CUET PG Mathematics Mock Tests

To determine if a non-empty subset $S$ of a ring $R$ is a subring, we must verify two conditions: 1. Closure under subtraction: For any $A, B \in S$, $A-B \in S$. 2. Closure under multiplication: For any $A, B \in S$, $A \cdot B \in S$. (The additive identity $0_R$ being in $S$ is implicitly covered by the first condition if $S$ is non-empty, as $A-A=0_R$.) Let's examine each option: **1.** **The set of all invertible matrices in $M_2(\mathbb{Z})$.** The additive identity of $M_2(\mathbb{Z})$ is the zero matrix: The $\det$ of the zero matrix is $0$, so it is not invertible. A subring must contain the additive identity. Therefore, this set is not a subring. **2.** **The set of all matrices in $M_2(\mathbb{Z})$ whose entries are all multiples of 3.** Let $S$ be this set. We can write $S$ as: * **Non-empty:** The zero matrix $0 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ has all entries (0) as multiples of 3, so $0 \in S$. * **Closure under subtraction:** Let $A, B \in S$. Then their difference is: All entries of $A-B$ are multiples of 3, so $A-B \in S$. * **Closure under multiplication:** Let $A, B \in S$. Their product is: Each entry in the product is a multiple of 3. So $A \cdot B \in S$. Since all conditions are satisfied, this set is a subring. **3.** **The set of all matrices in $M_2(\mathbb{Z})$ with determinant 0.** Consider: Both $A$ and $B$ have $\det(A) = 0$ and $\det(B) = 0$. However, their sum is: The $\det(A+B) = 1$. Since $A+B$ is not in the set, the set is not closed under addition (and thus not under subtraction). Therefore, it is not a subring. **4.** **The set of all matrices in $M_2(\mathbb{Z})$ whose diagonal entries are zero.** Let $S$ be this set. We can write $S$ as: Consider: Both $A$ and $B$ are in $S$. Their product is: The diagonal entries of $A \cdot B$ are $4$ and $6$, which are not zero. Since $A \cdot B$ is not in $S$, the set is not closed under multiplication. Therefore, it is not a subring. The final answer is $\boxed{\text{The set of all matrices in } M_2(\mathbb{Z}) \text{ whose entries are all multiples of 3.}}$

1. The given differential equation is of the form where: 2. First, we check for exactness by calculating the partial derivatives: Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact. 3. Next, we look for an integrating factor. We check the two standard cases: **Case 1:** If $\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}$ is a function of $x$ only, say $f(x)$. We calculate the expression: Assuming $xy - 1 \neq 0$, this simplifies to: This is a function of $x$ only. Therefore, an integrating factor exists and is given by $e^{\int f(x) dx}$. Integrating factor (IF) calculation: **Case 2:** If $\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}$ is a function of $y$ only, say $g(y)$. We calculate the expression: This is not a function of $y$ only. Thus, the integrating factor is $\frac{1}{x^2}$. Answer: $\boxed{\frac{1}{x^2}}$

1. For the sequence $\left\langle \frac{\sin(n)}{n} \right\rangle$: We know that for all $n \in \mathbb{N}$. Dividing by $n$ (which is positive for $n \in \mathbb{N}$), we get: As $n \to \infty$, we have and By the Squeeze Theorem, Thus, this sequence converges to $0$. 2. For the sequence $\left\langle n^{1/n} \right\rangle$: Let Consider the natural logarithm: This is an indeterminate form of type $\frac{\infty}{\infty}$. We can apply L'Hôpital's Rule (considering $x^{1/x}$ as a continuous function): So, $\ln L = 0$, which implies $L = e^0 = 1$. Thus, this sequence converges to $1$. 3. For the sequence $\left\langle \left(1 + \frac{1}{n}\right)^n \right\rangle$: This is a standard limit definition of the mathematical constant $e$. Thus, this sequence converges to $e$. 4. For the sequence $\left\langle (n!)^{1/n} \right\rangle$: We can use Cauchy's second theorem on limits (also known as the Root Test for sequences), which states that if $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$ for a sequence $\langle a_n \rangle$ with $a_n > 0$, then $\lim_{n \to \infty} (a_n)^{1/n} = L$. Let $a_n = n!$. Then we consider the limit of the ratio: As $n \to \infty$, $(n+1) \to \infty$. Therefore, by Cauchy's second theorem on limits, Since the limit is $\infty$, the sequence is divergent. Therefore, the divergent sequence is $\left\langle (n!)^{1/n} \right\rangle$. Answer: \boxed{\left\langle (n!)^{1/n} \right\rangle}

**Step 1: Start with the inequality $|f(x) - L| < \epsilon$.** We are given $f(x) = x^2$, $L = 4$, and $\epsilon = 0.05$. So, we need to satisfy: **Step 2: Factor the expression.** The left side can be factored as a difference of squares: This can be written as: **Step 3: Bound the term $|x + 2|$.** In the $\epsilon - \delta$ definition, we are looking for a $\delta$ such that if $0 < |x - 2| < \delta$, then $|x^2 - 4| < \epsilon$. To bound $|x + 2|$, we typically restrict $\delta$ with an initial choice. The problem statement guides us to choose $\delta \le 1$. If $|x - 2| < \delta \le 1$, then: Adding 2 to all parts of the inequality gives: Now, we need to find bounds for $x + 2$. Adding 2 to all parts of the inequality $1 < x < 3$: From this, we can conclude that: **Step 4: Substitute the bound back into the inequality.** We have $|x - 2| |x + 2| < 0.05$. Using $|x + 2| < 5$, we can write: **Step 5: Isolate $|x - 2|$.** Divide by 5: **Step 6: Determine the value of $\delta$.** From the previous step, we see that if $|x - 2| < 0.01$, then $|x^2 - 4| < 0.05$, provided our initial assumption that $|x + 2| < 5$ holds. This assumption holds if $\delta \le 1$. Therefore, we choose $\delta$ to be the minimum of our initial bound (1) and the value derived from $\epsilon$ (0.01): Thus, the largest possible value for $\delta$ is $0.01$. Answer: $\boxed{0.01}$

**Step 1: Identify $P$ and $Q$ from the line integral.** Given the line integral in the form $\oint_C (P\,dx + Q\,dy)$, we identify the functions $P(x,y)$ and $Q(x,y)$: **Step 2: Compute the partial derivatives required for Green's Theorem.** Green's Theorem states that for a positively oriented simple closed curve $C$ bounding a region $R$: We calculate the necessary partial derivatives: **Step 3: Apply Green's Theorem.** Substitute the partial derivatives into the Green's Theorem formula: **Step 4: Define the region $R$ and set up the double integral.** The region $R$ is bounded by the curves $y=x$ and $y=x^2$. To find the points of intersection, we set the equations equal to each other: This gives intersection points at $x=0$ and $x=1$. For $x \in [0,1]$, the line $y=x$ is above the parabola $y=x^2$. Thus, the region $R$ can be described by the inequalities: Now, we can set up the double integral: **Step 5: Evaluate the inner integral with respect to $y$.** **Step 6: Evaluate the outer integral with respect to $x$.** Now, integrate the result from Step 5 with respect to $x$ from $0$ to $1$: Thus, the value of the line integral is $\frac{1}{30}$. Answer: $\boxed{\frac{1}{30}}$