Let \(p\) and \(q\) be real numbers such that \(p^2 + q^2 = 1\). The eigenvalues of the matrix \(A = \begin{pmatrix} p & q \\ q & -p \end{pmatrix}\) are:
To find the eigenvalues \(\lambda\), we must solve the characteristic equation \(\det(A - \lambda I) = 0\).
\[A - \lambda I = \begin{pmatrix} p-\lambda & q \\ q & -p-\lambda \end{pmatrix}\]
The characteristic equation is: \[\det(A - \lambda I) = (p-\lambda)(-p-\lambda) - q^2 = 0\] \[\implies -(p^2 - \lambda^2) - q^2 = 0\] \[\implies \lambda^2 - p^2 - q^2 = 0\] \[\implies \lambda^2 - (p^2 + q^2) = 0\]
Given the condition \(p^2 + q^2 = 1\): \[\lambda^2 - 1 = 0\] \[\implies \lambda^2 = 1\] \[\implies \lambda = \pm 1\]
The eigenvalues are \(1\) and \(-1\). Ans- B
Let \(p(z) = z^3 + (1+j)z^2 + (2+j)z + 3\), where \(z\) is a complex number. Which one of the following is true?
The given polynomial is \(p(z) = z^3 + (1+j)z^2 + (2+j)z + 3\). The coefficients are complex: \(a_2 = 1+j\), \(a_1 = 2+j\).
Analysis of Options:
Option A (\(\overline{p(z)} = p(\overline{z})\)): This property holds only if all coefficients of the polynomial are real. Since the coefficients \(1+j\) and \(2+j\) are complex, this statement is false.
Option C (Conjugate Pairs): The property that complex roots come in conjugate pairs also holds only for polynomials with real coefficients. Since the coefficients are complex, this statement is false.
Option B (Sum of Roots): The sum of the roots is given by the formula \(-\frac{a_2}{a_3}\). \[\text{Sum of roots} = -(1+j) = -1 - j\] This is a complex number, not a real number. This statement is false.
Option D (All roots cannot be real): If all roots (\(r_1, r_2, r_3\)) were real, then the polynomial would be: \[p(z) = (z-r_1)(z-r_2)(z-r_3)\] When expanded, all coefficients must be real (since \(r_1, r_2, r_3\) are real). The given polynomial has complex coefficients (\(1+j\) and \(2+j\)). Therefore, it is impossible for all three roots to be real. This statement is true.
Ans- D
For real numbers, \(x\) and \(y\), with \(y = 3x^2 + 3x + 1\), the maximum and minimum value of \(y\) for \(x \in [-2, 0]\) are respectively, .
To find the maximum and minimum values of \(y = 3x^2 + 3x + 1\) on the closed interval \(x \in [-2, 0]\), we evaluate \(y\) at the critical points and the endpoints.
1. Find Critical Points: Take the first derivative and set it to zero: \[\frac{dy}{dx} = 6x + 3\] \[6x + 3 = 0 \implies x = -\frac{1}{2}\] Since \(x = -\frac{1}{2}\) is within the interval \([-2, 0]\), it is a critical point.
2. Evaluate \(y\) at Critical Point and Endpoints:
At \(x = -2\) (Endpoint): \[y(-2) = 3(-2)^2 + 3(-2) + 1 = 12 - 6 + 1 = 7\]
At \(x = 0\) (Endpoint): \[y(0) = 3(0)^2 + 3(0) + 1 = 1\]
At \(x = -\frac{1}{2}\) (Critical Point): \[y\left(-\frac{1}{2}\right) = 3\left(-\frac{1}{2}\right)^2 + 3\left(-\frac{1}{2}\right) + 1 = \frac{3}{4} - \frac{3}{2} + 1 = \frac{3 - 6 + 4}{4} = \frac{1}{4}\]
3. Conclusion: Comparing the values \(\{7, 1, \frac{1}{4}\}\): \[\text{Maximum value} = 7\] \[\text{Minimum value} = \frac{1}{4}\]
The maximum and minimum values are respectively \(7\) and \(\frac{1}{4}\). Ans- A
Which one of the following vector functions represents a magnetic field \(\vec{B}\)? (\(\hat{X}, \hat{Y}\) and \(\hat{Z}\) are unit vectors along x-axis, y-axis, and z-axis, respectively)
A vector function \(\vec{B} = B_x \hat{X} + B_y \hat{Y} + B_z \hat{Z}\) represents a magnetic field only if its divergence is zero, i.e., \(\nabla \cdot \vec{B} = 0\). The divergence is calculated as: \[\nabla \cdot \vec{B} = \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}\]
1. Option A: \(\vec{B} = 10x \hat{X} - 30z \hat{Y} + 20y \hat{Z}\) \[\nabla \cdot \vec{B} = \frac{\partial (10x)}{\partial x} + \frac{\partial (-30z)}{\partial y} + \frac{\partial (20y)}{\partial z} = 10 + 0 + 0 = 10 \ne 0\] Option A is incorrect.
2. Option B: \(\vec{B} = 10y \hat{X} + 20x \hat{Y} - 10z \hat{Z}\) \[\nabla \cdot \vec{B} = \frac{\partial (10y)}{\partial x} + \frac{\partial (20x)}{\partial y} + \frac{\partial (-10z)}{\partial z} = 0 + 0 - 10 = -10 \ne 0\] Option B is incorrect.
3. Option C: \(\vec{B} = 10x \hat{X} + 20y \hat{Y} - 30z \hat{Z}\) \[\nabla \cdot \vec{B} = \frac{\partial (10x)}{\partial x} + \frac{\partial (20y)}{\partial y} + \frac{\partial (-30z)}{\partial z} = 10 + 20 - 30 = 0\] Since \(\nabla \cdot \vec{B} = 0\), this function represents a magnetic field. Option C is correct.
4. Option D: \(\vec{B} = 10z \hat{X} + 20y \hat{Y} - 30x \hat{Z}\) \[\nabla \cdot \vec{B} = \frac{\partial (10z)}{\partial x} + \frac{\partial (20y)}{\partial y} + \frac{\partial (-30x)}{\partial z} = 0 + 20 + 0 = 20 \ne 0\] Option D is incorrect.
Ans- C
Suppose the circles \(x^2 + y^2 = 1\) and \((x-1)^2 + (y-1)^2 = r^2\) intersect each other orthogonally at the point \((u, v)\). Then \(u + v =\) .
Two curves intersect orthogonally if the product of their slopes at the intersection point is \(-1\). Let \(m_1\) be the slope of Curve 1 and \(m_2\) be the slope of Curve 2. The condition is \(m_1 m_2 = -1\).
Curve 1: \(x^2 + y^2 = 1\) Differentiating with respect to \(x\): \[2x + 2y \frac{dy}{dx} = 0 \implies m_1 = \frac{dy}{dx} = -\frac{x}{y}\]
Curve 2: \((x-1)^2 + (y-1)^2 = r^2\) Differentiating with respect to \(x\): \[2(x-1) + 2(y-1) \frac{dy}{dx} = 0 \implies m_2 = \frac{dy}{dx} = -\frac{2(x-1)}{2(y-1)} = \frac{1-x}{y-1}\]
At the intersection point \((u, v)\), the slopes are: \[m_1 = -\frac{u}{v}\] \[m_2 = \frac{1-u}{v-1}\]
Applying the orthogonality condition \(m_1 m_2 = -1\): \[\left(-\frac{u}{v}\right) \left(\frac{1-u}{v-1}\right) = -1\] \[\frac{-u(1-u)}{v(v-1)} = -1\] \[u(1-u) = v(v-1)\] \[u - u^2 = v^2 - v\] \[u + v = u^2 + v^2\]
The intersection point \((u, v)\) must lie on Curve 1: \[u^2 + v^2 = 1\]
Substituting \(u^2 + v^2 = 1\) into the result from the orthogonality condition: \[u + v = 1\]
The vector function expressed by \[\vec{F} = \vec{a}_x (5y - k_1 z) + \vec{a}_y (3z + k_2 x) + \vec{a}_z (k_3 y - 4x)\] represents a conservative field, where \(\vec{a}_x, \vec{a}_y, \vec{a}_z\) are unit vectors along \(x, y\) and \(z\) directions, respectively. The values of constants \(k_1, k_2, k_3\) are given by:
For a vector field \(\vec{F}\) to be conservative (or irrotational), its curl must be the zero vector: \(\nabla \times \vec{F} = 0\).
The components of \(\vec{F}\) are: \(F_1 = 5y - k_1 z\), \(F_2 = 3z + k_2 x\), and \(F_3 = k_3 y - 4x\).
Setting the components of \(\nabla \times \vec{F}\) to zero: \[\nabla \times \vec{F} = \vec{a}_x \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) + \vec{a}_y \left( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} \right) + \vec{a}_z \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) = 0\]
\(\vec{a}_x\) component: \[\frac{\partial (k_3 y - 4x)}{\partial y} - \frac{\partial (3z + k_2 x)}{\partial z} = 0\] \[k_3 - 3 = 0 \implies k_3 = 3\]
\(\vec{a}_y\) component: \[\frac{\partial (5y - k_1 z)}{\partial z} - \frac{\partial (k_3 y - 4x)}{\partial x} = 0\] \[-k_1 - (-4) = 0 \implies -k_1 + 4 = 0 \implies k_1 = 4\]
\(\vec{a}_z\) component: \[\frac{\partial (3z + k_2 x)}{\partial x} - \frac{\partial (5y - k_1 z)}{\partial y} = 0\] \[k_2 - 5 = 0 \implies k_2 = 5\]
The required values are \(k_1 = 4, k_2 = 5, k_3 = 3\). Ans- C
In the open interval \((0, 1)\), the polynomial \(p(x) = x^4 - 4x^3 + 2\) has
We use the Intermediate Value Theorem (IVT) and analyze the function’s derivative to find the number of real roots in the open interval \((0, 1)\).
1. Apply IVT: Evaluate the polynomial \(p(x) = x^4 - 4x^3 + 2\) at the endpoints of the interval: \[p(0) = 0^4 - 4(0)^3 + 2 = 2\] \[p(1) = 1^4 - 4(1)^3 + 2 = 1 - 4 + 2 = -1\]
Since \(p(0) = 2\) (positive) and \(p(1) = -1\) (negative), and \(p(x)\) is continuous, the Intermediate Value Theorem guarantees that there is at least one root in the interval \((0, 1)\) where \(p(x)=0\).
2. Analyze Monotonicity using the Derivative: To determine if the root is unique, we find the first derivative: \[p'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)\] For \(x \in (0, 1)\):
\(4x^2\) is always positive.
\((x-3)\) is always negative (since \(x\) is between 0 and 1).
Thus, \(p'(x) = 4x^2(x-3) < 0\) for all \(x \in (0, 1)\).
Since the derivative \(p'(x)\) is strictly negative on the interval \((0, 1)\), the function \(p(x)\) is strictly decreasing on this interval. A strictly monotonic function can cross the x-axis at most once.
Therefore, the polynomial \(p(x)\) has exactly one real root in the open interval \((0, 1)\). Ans- A
Suppose the probability that a coin toss shows "head" is \(p\), where \(0 < p < 1\). The coin is tossed repeatedly until the first "head" appears. The expected number of tosses required is
Let \(X\) be the random variable representing the number of tosses required until the first "head" appears. This follows a Geometric Distribution with probability of success \(p\).
The probability mass function is \(P(X=k) = (1-p)^{k-1} p\), where \(k = 1, 2, 3, \ldots\)
The expected value \(E[X]\) is defined as: \[E[X] = \sum_{k=1}^{\infty} k \cdot P(X=k) = \sum_{k=1}^{\infty} k \cdot (1-p)^{k-1} p\] \[E[X] = p \sum_{k=1}^{\infty} k \cdot (1-p)^{k-1}\]
Let \(a = 1-p\). The summation becomes: \[\sum_{k=1}^{\infty} k a^{k-1} = 1 + 2a + 3a^2 + 4a^3 + \ldots\] This is the derivative of the geometric series formula with respect to \(a\): \[\sum_{k=0}^{\infty} a^k = 1 + a + a^2 + a^3 + \ldots = \frac{1}{1-a} \quad \text{for } |a|<1\] Differentiating both sides with respect to \(a\): \[\sum_{k=1}^{\infty} k a^{k-1} = \frac{d}{da} \left( \frac{1}{1-a} \right) = -1(1-a)^{-2}(-1) = \frac{1}{(1-a)^2}\]
Substitute this back into the expected value formula: \[E[X] = p \cdot \frac{1}{(1-a)^2}\] Substitute \(a = 1-p\) back: \[E[X] = p \cdot \frac{1}{(1 - (1-p))^2} = p \cdot \frac{1}{p^2} = \frac{1}{p}\]
Thus, the expected number of tosses required is \(\frac{1}{p}\). Ans- C
The value of the following complex integral, with \(C\) representing the unit circle centered at origin in the counterclockwise sense, is: \[\oint_C \frac{z^2 + 1}{z^2 - 2z} dz\]
The complex integral is evaluated using the Residue Theorem: \[\oint_C f(z) dz = 2\pi i \sum (\text{Residues of } f(z) \text{ inside } C)\] The contour \(C\) is the unit circle \(|z|=1\), centered at the origin.
1. Find the Poles of \(f(z)\): The integrand is \(f(z) = \frac{z^2 + 1}{z^2 - 2z} = \frac{z^2 + 1}{z(z-2)}\). The poles are the roots of the denominator: \(z(z-2) = 0\), so the poles are at \(z_1 = 0\) and \(z_2 = 2\). Both are simple poles (order \(n=1\)).
2. Check which Poles are Inside \(C\):
\(z_1 = 0\): \(|0| < 1\). This pole is inside \(C\).
\(z_2 = 2\): \(|2| > 1\). This pole is outside \(C\).
We only need to calculate the residue at \(z=0\).
3. Calculate the Residue at \(z=0\) (Simple Pole): \[\text{Res}_{z=0} f(z) = \lim_{z \to 0} [(z - 0) f(z)]\] \[\text{Res}_{z=0} f(z) = \lim_{z \to 0} \left[ z \cdot \frac{z^2 + 1}{z(z - 2)} \right] = \lim_{z \to 0} \left[ \frac{z^2 + 1}{z - 2} \right]\] \[\text{Res}_{z=0} f(z) = \frac{0^2 + 1}{0 - 2} = -\frac{1}{2}\]
4. Apply the Residue Theorem: \[\oint_C \frac{z^2 + 1}{z^2 - 2z} dz = 2\pi i \cdot \text{Res}_{z=0} f(z)\] \[\oint_C \frac{z^2 + 1}{z^2 - 2z} dz = 2\pi i \left( -\frac{1}{2} \right) = -\pi i\] Ans- C