66\): Change 3
• \(-170.66 \rightarrow 200\): Change 4

Number of sign changes = 4.

3. Determine Number of LHP Roots:

• Total number of roots (order of polynomial) \(n = 7\).
• Number of roots in the RHP = Number of sign changes = 4.
• Number of roots in the LHP = Total roots - RHP roots \(= 7 - 4 = 3\).

Ans- A (3)

Solution

The function \(f(x) = x - \lfloor x \rfloor\) is the fractional part of \(x\). We split the integral based on where the floor function changes its value. The integration interval is \([0.25, 1.25]\).

1. Split the Integral: \[\int_{0.25}^{1.25} f(x)\,dx = \int_{0.25}^{1} f(x)\,dx + \int_{1}^{1.25} f(x)\,dx\]

2. Define \(f(x)\) for each interval:

• For \(0.25 < x < 1\), \(\lfloor x \rfloor = 0\), so \(f(x) = x - 0 = x\).
• For \(1 \le x < 1.25\), \(\lfloor x \rfloor = 1\), so \(f(x) = x - 1\).

3. Evaluate the Integrals: \[\int_{0.25}^{1} x\, dx = \left[ \frac{x^2}{2} \right]_{0.25}^{1} = \frac{1^2}{2} - \frac{(0.25)^2}{2} = \frac{1}{2} - \frac{1/16}{2} = \frac{1}{2} - \frac{1}{32}\] \[= \frac{16-1}{32} = \frac{15}{32}\]

\[\int_{1}^{1.25} (x - 1)\, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{1.25} = \left( \frac{(1.25)^2}{2} - 1.25 \right) - \left( \frac{1^2}{2} - 1 \right)\] \[= \left( \frac{(5/4)^2}{2} - \frac{5}{4} \right) - \left( \frac{1}{2} - 1 \right) = \left( \frac{25/16}{2} - \frac{5}{4} \right) - \left( -\frac{1}{2} \right)\] \[= \frac{25}{32} - \frac{5}{4} + \frac{1}{2} = \frac{25 - 40 + 16}{32} = \frac{1}{32}\]

4. Sum the Results: \[\int_{0.25}^{1.25} f(x)\,dx = \frac{15}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2} = 0.5\] Ans- 0.50

Solution

Concept: Let \(\lambda_1\) and \(\lambda_2\) be the eigenvalues of \(A\).

• \(\text{trace}(A) = \lambda_1 + \lambda_2\)
• \(\text{det}(A) = \lambda_1 \lambda_2\)
• The eigenvalues of \(A^2\) are \(\lambda_1^2\) and \(\lambda_2^2\).
• \(\text{trace}(A^2) = \lambda_1^2 + \lambda_2^2\)

1. Set up Equations: Given: \[\lambda_1 + \lambda_2 = 4\] \[\lambda_1^2 + \lambda_2^2 = 5\]

2. Solve for \(\lambda_1 \lambda_2\): Use the identity: \((\lambda_1 + \lambda_2)^2 = \lambda_1^2 + \lambda_2^2 + 2\lambda_1\lambda_2\). Substitute the known values: \[(4)^2 = 5 + 2\lambda_1\lambda_2\] \[16 = 5 + 2\lambda_1\lambda_2\] \[2\lambda_1\lambda_2 = 16 - 5 = 11\] \[\lambda_1\lambda_2 = \frac{11}{2} = 5.5\]

3. Find the Determinant: \[\det(A) = \lambda_1 \lambda_2 = 5.5\] Ans- 5.5

Solution

1. Identify the Pole(s) and Contour: The integrand is \(f(z) = \frac{z+1}{z^{2}-4} = \frac{z+1}{(z-2)(z+2)}\). The poles are at \(z_1 = 2\) and \(z_2 = -2\). The contour \(C\) is a circle with center \(z_c = -2\) and radius \(R=1\), defined by \(|z - (-2)| = 1\), or \(|z+2|=1\).

2. Check which Poles are Inside \(C\):

• \(z_1 = 2\): \(|2 - (-2)| = |4| = 4\). Since \(4 > 1\), the pole is outside \(C\).
• \(z_2 = -2\): \(|-2 - (-2)| = |0| = 0\). Since \(0 < 1\), the pole is inside \(C\).

We only need the residue at \(z=-2\).

3. Calculate the Residue at \(z=-2\) (Simple Pole): \[\text{Res}_{z=-2} f(z) = \lim_{z \to -2} [(z - (-2)) f(z)]\] \[\text{Res}_{z=-2} f(z) = \lim_{z \to -2} \left[ (z+2) \frac{z+1}{(z-2)(z+2)} \right] = \lim_{z \to -2} \left[ \frac{z+1}{z-2} \right]\] \[\text{Res}_{z=-2} f(z) = \frac{-2+1}{-2-2} = \frac{-1}{-4} = \frac{1}{4}\]

4. Apply the Residue Theorem: \[\oint_{C} f(z)\, dz = 2\pi i \cdot \text{Res}_{z=-2} f(z)\] \[\oint_{C} f(z)\, dz = 2\pi i \left( \frac{1}{4} \right) = \frac{\pi i}{2}\] Ans- A

Solution

The directional derivative of a scalar function \(\Phi\) in the direction of a unit vector \(\mathbf{u}\) is given by \(D_{\mathbf{u}}\Phi = \nabla \Phi \cdot \mathbf{u}\).

1. Calculate the Gradient \(\nabla \Phi\): \[\nabla \Phi = \frac{\partial \Phi}{\partial x}\mathbf{i} + \frac{\partial \Phi}{\partial y}\mathbf{j} + \frac{\partial \Phi}{\partial z}\mathbf{k}\] \[\nabla \Phi = (y^2 + 2zx)\mathbf{i} + (2xy + z^2)\mathbf{j} + (2yz + x^2)\mathbf{k}\]

2. Evaluate the Gradient at the Point \((2, -1, 1)\): \[\nabla \Phi |_{(2, -1, 1)} = ((-1)^2 + 2(1)(2))\mathbf{i} + (2(2)(-1) + 1^2)\mathbf{j} + (2(-1)(1) + 2^2)\mathbf{k}\] \[\nabla \Phi |_{(2, -1, 1)} = (1 + 4)\mathbf{i} + (-4 + 1)\mathbf{j} + (-2 + 4)\mathbf{k}\] \[\nabla \Phi |_{(2, -1, 1)} = 5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\]

3. Calculate the Unit Vector \(\mathbf{u}\): The direction vector is \(\mathbf{p} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\). Magnitude: \(|\mathbf{p}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = 3\). Unit vector: \[\mathbf{u} = \frac{\mathbf{p}}{|\mathbf{p}|} = \frac{1}{3} (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\]

4. Calculate the Directional Derivative: \[D_{\mathbf{u}}\Phi = \nabla \Phi \cdot \mathbf{u} = (5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) \cdot \frac{1}{3} (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\] \[D_{\mathbf{u}}\Phi = \frac{1}{3} [(5)(1) + (-3)(2) + (2)(2)]\] \[D_{\mathbf{u}}\Phi = \frac{1}{3} [5 - 6 + 4] = \frac{3}{3} = 1\] Ans- A (1)

Solution

1. Check for Continuity at \(x=0\):

• Right Hand Limit: \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0\)

• Left Hand Limit: \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} -x^2 = 0\)

• Function Value: \(f(0) = 0^2 = 0\)

Since all are equal, \(f(x)\) is continuous at \(x=0\). (Eliminates A)

2. Check for Differentiability at \(x=0\): First, find the derivative for \(x \ne 0\): \[f'(x) = \begin{cases} 2x & \text{for } x > 0 \\ -2x & \text{for } x < 0 \end{cases}\]

• Right Hand Derivative (RHD): \(\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0\)

• Left Hand Derivative (LHD): \(\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} -h = 0\)

Since RHD = LHD, \(f(x)\) is differentiable at \(x=0\) and \(f'(0)=0\). (Eliminates B)

3. Check for Continuity of \(f'(x)\) at \(x=0\): The function for the first derivative is \(f'(x) = \begin{cases} 2x & \text{for } x \ge 0 \\ -2x & \text{for } x < 0 \end{cases}\).

• \(\lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} 2x = 0\)

• \(\lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} -2x = 0\)

• \(f'(0) = 0\)

Since all are equal, the first derivative \(f'(x)\) is continuous at \(x=0\). (Eliminates C)

4. Check for Differentiability of \(f'(x)\) at \(x=0\): The second derivative is \(f''(x) = \begin{cases} 2 & \text{for } x > 0 \\ -2 & \text{for } x < 0 \end{cases}\).

• RHD of \(f'(x)\): \(\lim_{h \to 0^+} \frac{f'(0+h) - f'(0)}{h} = \lim_{h \to 0^+} \frac{2h - 0}{h} = 2\)

• LHD of \(f'(x)\): \(\lim_{h \to 0^-} \frac{f'(0+h) - f'(0)}{h} = \lim_{h \to 0^-} \frac{-2h - 0}{h} = -2\)

Since RHD \(\ne\) LHD, the first derivative \(f'(x)\) is not differentiable at \(x=0\). Ans- D