\(ax^3 + bx^2 + cx + d\) is a polynomial on real \(x\) over real coefficients \(a, b, c, d\) wherein \(a \ne 0\). Which of the following statements is true?
Substituting \(x=0\) into the equation \(ax^3 + bx^2 + cx + d = 0\) gives: \[a(0)^3 + b(0)^2 + c(0) + d = 0 \implies d = 0\] Thus, \(x=0\) is a root if and only if \(d=0\). Ans- A
Which of the following is true for all possible non-zero choices of integers \(m, n\); \(m \ne n\), or all possible non-zero choices of real numbers \(p, q\); \(p \ne q\), as applicable?
Option 1 (A): \[\frac{1}{\pi} \int_0^{\pi} \sin m\theta \sin n\theta d\theta\] \[= \frac{1}{2\pi} \int_0^{\pi} \cos (m-n)\theta - \cos (m+n)\theta d\theta\] \[= \frac{1}{2\pi} \left[ \frac{\sin(m-n)\theta}{m-n} - \frac{\sin(m+n)\theta}{m+n} \right]_0^{\pi} = 0\] Therefore, Option (1) is correct.
Option 2 (B): \[\frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} \sin p\theta \sin q\theta d\theta\] \[= \frac{1}{4\pi} \int_{-\pi/2}^{\pi/2} \cos (p-q)\theta - \cos (p+q)\theta d\theta\] \[= \frac{1}{4\pi} \left[ \frac{\sin(p-q)\theta}{p-q} - \frac{\sin(p+q)\theta}{p+q} \right]_{-\pi/2}^{\pi/2}\] \[= \frac{1}{2\pi} \left[ \frac{\sin(p-q)\frac{\pi}{2}}{p-q} - \frac{\sin(p+q)\frac{\pi}{2}}{p+q} \right]\] The above expression is not necessarily zero. Therefore, Option (2) is incorrect.
Option 3 (C): \[\frac{1}{2\pi} \int_{-\pi}^{\pi} \sin p\theta \cos q\theta d\theta\] \[= \frac{1}{4\pi} \int_{-\pi}^{\pi} \sin (p+q)\theta + \sin (p-q)\theta d\theta\] \[= \frac{1}{4\pi} \left[ -\frac{\cos(p+q)\theta}{p+q} - \frac{\cos(p-q)\theta}{p-q} \right]_{-\pi}^{\pi}\] \[\begin{aligned} &= \frac{1}{4\pi} \left[ \left( -\frac{\cos((p+q)\pi)}{p+q} - \frac{\cos((p-q)\pi)}{p-q} \right) \right. \\ &\quad \left. - \left( -\frac{\cos(-(p+q)\pi)}{p+q} - \frac{\cos(-(p-q)\pi)}{p-q} \right) \right] = 0 \end{aligned}\] Therefore, Option (3) is correct.
Option 4 (D): \[\lim_{\alpha \to \infty} \frac{1}{2\alpha} \int_{-\alpha}^{\alpha} \sin p\theta \sin q\theta d\theta\] \[= \lim_{\alpha \to \infty} \frac{1}{4\alpha} \int_{-\alpha}^{\alpha} \cos (p-q)\theta - \cos (p+q)\theta d\theta\] \[= \lim_{\alpha \to \infty} \frac{1}{4\alpha} \left[ \frac{\sin(p-q)\theta}{p-q} - \frac{\sin(p+q)\theta}{p+q} \right]_{-\alpha}^{\alpha}\] \[\begin{aligned} &= \lim_{\alpha \to \infty} \frac{1}{4\alpha} \left[ 2\frac{\sin(p-q)\alpha}{p-q} - 2\frac{\sin(p+q)\alpha}{p+q} \right] \\ &= \lim_{\alpha \to \infty} \frac{1}{2} \left[ \frac{\sin((p-q)\alpha)}{\alpha(p-q)} - \frac{\sin((p+q)\alpha)}{\alpha(p+q)} \right] = 0 \end{aligned}\] (Since \(|\sin(k\alpha)| \le 1\), the terms approach 0 as \(\alpha \to \infty\).) Therefore, Option (4) is correct.
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 integrand is \(f(z) = \frac{z^2+1}{z(z-2)}\). The singularities (poles) are at \(z=0\) and \(z=2\).
The contour \(C\) is the unit circle, \(|z|=1\). The simple pole inside \(C\) is \(z_1 = 0\) (\(|0| < 1\)). The pole \(z_2 = 2\) is outside \(C\).
By the Cauchy Residue Theorem, \(\oint_{C} f(z) dz = 2\pi i \times \text{Res}(0)\).
The residue at \(z=0\) (a simple pole) is: \[\text{Res}(0) = \lim_{z\to 0} (z-0) f(z) = \lim_{z\to 0} z \left( \frac{z^2+1}{z(z-2)} \right)\] \[\text{Res}(0) = \lim_{z\to 0} \left( \frac{z^2+1}{z-2} \right) = \frac{0^2+1}{0-2} = -\frac{1}{2}\]
The value of the integral is: \[\oint_{C} \frac{z^2+1}{z^2 - 2z} dz = 2\pi i \times \left(-\frac{1}{2}\right) = -\pi i\] Ans- C
Consider the initial value problem below. The value of \(y\) at \(x = \ln 2\), (rounded off to 3 decimal places) is . \[\frac{dy}{dx} = 2x - y, \quad y(0) = 1\]
The given differential equation is a first-order linear equation: \[\frac{dy}{dx} + y = 2x\] Integrating Factor: \(\text{IF} = e^{\int 1 dx} = e^x\).
General Solution: \(y e^x = \int e^x (2x) dx = 2(x e^x - e^x) + C\) \[y = 2(x - 1) + C e^{-x}\]
Using the initial condition \(y(0) = 1\): \[1 = 2(0 - 1) + C e^{-0} \implies 1 = -2 + C \implies C = 3\]
Particular Solution: \[y = 2x - 2 + 3e^{-x}\]
Value at \(x = \ln 2\): \[y(\ln 2) = 2(\ln 2) - 2 + 3e^{-\ln 2}\] \[y(\ln 2) = 2\ln 2 - 2 + 3 \left(\frac{1}{2}\right)\] \[y(\ln 2) = 2\ln 2 - 0.5\] \[y \approx 2(0.69315) - 0.5 \approx 1.3863 - 0.5 \approx 0.8863\]
Rounded off to 3 decimal places, the value is \(0.886\).
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.
(The second derivative is \(\frac{d^2y}{dx^2} = 6 > 0\), confirming \(x = -\frac{1}{2}\) is a minimum, but we still must check endpoints for absolute max/min).
2. Evaluate \(y\) at Critical Point and Endpoints:
At \(x = -2\) (Endpoint): \[y(-2) = 3(-2)^2 + 3(-2) + 1 = 3(4) - 6 + 1 = 12 - 6 + 1 = 7\]
At \(x = 0\) (Endpoint): \[y(0) = 3(0)^2 + 3(0) + 1 = 1\]
At \(x = -\frac{1}{2}\) (Critical Point): \[y\left(-\frac{1}{2}\right) = 3\left(-\frac{1}{2}\right)^2 + 3\left(-\frac{1}{2}\right) + 1 = 3\left(\frac{1}{4}\right) - \frac{3}{2} + 1\] \[y\left(-\frac{1}{2}\right) = \frac{3}{4} - \frac{6}{4} + \frac{4}{4} = \frac{3 - 6 + 4}{4} = \frac{1}{4}\]
3. Conclusion: Comparing the values \(\{7, 1, \frac{1}{4}\}\): \[\text{Maximum value} = \max\{7, 1, \frac{1}{4}\} = 7\] \[\text{Minimum value} = \min\{7, 1, \frac{1}{4}\} = \frac{1}{4}\]
The maximum and minimum values are respectively \(7\) and \(\frac{1}{4}\). Ans- A
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 \(\vec{F} = F_1 \vec{a}_x + F_2 \vec{a}_y + F_3 \vec{a}_z\) to be conservative (or irrotational), its curl must be the zero vector: \(\nabla \times \vec{F} = 0\).
The components of \(\vec{F}\) are: \[F_1 = 5y - k_1 z\] \[F_2 = 3z + k_2 x\] \[F_3 = k_3 y - 4x\]
The curl is calculated as: \[\nabla \times \vec{F} = \begin{vmatrix} \vec{a}_x & \vec{a}_y & \vec{a}_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_1 & F_2 & F_3 \end{vmatrix} = 0\]
Expanding the determinant and setting each component to zero:
\(\vec{a}_x\) component: \[\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} = 0\] \[\frac{\partial}{\partial y}(k_3 y - 4x) - \frac{\partial}{\partial z}(3z + k_2 x) = 0\] \[k_3 - 3 = 0 \implies k_3 = 3\]
\(\vec{a}_y\) component (must be zero): \[\frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} = 0\] \[\frac{\partial}{\partial z}(5y - k_1 z) - \frac{\partial}{\partial x}(k_3 y - 4x) = 0\] \[-k_1 - (-4) = 0 \implies -k_1 + 4 = 0 \implies k_1 = 4\]
\(\vec{a}_z\) component: \[\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 0\] \[\frac{\partial}{\partial x}(3z + k_2 x) - \frac{\partial}{\partial y}(5y - k_1 z) = 0\] \[k_2 - 5 = 0 \implies k_2 = 5\]
The required values are \(k_1 = 4, k_2 = 5, k_3 = 3\). Ans- C
The number of purely real elements in a lower triangular representation of the given \(3 \times 3\) matrix obtained through the given decomposition is . \[\begin{pmatrix} 2 & 3 & 3 \\ 3 & 2 & 1 \\ 3 & 1 & 7 \end{pmatrix} = \begin{pmatrix} a_{11} & 0 & 0 \\ a_{21} & a_{22} & 0 \\ a_{31} & a_{23} & a_{33} \end{pmatrix} \begin{pmatrix} a_{11} & 0 & 0 \\ a_{21} & a_{22} & 0 \\ a_{31} & a_{23} & a_{33} \end{pmatrix}^T\]
The given decomposition is \(A = L L^T\), where \(A\) is a \(3 \times 3\) matrix and \(L\) is a \(3 \times 3\) lower triangular matrix.
The lower triangular matrix \(L\) is a \(3 \times 3\) matrix of the form: \[L = \begin{pmatrix} l_{11} & 0 & 0 \\ l_{21} & l_{22} & 0 \\ l_{31} & l_{32} & l_{33} \end{pmatrix}\]
The total number of elements in the \(3 \times 3\) lower triangular matrix \(L\) is \(3 \times 3 = 9\).
The number of non-zero elements in \(L\) (which must be calculated during the decomposition) is \(1 + 2 + 3 = 6\). The remaining 3 elements are zero.
The question asks for the number of "purely real elements" in the lower triangular matrix. Since the input matrix \(A\) is real and the decomposition can be carried out using real arithmetic (implied by the solution procedure in the images), all elements of the resulting lower triangular matrix \(L\) are real.
Therefore, the number of purely real elements in the \(3 \times 3\) lower triangular matrix is the total number of elements: \(3 \times 3 = 9\). Ans- D