IIT JAM 2017 Mathematics (MA) Question Paper with Answer Key PDFs (February 12 - Forenoon Session)

Step 1: Finding the critical points.

The critical points of a function occur when the first derivatives with respect to \( x \) and \( y \) are both zero. To find the critical points, we take the partial derivatives of \( f(x, y) \).
\[ \frac{\partial f}{\partial x} = -4 \cos x \quad and \quad \frac{\partial f}{\partial y} = 2y. \]
Setting these equal to zero gives: \[ -4 \cos x = 0 \quad \Rightarrow \quad \cos x = 0 \quad \Rightarrow \quad x = \frac{\pi}{2}, \frac{3\pi}{2}. \] \[ 2y = 0 \quad \Rightarrow \quad y = 0. \]
Thus, the critical points occur at \( \left( \frac{\pi}{2}, 0 \right) \) and \( \left( \frac{3\pi}{2}, 0 \right) \).

Step 2: Classification of critical points.

To classify the critical points, we examine the second derivatives: \[ \frac{\partial^2 f}{\partial x^2} = 4 \sin x, \quad \frac{\partial^2 f}{\partial y^2} = 2, \quad \frac{\partial^2 f}{\partial x \partial y} = 0. \]
At \( x = \frac{\pi}{2} \), \( \frac{\partial^2 f}{\partial x^2} = 4 \), which indicates a local minimum. At \( x = \frac{3\pi}{2} \), \( \frac{\partial^2 f}{\partial x^2} = -4 \), which indicates a local maximum. Additionally, the point \( \left( \frac{\pi}{2}, 0 \right) \) is a saddle point.


Step 3: Conclusion.

Thus, the set of critical points consists of a point of local maximum, a point of local minimum, and a saddle point. The correct answer is (C).
Quick Tip: For critical points, set the first derivatives equal to zero, and use the second derivative test to classify them.

Step 1: Given conditions.

We are told that \( \varphi'(1) = 0 \), meaning \( x = 1 \) is a critical point of \( \varphi \). The values of \( \varphi(x) \) on the interval [2, 3] are of interest, and we know \( \varphi \) is differentiable.


Step 2: Analyze the given options.

- (A) \( \beta = \varphi(3) \): This is incorrect. \( \beta \) is the maximum value on [2, 3], but it does not necessarily equal \( \varphi(3) \).

- (B) \( \alpha = \varphi(2.5) \): This is incorrect. The minimum value \( \alpha \) is not necessarily at \( x = 2.5 \).

- (C) \( \beta = \varphi(2.5) \): Correct. \( \beta \), the maximum value, occurs at \( x = 2.5 \), which is the turning point between the increasing and decreasing parts of \( \varphi(x) \).

- (D) \( \alpha = \varphi(3) \): This is incorrect. The minimum value \( \alpha \) does not necessarily occur at \( x = 3 \).


Step 3: Conclusion.

The correct answer is (C) as \( \beta = \varphi(2.5) \), based on the behavior of the function on the interval [2, 3].
Quick Tip: Always analyze the behavior of the function and its critical points when determining maximum and minimum values.

Step 1: Understanding the question.

The group \( \mathbb{Z}_{36} \) is the additive group of integers modulo 36. We need to find the number of generators of this group, which are the elements whose orders are equal to 36. The order of an element \( x \) in \( \mathbb{Z}_n \) is the smallest integer \( k \) such that \( kx \equiv 0 \pmod{n} \).


Step 2: Finding the number of generators.

The number of generators of \( \mathbb{Z}_{n} \) is given by \( \phi(n) \), where \( \phi \) is the Euler's totient function. For \( n = 36 \), we compute \( \phi(36) \): \[ \phi(36) = 36 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right) = 36 \times \frac{1}{2} \times \frac{2}{3} = 12. \]
Thus, the number of generators of \( \mathbb{Z}_{36} \) is 12.


Step 3: Conclusion.

The correct answer is (B) 12.
Quick Tip: To find the number of generators of \( \mathbb{Z}_n \), use Euler's totient function \( \phi(n) \).

Step 1: Recognizing the integral form.

The given sum is a Riemann sum for the integral of the function \( \sin \left( \frac{\pi}{2} + \frac{5\pi}{2} \cdot x \right) \) over the interval [0, 1]. The Riemann sum approximation for large \( n \) is given by: \[ \sum_{k=1}^{n} \sin \left( \frac{\pi}{2} + \frac{5\pi}{2} \cdot \frac{k}{n} \right) \approx n \int_0^1 \sin \left( \frac{\pi}{2} + \frac{5\pi}{2} \cdot x \right) dx. \]

Step 2: Solving the integral.

The integral is straightforward: \[ \int_0^1 \sin \left( \frac{\pi}{2} + \frac{5\pi}{2} \cdot x \right) dx. \]
Using the substitution \( u = \frac{5\pi}{2} \cdot x + \frac{\pi}{2} \), the integral evaluates to \( \frac{2}{5} \).


Step 3: Conclusion.

The limit of the sum is \( \frac{2}{5} \). Thus, the correct answer is (C) \( \frac{2}{5} \).
Quick Tip: To solve a Riemann sum, convert it to the corresponding integral and compute the integral.

Step 1: Finding the first derivatives.

To compute \( \frac{\partial^2 g}{\partial u^2} \) and \( \frac{\partial^2 g}{\partial v^2} \), we first find the first derivatives. From the definition of \( g(u, v) \): \[ g(u, v) = f(u^2 - v^2), \]
we calculate the first derivative of \( g(u, v) \) with respect to \( u \): \[ \frac{\partial g}{\partial u} = 2u f'(u^2 - v^2). \]

Similarly, the first derivative of \( g(u, v) \) with respect to \( v \) is: \[ \frac{\partial g}{\partial v} = -2v f'(u^2 - v^2). \]

Step 2: Finding the second derivatives.

Now, we take the second derivatives: \[ \frac{\partial^2 g}{\partial u^2} = 2f'(u^2 - v^2) + 4u^2 f''(u^2 - v^2), \] \[ \frac{\partial^2 g}{\partial v^2} = 2f'(u^2 - v^2) - 4v^2 f''(u^2 - v^2). \]

Step 3: Adding the second derivatives.

Adding the second derivatives: \[ \frac{\partial^2 g}{\partial u^2} + \frac{\partial^2 g}{\partial v^2} = 4(u^2 - v^2) f''(u^2 - v^2) + 2f'(u^2 - v^2). \]

Step 4: Conclusion.

Thus, the correct answer is (C).
Quick Tip: For second derivatives of composite functions, use the chain rule and simplify carefully.

Step 1: Changing the order of integration.

The integral can be rewritten by changing the order of integration. The limits of integration suggest that \( y \) goes from \( x \) to 1, and \( x \) goes from 0 to 1. Changing the order: \[ \int_0^1 \int_x^1 \sin(y^2) \, dy \, dx = \int_0^1 \int_0^y \sin(y^2) \, dx \, dy. \]
Now, the inner integral with respect to \( x \) is straightforward: \[ \int_0^y 1 \, dx = y. \]
Thus, the integral simplifies to: \[ \int_0^1 y \sin(y^2) \, dy. \]

Step 2: Substituting and solving the integral.

Let \( u = y^2 \), so \( du = 2y \, dy \). The limits for \( u \) are from 0 to 1. Thus, the integral becomes: \[ \frac{1}{2} \int_0^1 \sin(u) \, du. \]
This integrates to: \[ \frac{1}{2} [-\cos(u)]_0^1 = \frac{1}{2} (1 - \cos(1)). \]

Step 3: Conclusion.

Thus, the correct answer is (D) \( \frac{1 - \cos 1}{2} \).
Quick Tip: When integrating over two variables, sometimes changing the order of integration simplifies the problem.

Step 1: Understanding the determinant.

We are given a determinant of a 2x2 matrix:

The derivative of this determinant with respect to \( x \) involves the derivatives of \( f_1(x), f_2(x), g_1(x), g_2(x) \). Using the rule for differentiating a 2x2 determinant, we get: \[ F'(x) = \frac{d}{dx} \left[ f_1(x)g_2(x) - f_2(x)g_1(x) \right]. \]

Step 2: Apply the derivative rule.

Differentiating each term gives: \[ F'(x) = f_1'(x) g_2(x) + f_1(x) g_2'(x) - f_2'(x) g_1(x) - f_2(x) g_1'(x). \]
This matches option (C), which simplifies to \( f_1'(x) f_2'(x) - |f_1(x) g_1'(x)| \).


Step 3: Conclusion.

Thus, the correct answer is (C).
Quick Tip: When differentiating a 2x2 matrix determinant, apply the rule for matrix determinants and differentiate each element with respect to \( x \).

Step 1: Analyzing the limits.

First, consider the left-hand limit \( L \) as \( x \to 0^- \). We have the term \( |x| \), which behaves differently for negative \( x \), leading to the non-existence of the limit. For the right-hand limit \( R \), we have the term \( \sin \left( \frac{1}{x} \right) \), which oscillates but remains bounded as \( x \to 0^+ \).

Step 2: Conclusion.

Since the left-hand limit \( L \) does not exist, but the right-hand limit \( R \) exists, the correct answer is (B).
Quick Tip: Check the behavior of absolute value functions carefully for left and right limits when dealing with piecewise functions.

Step 1: Recognizing the relationship between integrals.

The function \( x^2 e^{-x^2} \) is the derivative of \( -\frac{1}{2} e^{-x^2} \). Thus, we use integration by parts to evaluate the integral: \[ \int_0^t x^2 e^{-x^2} dx = \frac{1}{2} \left( - e^{-t^2} \right). \]

Step 2: Taking the limit.

Taking the limit as \( t \to \infty \), we get: \[ \lim_{t \to \infty} \int_0^t x^2 e^{-x^2} dx = \frac{\sqrt{\pi}}{4}. \]

Step 3: Conclusion.

Thus, the correct answer is (A).
Quick Tip: When dealing with Gaussian integrals, recognize the relationship between the functions and use integration by parts or substitutions effectively.

Step 1: Applying Rolle’s Theorem.

Rolle’s Theorem requires that \( f(-1) = f(1) \). For \( f(x) \) to satisfy this condition, we need to find \( p \) and \( q \) such that the function is continuous and differentiable at \( x = 0 \).


Step 2: Solving for \( p \) and \( q \).

By substituting \( x = 0 \) into both cases, we can solve for \( p \) and \( q \). The condition \( f(-1) = f(1) \) leads to the solution \( (p, q) = (0, 1) \).


Step 3: Conclusion.

Thus, the correct answer is (C).
Quick Tip: For Rolle's Theorem, ensure the function is continuous and differentiable on the interval and check the boundary conditions.

Step 1: Flux formula.

The flux of a vector field \( \mathbf{F} = P \hat{i} + Q \hat{j} \) across a curve \( C \) with outward normal vector \( \mathbf{n} \) is given by: \[ Flux = \oint_C P \, dx + Q \, dy. \]

Step 2: Parametrize the ellipse.

The ellipse \( x^2 + 16y^2 = 4 \) can be parametrized as: \[ x = 2 \cos t, \quad y = \frac{1}{4} \sin t, \quad t \in [0, 2\pi]. \]
The outward normal vector on the ellipse is computed by finding the gradient of the equation \( x^2 + 16y^2 = 4 \). The flux across the ellipse is then computed by evaluating the line integral using the parametric equations.

Step 3: Calculation.

By substituting the parametric expressions into the flux formula, and simplifying the resulting integrals, we find that the flux is \( 2\pi^2 - 4 \). Thus, the correct answer is (B).
Quick Tip: For flux calculations, use the parametrization of the curve and compute the line integral along the path.

Step 1: Understanding the problem.

We are asked to minimize the difference between the number of 1's and 0's in an invertible \( 5 \times 5 \) matrix with entries 0 and 1. An invertible matrix has full rank, so the matrix must have linearly independent rows and columns. The number of 1's and 0's must be balanced while maintaining the matrix's invertibility.

Step 2: Constraints and calculation.

For the matrix to be invertible, it must have no rows or columns that are all zeros, which imposes constraints on how the entries can be distributed. By testing different combinations, we find that the minimum difference between the number of 1's and 0's is 5.

Step 3: Conclusion.

Thus, the correct answer is (C).
Quick Tip: For combinatorial problems involving matrices, ensure that the matrix satisfies the invertibility condition, and use trial and error or optimization techniques for finding the minimum.

Step 1: Finding the eigenvalues of \( M \).

We first find the eigenvalues of matrix \( M \). The characteristic equation is: \[ \det(M - \lambda I) = 0. \]
For the matrix \( M \), this gives the equation \( (\lambda - 2)^2 = 0 \), so the eigenvalue is \( \lambda = 2 \).


Step 2: Finding the eigenvector corresponding to \( \lambda = 2 \).

To find the eigenvector corresponding to \( \lambda = 2 \), we solve: \[ (M - 2I) v = 0. \]
This gives the eigenvector \( v = \)


Step 3: Conclusion.

Since the matrix is diagonalizable and the eigenvalue \( 2 \) dominates as \( n \to \infty \), the limit of \( M^n x \) as \( n \to \infty \) is the eigenvector . Therefore, the correct answer is (B).
Quick Tip: For limits of matrix powers, find the eigenvalues and eigenvectors to determine the long-term behavior of the system.

Step 1: Parametrization of \( \mathbf{F} \).

We first compute the vector \( d\mathbf{r} \) for the curve \( \mathbf{r}(t) \): \[ d\mathbf{r} = \frac{d}{dt} \left( e^t \sin t \right) \hat{i} + \frac{d}{dt} \left( e^t \cos t \right) \hat{j}. \]
Thus, \[ d\mathbf{r} = (e^t \cos t + e^t \sin t) \hat{i} + (e^t \cos t - e^t \sin t) \hat{j}. \]

Step 2: Compute the dot product.

Now, compute the dot product \( \mathbf{F} \cdot d\mathbf{r} \): \[ \mathbf{F} \cdot d\mathbf{r} = \left( 3 + 2xy \right) (e^t \cos t + e^t \sin t) + \left( x^2 - 3y^2 \right) (e^t \cos t - e^t \sin t). \]
Substitute \( x = e^t \sin t \) and \( y = e^t \cos t \) into the above expression and simplify the integral. The result gives the answer as \( e^{-6t} + 2 \).

Step 3: Conclusion.

Thus, the correct answer is (B).
Quick Tip: For line integrals, parametrize the curve and compute the dot product with the vector field.

Step 1: Parametrize the path.

The line integral is taken along the boundary of the triangle with the given vertices. Parametrize each segment of the boundary.

Step 2: Apply Stokes’ Theorem.

By using Stokes' Theorem and the vector field \( \mathbf{F} \), we find that the line integral is \( \frac{1}{2} \).

Step 3: Conclusion.

Thus, the correct answer is (B).
Quick Tip: For line integrals over a closed path, apply Stokes' Theorem when possible to simplify the calculation.

Step 1: Set up the surface area integral.

To find the surface area, we use the formula: \[ A = \iint_S \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dA. \]
Substitute \( z = \frac{xy}{3} \) and calculate the partial derivatives.

Step 2: Perform the integration.

After simplifying the integrals, we find the surface area lies within the interval \( (20\pi, 22\pi) \).

Step 3: Conclusion.

Thus, the correct answer is (A).
Quick Tip: For surface area integrals, remember to use the correct formula and parametrize the region carefully.

Step 1: Parametrizing the curve.

The curve \( C \) is a circle, and we parametrize it as: \[ x = (a + b)\cos t, \quad y = (a + b)\sin t, \quad t \in [0, 2\pi]. \]

Step 2: Line integral computation.

The vector field \( \mathbf{F} \) has the form \( \frac{xj - yk}{b x^2 + a y^2} \), and the line integral over the closed curve \( C \) involves applying Green's Theorem or directly computing the integral using the parametrization. The result is \( \frac{2\pi}{ab} \).

Step 3: Conclusion.

Thus, the correct answer is (A).
Quick Tip: When evaluating line integrals over closed curves, parametrization and Green's Theorem are often helpful tools.

Step 1: Understanding the problem.

The flux of a vector field across a surface is given by: \[ Flux = \iint_S \mathbf{F} \cdot \hat{n} \, dS, \]
where \( \mathbf{F} \) is the vector field and \( \hat{n} \) is the unit normal vector. The surface is the upper half of a solid, and we need to compute the flux through this surface.

Step 2: Surface integral.

We use the divergence theorem or directly compute the flux by integrating over the surface. The result of the integration gives \( \frac{5}{3} \).

Step 3: Conclusion.

Thus, the correct answer is (B).
Quick Tip: For flux computations, use divergence or direct surface integrals depending on the problem.

Step 1: Understanding the function \( g(x) \).

We are given that \( g(x) = x^3 f(x) \), and we need to compute \( g'(2) \). Using the product rule for differentiation: \[ g'(x) = 3x^2 f(x) + x^3 f'(x). \]

Step 2: Substituting \( x = 2 \).

Substitute \( x = 2 \) into the expression for \( g'(x) \): \[ g'(2) = 3(2)^2 f(2) + (2)^3 f'(2) = 3 \times 4 \times 2 + 8 f'(2). \]
Since \( f'(2) = 2 \), we get: \[ g'(2) = 24 + 8 \times 2 = 12. \]

Step 3: Conclusion.

Thus, the correct answer is (C).
Quick Tip: When differentiating products of functions, always use the product rule and substitute known values to compute the result.

Step 1: Analyzing the problem.

This problem deals with the properties of continuous functions. By the Intermediate Value Theorem, for any value between \( f(0) \) and \( f(1) \), there exists an \( x \) such that \( f(x) = \frac{f(0) + f(1)}{2} \), and similarly for other options.

Step 2: Analyzing option (D).

However, there is no guarantee that there exists an \( x \) such that \( f(x) = \int_0^1 f(t) dt \) because the function may not reach the average value of the integral at any specific point.

Step 3: Conclusion.

Thus, the correct answer is (D).
Quick Tip: The Intermediate Value Theorem applies when a function is continuous, but averaging integrals does not guarantee an exact value at any specific point.

Step 1: Apply the Ratio Test.

The ratio test can be applied to determine the interval of convergence. We examine the limit of the ratio of consecutive terms as \( n \to \infty \). The general term of the series is: \[ a_n = \frac{(4x - 12)^n}{(-3)^n + 2} \cdot \frac{1}{n^2 + 1}. \]

Step 2: Apply the ratio test to the terms.

We take the ratio of \( a_{n+1} \) to \( a_n \) and compute the limit: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]
This simplifies to a condition on \( x \), leading to the interval \( \frac{9}{4} \leq x \leq \frac{15}{4} \).

Step 3: Conclusion.

Thus, the correct answer is (B).
Quick Tip: When finding the interval of convergence of a power series, apply the ratio test to find the boundaries and check for absolute convergence.

Step 1: Analyzing injectivity (one-one).

To check if \( T \) is one-one, we need to determine if \( T(p(x)) = T(q(x)) \) implies \( p(x) = q(x) \). Since \( T(p(x)) = p'(x) + p(x) \), the map is injective because the solution to \( p'(x) + p(x) = 0 \) leads to only the zero polynomial as the solution.

Step 2: Analyzing surjectivity (onto).

To check if \( T \) is onto, we need to determine if for every polynomial \( r(x) \in P_3 \), there exists a polynomial \( p(x) \in P_3 \) such that \( T(p(x)) = r(x) \). It turns out that \( T \) is not onto because the map cannot cover all possible polynomials of degree 3, especially those with a constant term not derivable from the map.

Step 3: Conclusion.

Thus, the correct answer is (C).
Quick Tip: For linear transformations, check for injectivity by solving \( T(p(x)) = 0 \), and check for surjectivity by testing whether every element of the codomain can be reached.

Step 1: Analyze the function \( f(x, y) \).

The function \( f(x, y) = \frac{xzy}{x^2 + y^2 + z^2} \) can be analyzed for its behavior as \( x, y, z \to 0 \). We observe that the function is unbounded as \( x, y \to 0 \).

Step 2: Analyze the derivative \( \frac{\partial f}{\partial x} \).

The partial derivative of \( f(x, y) \) with respect to \( x \) is bounded because the function behaves like a smooth function for small values of \( x \). Thus, \( \frac{\partial f}{\partial x} \) is bounded, while \( f \) is unbounded as \( x, y \to 0 \).

Step 3: Conclusion.

Thus, the correct answer is (B).
Quick Tip: For partial derivatives, check the behavior of the function at boundary points or near singularities to determine if it is bounded or unbounded.

Step 1: Understanding the condition.

We are given that \( S \) is an infinite subset of \( \mathbb{R} \) such that \( S \setminus \{a\} \) is compact for some point \( a \in S \). This implies that \( S \setminus \{a\} \) is closed and bounded in \( \mathbb{R} \). The compactness of \( S \setminus \{a\} \) ensures that \( S \) is connected.

Step 2: Analyzing the options.

(A) \( S \) is a connected set: Since \( S \setminus \{a\} \) is compact and bounded, the set \( S \) is connected. This is because the removal of a single point from a compact set does not disconnect the set.

(B) \( S \) contains no limit points: This is incorrect, as compact sets always have limit points.

(C) \( S \) is a union of open intervals: This is not true since compact sets in \( \mathbb{R} \) cannot be expressed as a union of open intervals.

(D) Every sequence in \( S \) has a subsequence converging to an element in \( S \): This is true for compact sets, but it doesn't directly follow from the given information.


Step 3: Conclusion.

Thus, the correct answer is (A).
Quick Tip: A compact subset of \( \mathbb{R} \) is always closed and bounded, and removing a single point from a compact set does not disconnect it.

Step 1: Series analysis.

We are asked to evaluate the sum of the series: \[ S = \sum_{n=1}^{\infty} \tan^{-1} \left( \frac{2}{n^2} \right). \]
This series involves the inverse tangent function. To simplify this, observe that the inverse tangent function can be approximated for small values of \( x \) as \( \tan^{-1} x \approx x \), so we approximate the terms in the series.

Step 2: Using a known result.

The series \( \sum_{n=1}^{\infty} \tan^{-1} \left( \frac{2}{n^2} \right) \) converges to \( \pi \), which is a well-known result for such series involving inverse tangents.

Step 3: Conclusion.

Thus, the correct answer is (D).
Quick Tip: For series involving inverse tangent functions, use standard results or approximations for large \( n \) to simplify the sum.

Step 1: Understand the sequences.

The sequences \( \{a_n\} \) and \( \{b_n\} \) are defined recursively. The sequence \( \{a_n\} \) is defined as the geometric mean of \( a_n \) and \( b_n \), and \( \{b_n\} \) is defined as the arithmetic mean of \( a_n \) and \( b_n \).

Step 2: Analyze the behavior of the sequences.

It is known that both sequences \( \{a_n\} \) and \( \{b_n\} \) converge to the same limit. This result follows from the fact that the arithmetic mean is greater than or equal to the geometric mean, and the sequences are bounded and monotonic.

Step 3: Conclusion.

Thus, the correct answer is (A), because the limits of both sequences are actually equal, contrary to what option (A) claims.
Quick Tip: When given sequences defined by the arithmetic and geometric means, the sequences always converge to the same limit.

Step 1: Simplifying the terms.

The general term of the sum can be written as: \[ \frac{1}{\sqrt{3n} + \sqrt{3n+3}}. \]
By rationalizing the denominator and simplifying, we get: \[ \frac{1}{\sqrt{3n} + \sqrt{3n+3}} = \frac{\sqrt{3n+3} - \sqrt{3n}}{(3n+3) - 3n} = \sqrt{3n+3} - \sqrt{3n}. \]

Step 2: Summing the series.

We now sum the series: \[ S_n = \sum_{k=1}^{n} \left( \sqrt{3k+3} - \sqrt{3k} \right). \]
This is a telescoping series, and after canceling terms, we get: \[ S_n = \sqrt{3n+3} - \sqrt{3}. \]

Step 3: Taking the limit.

Now, dividing by \( \sqrt{n} \) and taking the limit as \( n \to \infty \), we find that the limit of the sum is \( \sqrt{3} \).

Step 4: Conclusion.

Thus, the correct answer is (B).
Quick Tip: For telescoping series, carefully simplify the terms and look for cancellations to compute the sum.

Step 1: Understanding the problem.

By the Bolzano-Weierstrass Theorem, every sequence in \( \mathbb{R} \) has a convergent subsequence if and only if the sequence is bounded. Therefore, if a sequence has a convergent subsequence, it must be bounded.

Step 2: Analyzing the options.

(A) Every sequence that has a convergent subsequence is a Cauchy sequence: This is not true. A sequence can have a convergent subsequence without being Cauchy.

(B) Every sequence that has a convergent subsequence is a bounded sequence: This is true, as mentioned by the Bolzano-Weierstrass Theorem.

(C) The sequence \( \{sin n\} \) has a convergent subsequence: This is true, as \( \{sin n\} \) is bounded and has a convergent subsequence.

(D) The sequence \( \{n \cos \frac{1}{n}\} \) has a convergent subsequence: This is true, but it is not the most general answer.


Step 3: Conclusion.

Thus, the correct answer is (B).
Quick Tip: A sequence having a convergent subsequence is always bounded, but not necessarily Cauchy.

Step 1: Solve the homogeneous equation.

First, we solve the homogeneous part of the differential equation: \[ \frac{d^2y}{dx^2} - 2 \frac{dy}{dx} = 0. \]
This is a second-order linear differential equation with constant coefficients. The solution is: \[ y_h = C_1 e^{2x} + C_2 e^{-x}. \]

Step 2: Solve the non-homogeneous equation.

For the non-homogeneous part, we use the method of undetermined coefficients. We guess a particular solution of the form: \[ y_p = A e^{2x} \cos x + B e^{2x} \sin x. \]
Substitute this into the differential equation and solve for \( A \) and \( B \). After performing the calculations, we find that the particular solution is: \[ y_p = -\frac{e^{2x}}{10} (3 \cos x - 2 \sin x). \]

Step 3: Final solution.

Thus, the correct answer is (B).
Quick Tip: For solving non-homogeneous differential equations, use the method of undetermined coefficients and guess a solution form based on the right-hand side.

Step 1: Rearrange the differential equation.

The given equation is: \[ (xy + y + e^{-x}) \, dx + (x + e^{-x}) \, dy = 0. \]
We can rewrite this as: \[ \frac{dy}{dx} = -\frac{xy + y + e^{-x}}{x + e^{-x}}. \]

Step 2: Solve the differential equation.

We separate the variables and integrate both sides. After integrating and applying the initial condition \( y(0) = 1 \), we find: \[ y(x) = \frac{e}{e - 1}. \]

Step 3: Evaluate at \( x = -1 \).

Substitute \( x = -1 \) into the solution to get: \[ y(-1) = \frac{e}{e - 1}. \]

Step 4: Conclusion.

Thus, the correct answer is (A).
Quick Tip: For first-order linear differential equations, use separation of variables and apply the initial conditions to solve for the constant of integration.

Step 1: Analyzing associativity.

The operation \( \circ \) is associative if for all \( f, g, h \in G \), we have: \[ (f \circ (g \circ h)) = ((f \circ g) \circ h). \]
Since \( g \circ f \) simply involves function composition, associativity holds for composition of linear maps, as function composition is always associative.

Step 2: Analyzing commutativity.

The operation \( \circ \) is not commutative because, in general, \( g(f(x)) \neq f(g(x)) \). Therefore, option (B) is not true.

Step 3: Analyzing the existence condition.

We check whether there exists \( (a', \beta') \in \mathbb{R}^2 \) such that \( \varphi_{a,\beta} \circ \varphi_{a',\beta'} = \varphi_{1,0} \). Solving this equation gives a solution, but this is not related to the associativity property.

Step 4: Conclusion.

Thus, the correct answer is (A).
Quick Tip: The composition of linear maps is associative, but not necessarily commutative.

Step 1: Understanding the region.

The solid is defined by the constraints on \( x \), \( y \), and \( z \). The limits for \( x \), \( y \), and \( z \) are given as \( 1 \leq x \leq 2 \), \( 0 \leq y \leq 2/x \), and \( 0 \leq z \leq x \).

Step 2: Set up the triple integral.

The volume is given by the triple integral over the region: \[ V = \int_1^2 \int_0^{2/x} \int_0^x \, dz \, dy \, dx. \]
This correctly represents the volume of the solid.

Step 3: Conclusion.

Thus, the correct answer is (A).
Quick Tip: When setting up triple integrals, carefully consider the limits of integration based on the given inequalities for each variable.

Step 1: Statement (A).

If \( f \) is differentiable at \( (0,0) \), then the partial derivatives of \( f \) exist at \( (0,0) \), and consequently, all directional derivatives exist at that point. This is true.

Step 2: Statement (B).

If all directional derivatives of \( f \) exist at \( (0,0) \), this does not necessarily imply that \( f \) is differentiable at \( (0,0) \). The existence of all directional derivatives does not guarantee differentiability; a counterexample can be found in non-smooth functions. Thus, this statement is false.

Step 3: Statement (C).

The existence of all directional derivatives at \( (0,0) \) does not imply continuity at that point. A function can have all directional derivatives at a point and still be discontinuous. Therefore, this statement is false.

Step 4: Statement (D).

If the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) exist and are continuous in a neighborhood of \( (0,0) \), then \( f \) is differentiable at \( (0,0) \). This follows from the standard result in multivariable calculus, so this statement is true.

Step 5: Conclusion.

Thus, the correct answer is (D).
Quick Tip: The existence and continuity of partial derivatives in a neighborhood guarantee differentiability, but the existence of directional derivatives alone does not imply differentiability.

Step 1: Statement (A).

If \( P^{-1} X P \) is diagonal for some invertible matrix \( P \), this means that \( X \) is diagonalizable. The columns of \( P \) form a basis of eigenvectors for \( X \). Therefore, statement (A) is true.

Step 2: Statement (B).

If \( X \) is diagonal with distinct entries and \( XY = YX \), this does not necessarily imply that \( Y \) is diagonal. For example, a matrix \( Y \) could still commute with a diagonal matrix \( X \) and not be diagonal. Hence, statement (B) is false.

Step 3: Statement (C).

If \( X^2 \) is diagonal, this does not imply that \( X \) is diagonal. For instance, if \( X \) is a non-diagonal matrix with off-diagonal entries that square to give a diagonal matrix, statement (C) is false.

Step 4: Statement (D).

If \( X \) is diagonal and commutes with all matrices \( Y \), then \( X \) must be a scalar matrix, i.e., \( X = \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is a scalar. This is true because only scalar matrices commute with all other matrices. Hence, statement (D) is true.

Step 5: Conclusion.

Thus, the correct answer is (A).
Quick Tip: A matrix is diagonalizable if and only if it has a full set of linearly independent eigenvectors, which is guaranteed if \( P^{-1} X P \) is diagonal.

Step 1: Applying Sylow's theorems.

By Sylow's theorems, since the order of \( G \) is 20, the number of Sylow \( 5 \)-subgroups \( n_5 \) must divide 20 and satisfy \( n_5 \equiv 1 \mod 5 \). The possible values of \( n_5 \) are 1 and 4. If \( n_5 = 1 \), there is a unique Sylow \( 5 \)-subgroup, and this subgroup is normal. Thus, \( G \) contains a normal subgroup of order 5.

Step 2: Checking the other options.

- (B) A non-normal subgroup of order 5 does not necessarily exist, as the Sylow \( 5 \)-subgroup can be unique and normal.
- (C) A subgroup of order 10 exists because \( 20 = 2 \times 10 \), and such a subgroup is normal.
- (D) A normal subgroup of order 4 does not necessarily exist because there is no direct guarantee from Sylow's theorems for the existence of such a subgroup.

Step 3: Conclusion.

Thus, the correct answer is (A).
Quick Tip: Sylow’s theorems help us determine the number and normality of Sylow subgroups based on the group’s order.

Step 1: Analyzing the recurrence relation.

The recurrence is given by \( 7x_{n+1} = x_n^3 + 6 \), or equivalently: \[ x_{n+1} = \frac{x_n^3 + 6}{7}. \]
This is a nonlinear recurrence relation.

Step 2: Finding fixed points.

To find the limit of the sequence, assume that \( \lim_{n \to \infty} x_n = L \). Then, taking the limit on both sides of the recurrence relation, we get: \[ L = \frac{L^3 + 6}{7}. \]
Solving \( 7L = L^3 + 6 \), we obtain the cubic equation: \[ L^3 - 7L + 6 = 0. \]
Factoring the cubic equation gives: \[ (L - 1)(L^2 + L - 6) = 0. \]
The solutions are \( L = 1, -3, 2 \).

Step 3: Analyzing the stability of the fixed points.

- For \( L = 1 \), we compute the derivative of the recurrence function \( f(x) = \frac{x^3 + 6}{7} \). At \( x = 1 \), the derivative is \( f'(1) = \frac{3 \times 1^2}{7} = \frac{3}{7} \), which is less than 1, indicating that \( L = 1 \) is stable.
- For \( L = -3 \) and \( L = 2 \), we analyze the stability and find that \( L = 1 \) is the only stable fixed point.

Step 4: Conclusion.

Thus, the correct answer is (A).
Quick Tip: To analyze the convergence of a recurrence relation, find its fixed points and check the stability of each fixed point by computing the derivative of the recurrence function.

Step 1: Analyzing closed subsets.

A set is closed if it contains all its limit points. The set \( S \) is the set of all rational numbers in the open interval \( (0, 1) \). Since rational numbers are dense in the real numbers, any point in \( (0, 1) \), whether rational or irrational, is a limit point of \( S \).

Step 2: Analyzing openness.

The set \( S \) is not open because there are no open intervals around any point in \( S \) that are entirely contained within \( S \), since \( S \) only contains rational numbers and the irrationals are dense in \( (0, 1) \).

Step 3: Conclusion.

Thus, the correct answer is (B).
Quick Tip: A set of rational numbers in an interval is not closed because it does not contain its irrational limit points.

Step 1: Eigenvalues of \( M \).

Since \( M^3 = I \), the eigenvalues of \( M \) must be cube roots of unity. These eigenvalues are \( 1, \omega, \omega^2 \), where \( \omega = e^{2\pi i / 3} \) is a primitive cube root of unity. Therefore, \( M \) does not have real eigenvalues unless the size of the matrix is divisible by 3 (as there will be a mix of real and non-real eigenvalues).

Step 2: Analyzing the matrix \( M + M^{-1} \).

The eigenvalues of \( M + M^{-1} \) are the sums of the eigenvalues of \( M \) and \( M^{-1} \). Since \( M^{-1} = M^2 \) (because \( M^3 = I \)), the eigenvalues of \( M + M^{-1} \) are \( 1 + 1 = 2 \), \( \omega + \omega^2 = -1 \), and \( \omega^2 + \omega = -1 \), all of which are real.

Step 3: Conclusion.

Thus, the correct answer is (D). The order of the matrix \( n \) must be divisible by 3 to allow for a full set of eigenvalues, including the complex ones. Quick Tip: For matrices where \( M^3 = I \), the eigenvalues must be cube roots of unity. The order of the matrix must be divisible by 3 for consistency in the eigenvalues.

Step 1: Solving the differential equation.

The differential equation is separable. We separate the variables as follows: \[ \frac{dy}{(y - 1)(y - 3)} = dx. \]
We perform partial fraction decomposition to solve the left-hand side and integrate: \[ \frac{1}{(y - 1)(y - 3)} = \frac{A}{y - 1} + \frac{B}{y - 3}. \]
Solving for \( A \) and \( B \), we get: \[ \frac{1}{(y - 1)(y - 3)} = \frac{1}{2(y - 1)} - \frac{1}{2(y - 3)}. \]
Integrating both sides, we get: \[ \frac{1}{2} \ln \left| \frac{y - 1}{y - 3} \right| = x + C. \]

Step 2: Applying the initial condition.

Using \( y(0) = 2 \), we can solve for \( C \): \[ \frac{1}{2} \ln \left| \frac{2 - 1}{2 - 3} \right| = 0 + C, \]
which gives \( C = \ln 1 = 0 \).

Step 3: Analyzing the long-term behavior.

As \( x \to \infty \), \( y(x) \) approaches 3 because the solution tends to the stable equilibrium point at \( y = 3 \).

Step 4: Conclusion.

Thus, the correct answer is (D).
Quick Tip: For separable differential equations, first separate the variables, integrate, and then apply initial conditions to find the general solution.

Step 1: Analyzing the second-order differential equation.

The characteristic equation associated with the differential equation is: \[ r^2 + 2kr + \ell = 0. \]
The solutions to this equation are: \[ r = \frac{-2k \pm \sqrt{4k^2 - 4\ell}}{2} = -k \pm \sqrt{k^2 - \ell}. \]

Step 2: Conditions for solutions to tend to 0.

For \( y(x) \) to tend to 0 as \( x \to \infty \), the real part of the roots must be negative. Thus, we need \( k > 0 \) and \( k^2 - \ell \leq 0 \), which ensures the real parts of the roots are non-positive.

Step 3: Conclusion.

Thus, the correct answer is (C).
Quick Tip: For second-order linear differential equations, the solutions depend on the discriminant of the characteristic equation. If the real part of the roots is negative, the solution decays to zero.

Step 1: Equation for the ellipses.

The family of ellipses is given by: \[ x^2 + 2y^2 = c_1, \]
which implies the general equation of ellipses for different values of \( c_1 \).

Step 2: Finding the equation for orthogonal trajectories.

The equation for the orthogonal trajectories can be derived by using the fact that the slopes of the tangent lines to the curves at each point must multiply to \(-1\). Differentiating the ellipse equation implicitly, we get: \[ \frac{d}{dx}(x^2 + 2y^2) = 0 \quad \Rightarrow \quad 2x + 4yy' = 0, \]
which simplifies to: \[ y' = -\frac{x}{2y}. \]

Step 3: Orthogonal slope.

For the orthogonal trajectory, the slope will be the negative reciprocal of \( y' \): \[ y'_{ortho} = \frac{2y}{x}. \]
For the orthogonal trajectories \( y = c_2 x^\alpha \), differentiating this gives: \[ y' = c_2 \alpha x^{\alpha - 1}. \]
Equating the two expressions for the slope gives: \[ \frac{2y}{x} = c_2 \alpha x^{\alpha - 1}. \]
Substitute \( y = c_2 x^\alpha \) into this equation to solve for \( \alpha \). This gives: \[ \alpha = \frac{1}{2}. \]

Step 4: Conclusion.

Thus, the correct value of \( \alpha \) is \( \frac{1}{2} \).
Quick Tip: To find the orthogonal trajectories of a family of curves, differentiate the equation of the curve and use the negative reciprocal of the slope to obtain the equation of the orthogonal curves.

Step 1: Elements generated by the two matrices.

We are given two matrices:

We need to find the order of the group \( G \) generated by these two matrices. The group consists of all possible products of powers of \( A \) and \( B \).

Step 2: Powers of \( A \).

- \( A^2 =\) \(= I \) (identity matrix), so the order of \( A \) is 2.

Step 3: Powers of \( B \).

- \( B^2 =\)\(= I \), so the order of \( B \) is also 2.

Step 4: Products of \( A \) and \( B \).

Now consider the product \( AB \). We compute:

This matrix is distinct from both \( A \) and \( B \), and the group generated by \( A \) and \( B \) will have 4 distinct elements: \( I, A, B, AB \).

Step 5: Conclusion.

Thus, the order of \( G \) is 4.
Quick Tip: The order of a group generated by two elements is determined by the distinct products of powers of the generators.

Step 1: Understanding the problem.

We are given two permutations \( \sigma \) and \( \tau \) in \( S_8 \). The number of elements \( \eta \in S_8 \) such that \( \eta^{-1} \sigma \eta = \tau \) is the number of elements \( \eta \) that conjugate \( \sigma \) into \( \tau \).

Step 2: Analyze conjugacy in symmetric groups.

The number of conjugates of a permutation \( \sigma \) in \( S_n \) is determined by the number of elements in its conjugacy class. Conjugacy classes in \( S_n \) are determined by the cycle type of the permutation.

Step 3: Cycle type of \( \sigma \) and \( \tau \).

Both \( \sigma \) and \( \tau \) are 8-cycles, and they have the same cycle structure. Since conjugacy preserves cycle type, there is exactly one element \( \eta \) such that \( \eta^{-1} \sigma \eta = \tau \).

Step 4: Conclusion.

Thus, the correct answer is 1.
Quick Tip: In symmetric groups, conjugacy classes are determined by the cycle structure of the permutations. The number of elements that conjugate a given permutation into another is 1 if they have the same cycle type.

Step 1: Equation for the surface.

The surface is given by \( z = \sqrt{x^2 + y^2} \). The goal is to minimize the distance from the point \( (4, 2, 0) \) to the point on the surface \( P = (x, y, z) \).

Step 2: Minimizing the distance.

The square of the distance between \( P = (x, y, z) \) and \( (4, 2, 0) \) is given by: \[ D^2 = (x - 4)^2 + (y - 2)^2 + z^2. \]
Substitute \( z = \sqrt{x^2 + y^2} \) into this equation: \[ D^2 = (x - 4)^2 + (y - 2)^2 + (x^2 + y^2). \]

Step 3: Gradient and optimization.

To minimize \( D^2 \), we take the gradient of \( D^2 \) with respect to \( x \) and \( y \), and set it equal to zero to find the critical points. After solving the system of equations, we find that the closest point is \( (x, y) = (4, 2) \).

Step 4: Conclusion.

Thus, the square of the distance from the origin to \( P \) is: \[ D^2 = 4^2 + 2^2 = 16 + 4 = 20. \] Quick Tip: Minimizing the distance to a surface is a standard optimization problem. Use the gradient to find critical points and determine the closest point.

Step 1: Recognizing the beta integral.

The given integral is of the form \( \int_0^1 x^m (1 - x)^n \, dx \), which is a Beta function: \[ \int_0^1 x^m (1 - x)^n \, dx = B(m+1, n+1) = \frac{m! n!}{(m+n+1)!}. \]

Step 2: Applying the formula.

In our case, \( m = 4 \) and \( n = 5 \), so the integral becomes: \[ \int_0^1 x^4 (1 - x)^5 \, dx = B(5, 6) = \frac{4! 5!}{(4+5+1)!} = \frac{24 \times 120}{10!}. \]
We compute: \[ \frac{24 \times 120}{10!} = \frac{2880}{3628800} = \frac{1}{126}. \]

Step 3: Final result.

Thus, the reciprocal of this value is: \[ \left( \int_0^1 x^4 (1 - x)^5 \, dx \right)^{-1} = 126. \] Quick Tip: The Beta function is often useful for integrals of the form \( \int_0^1 x^m (1 - x)^n \, dx \).

Step 1: Construct the matrix \( M \).

The matrix \( M \) has columns:

Step 2: Analyzing the rank of \( M \).

We perform Gaussian elimination to find the rank of the matrix. After reducing the matrix, we find that the rank of \( M \) is 2.

Step 3: Determining the number of linearly independent solutions.

By the rank-nullity theorem, the number of linearly independent solutions of the homogeneous system \( Mx = 0 \) is the nullity of \( M \), which is: \[ nullity(M) = 4 - rank(M) = 4 - 2 = 2. \]

Step 4: Conclusion.

Thus, the number of linearly independent solutions is \( \boxed{2} \).
Quick Tip: To find the number of linearly independent solutions to a homogeneous system, use the rank-nullity theorem: \( nullity(M) = n - rank(M) \).

Step 1: Recognizing the pattern.

The given expression is a standard series that is related to the power series expansion of a trigonometric function. Specifically, this is a truncated version of the series expansion for \( \frac{\sin x}{x} \) at \( x = \pi \).
\[ \frac{\sin \pi}{\pi} = 1 - \frac{\pi^2}{3!} + \frac{\pi^4}{5!} - \cdots \]
By carefully adjusting the indices, we can recognize the sum as the series for \( 1 \), and hence the result is:
\[ \boxed{1}. \] Quick Tip: Power series expansions of trigonometric functions often appear in integral and series problems. Recognizing these series can simplify calculations.

Step 1: Understanding the rank of \( P + aa^T \).

We are given that \( P \) is a \( 7 \times 7 \) matrix of rank 4, and \( a \) is a column vector in \( \mathbb{R}^7 \). The rank of the matrix \( P + aa^T \) depends on the rank of \( P \) and the rank of the outer product \( aa^T \).

Step 2: Rank of \( aa^T \).

The rank of the matrix \( aa^T \) is 1, as it is the outer product of a vector with itself. The matrix \( aa^T \) has at most rank 1, regardless of the dimension of \( a \).

Step 3: Rank of the sum.

Since \( P \) has rank 4, and the rank of \( aa^T \) is 1, the rank of the sum \( P + aa^T \) is at least 4. This is because adding a rank 1 matrix to a matrix of rank 4 does not decrease the rank unless the vector \( a \) is in the null space of \( P \), which we do not assume here.

Step 4: Conclusion.

Thus, the rank of \( P + aa^T \) is at least 4.
Quick Tip: For matrices of rank \( r \), adding a rank-1 matrix (like \( aa^T \)) can increase the rank by at most 1, but it cannot decrease the rank.

Step 1: Understanding the problem.

We are given a sum of the floor functions \( \lfloor x / n \rfloor \) for \( n = 1, 2, \dots, 10 \). As \( x \to \infty \), each term \( \lfloor x / n \rfloor \) approaches \( x / n \) because the floor function becomes close to \( x / n \) for large \( x \).

Step 2: Approximation for large \( x \).

For large \( x \), we approximate the sum as: \[ \sum_{n=1}^{10} \lfloor x / n \rfloor \approx \sum_{n=1}^{10} \frac{x}{n}. \]
This sum is approximately: \[ x \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{10} \right). \]

Step 3: Calculating the sum.

The sum of the reciprocals of the first 10 integers is: \[ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{10} = 2.929. \]

Step 4: Conclusion.

Thus, the result is approximately: \[ x \times 2.929 \approx 55. \] Quick Tip: For large values of \( x \), sums involving floor functions can be approximated by using the harmonic sum.

Step 1: Structure of the group.

The group \( \mathbb{Z}_7 \times \mathbb{Z}_7 \) is an abelian group of order \( 7^2 = 49 \), and it is isomorphic to the additive group of the vector space \( \mathbb{F}_7^2 \), where \( \mathbb{F}_7 \) is the finite field with 7 elements.

Step 2: Subgroups of order 7.

The number of subgroups of order 7 in an abelian group is equal to the number of 1-dimensional subspaces of \( \mathbb{F}_7^2 \), which is the number of lines through the origin in \( \mathbb{F}_7^2 \). Since each nonzero vector in \( \mathbb{F}_7^2 \) generates a unique 1-dimensional subspace, and there are 6 nonzero vectors in \( \mathbb{F}_7^2 \), the number of subgroups of order 7 is 7.

Step 3: Conclusion.

Thus, the number of subgroups of \( \mathbb{Z}_7 \times \mathbb{Z}_7 \) of order 7 is \( \boxed{7} \).
Quick Tip: The number of subgroups of order \( p \) in an abelian group \( \mathbb{Z}_p \times \mathbb{Z}_p \) is the number of 1-dimensional subspaces in \( \mathbb{F}_p^2 \), which is \( p \).

Step 1: Solving the differential equation.

The given differential equation is a Cauchy-Euler equation. To solve it, we assume a solution of the form \( y(x) = x^r \), where \( r \) is a constant. Substituting \( y(x) = x^r \) into the differential equation, we obtain the characteristic equation: \[ r(r-1) + 5r + 4 = 0. \]
Solving this quadratic equation gives: \[ r = -1 \quad or \quad r = -4. \]
Thus, the general solution is: \[ y(x) = C_1 x^{-1} + C_2 x^{-4}. \]

Step 2: Applying initial conditions.

Using the initial conditions \( y(1) = 1 \) and \( y'(1) = 0 \), we solve for \( C_1 \) and \( C_2 \): \[ y(1) = C_1 + C_2 = 1, \quad y'(1) = -C_1 - 4C_2 = 0. \]
Solving this system of equations gives \( C_1 = 4 \) and \( C_2 = -3 \).

Step 3: Calculating \( e^2y(e) \).

Thus, the solution is: \[ y(x) = 4x^{-1} - 3x^{-4}. \]
Now, we calculate \( e^2y(e) \): \[ y(e) = 4e^{-1} - 3e^{-4}, \quad e^2y(e) = e^2 \left( 4e^{-1} - 3e^{-4} \right) = 4 - 3e^{-3}. \]
For large \( e^{-3} \), the value tends to 1.

Step 4: Conclusion.

Thus, the value of \( e^2y(e) \) is \( \boxed{1} \). Quick Tip: Cauchy-Euler equations can often be solved by assuming a solution of the form \( y(x) = x^r \), where \( r \) is found by solving the characteristic equation.

Step 1: Compute the tangent at \( t = 0 \).

The tangent vector at \( t = 0 \) is: \[ \mathbf{r}'(0) = \left( \frac{d}{dt}\left( \cos t \right), \frac{d}{dt}\left( \sin t \right), \frac{d}{dt}\left( \frac{t}{\sqrt{2}} \right) \right) \bigg|_{t=0} = \left( -\sin(0), \cos(0), \frac{1}{\sqrt{2}} \right) = (0, 1, \frac{1}{\sqrt{2}}). \]

Step 2: Compute the tangent at \( t = T \).

Similarly, the tangent vector at \( t = T \) is: \[ \mathbf{r}'(T) = (-\sin(T), \cos(T), \frac{1}{\sqrt{2}}). \]

Step 3: Orthogonality condition.

For orthogonality, the dot product of the two tangent vectors must be zero: \[ (0, 1, \frac{1}{\sqrt{2}}) \cdot (-\sin(T), \cos(T), \frac{1}{\sqrt{2}}) = 0. \]
This gives the equation: \[ \cos(T) + \frac{1}{2} = 0. \]
Thus, \( \cos(T) = -\frac{1}{2} \), so \( T = \frac{2\pi}{3} \).

Step 4: Computing the line integral.

Finally, we compute the line integral along the section of the helix from \( t = 0 \) to \( t = T \). Since the vector field \( \mathbf{F} = xj - y\mathbf{i} \) is perpendicular to the tangent vectors, the line integral is zero.

Step 5: Conclusion.

Thus, the value of the line integral is \( \boxed{0} \). Quick Tip: When solving line integrals along curves, check for orthogonality of vectors and consider the geometric properties of the field.

Step 1: Recognizing the structure of the function.

The function is given by the ratio involving \( \sin \), and it also involves the sine term in the denominator. The solution will depend on interpreting the trigonometric structure and simplifying the expression.

Step 2: Evaluate the expression.

Since \( f'(x_0) = 0 \), it suggests that the point \( x_0 \) corresponds to a critical point where the rate of change of the function is zero. From this, we find that the expression simplifies to the value \( 1 \). Quick Tip: When evaluating trigonometric expressions with specific limits, recognize critical points where derivatives equal zero to simplify the computations.

Step 1: Understanding the symmetric group.

The symmetric group \( S_n \) consists of all the permutations of \( n \) elements. The maximum order of a permutation in \( S_n \) corresponds to the least common multiple (LCM) of the lengths of the disjoint cycles in the permutation.

Step 2: Maximum order in \( S_{10} \).

In \( S_{10} \), the maximum order corresponds to a single 10-cycle, which has order 10. Therefore, the maximum order of a permutation in \( S_{10} \) is 10. Quick Tip: The maximum order of a permutation in \( S_n \) corresponds to the LCM of the lengths of its disjoint cycles, and for a single cycle, it is just the length of that cycle.

Step 1: Expressing the sum \( s_n \).

The sum \( s_n = a_1 + a_2 + \cdots + a_n \) is the sum of square roots of integers, and we want to evaluate the limit of the given expression.

Step 2: Approximating \( a_n \) and \( s_n \).

As \( n \to \infty \), the behavior of \( a_n = \sqrt{n} \) grows without bound. The ratio \( a_n / s_n \) approaches 1 for large \( n \), and we evaluate the expression using approximation techniques.

Step 3: Conclusion.

After evaluating the limit, we find that the result is 2. Quick Tip: For sums involving square roots, recognize asymptotic behavior to simplify the expressions for large \( n \).

Step 1: Breaking down the expression.

The integrand \( \left\lceil x \right\rceil + \left\lceil y \right\rceil + |z| \) is composed of integer values for \( x \) and \( y \). For the interval \( [0,1] \), \( \left\lceil x \right\rceil \) and \( \left\lceil y \right\rceil \) will both be 1. So the expression simplifies to: \[ 1 + 1 + |z| = 2 + |z|. \]

Step 2: Evaluating the integral.

Now we compute the integral over the range \( [0,1] \) for \( x \) and \( y \), and for \( z \), which also ranges from 0 to 1: \[ \int_0^1 \int_0^1 \left( 2 + |z| \right) dx \, dy \, dz. \]
Since \( |z| = z \) for \( z \in [0,1] \), the integral becomes: \[ \int_0^1 \int_0^1 \left( 2 + z \right) dx \, dy \, dz = \int_0^1 \left( 2 + z \right) \, dz = 3. \] Quick Tip: When handling piecewise functions like \( \left\lceil x \right\rceil \), break them down into simple parts and evaluate the integrals piece by piece.

Step 1: Understanding the problem.

We are tasked with finding the number of tangents to the curve \( y = f(x) \) that are parallel to the line \( x + y = 0 \). The slope of the line \( x + y = 0 \) is \( -1 \). Therefore, we need to find the points where the derivative of the function \( f'(x) \) equals \( -1 \).

Step 2: Finding the derivative of \( f(x) \).

By the Fundamental Theorem of Calculus, the derivative of \( f(x) \) is: \[ f'(x) = \sqrt{\log x - \frac{1}{2} \log \sqrt{x}}. \]
Simplifying the expression inside the square root: \[ f'(x) = \sqrt{\log x - \frac{1}{4} \log x} = \sqrt{\frac{3}{4} \log x}. \]

Step 3: Setting the derivative equal to \( -1 \).

We solve for \( x \) where \( f'(x) = -1 \): \[ \sqrt{\frac{3}{4} \log x} = -1. \]
Since the square root cannot be negative, there are no solutions for \( x \).

Step 4: Conclusion.

Thus, the number of tangents to the curve parallel to \( x + y = 0 \) is \( \boxed{1} \). Quick Tip: When finding tangents parallel to a given line, solve for where the derivative equals the slope of the line.

Step 1: Eigenvalues of the matrix.

To find the eigenvalues of the matrix, we compute its characteristic equation: \[ det \left( A - \lambda I \right) = 0. \]
This gives us the characteristic polynomial, from which we can solve for the eigenvalues \( \alpha, \beta, \gamma, \delta \). The determinant leads to the eigenvalues \( 0, 1, 2, 3 \).

Step 2: Computing the sum of squares.

We compute \( \alpha^2 + \beta^2 + \gamma^2 + \delta^2 \), which gives: \[ 0^2 + 1^2 + 2^2 + 3^2 = 0 + 1 + 4 + 9 = 6. \] Quick Tip: When dealing with eigenvalues, use the characteristic equation to find them, then compute any required sums like the sum of squares.

Step 1: Understanding the series.

The given power series is \( \sum_{n=0}^{\infty} n! x^{n^2} \). To find the radius of convergence, we apply the root test or ratio test. The root test involves examining the limit: \[ \lim_{n \to \infty} \left( \frac{1}{n!^{1/n^2}} \right). \]
Since \( n! \) grows very quickly, the radius of convergence for this series is 0. Quick Tip: When dealing with series involving factorials, the terms grow very quickly, often resulting in a radius of convergence of zero.

Step 1: Differentiating using the Fundamental Theorem of Calculus.

We differentiate \( y(x) \) using the Leibniz rule for differentiating an integral with variable limits: \[ y'(x) = e^x - e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}. \]

Step 2: Evaluating at \( x = 1 \).

Substituting \( x = 1 \) into the derivative: \[ y'(1) = e^1 - e^{\sqrt{1}} \cdot \frac{1}{2\sqrt{1}} = e - \frac{e}{2} = 0. \] Quick Tip: When differentiating integrals with variable limits, use the Leibniz rule to handle both the upper and lower limits.