Step 1: Apply the given condition.
We are given that: \[ m^2 + p^2<100 \quad \text{and} \quad (m - p)^2 + 2mp<100 \] This simplifies to: \[ 2mp<100 - (m - p)^2 \quad \text{(Equation 1)} \] Step 2: Find the condition for maximum value of \( mp \).
When \( (m - p)^2 = 0 \), we get: \[ m = p \] Substituting \( m = p \) into Equation (1): \[ 2mp<100 - 0 \quad \Rightarrow \quad mp<50 \] Step 3: Maximize \( mp \).
Now, we know \( mp<50 \). The greatest integer value for \( mp \) is 49, where \( m = p = 7 \).
Step 4: Conclusion.
The greatest possible value of \( mp \) is 49.
Find the residue of \( (67 + 89 + 90 + 87) \pmod{11} \):
Match List-I with List-II and choose the correct option:
LIST-I (Infinite Series) | LIST-II (Nature of Series) |
---|---|
(A) \( 12 - 7 - 3 - 2 + 12 - 7 - 3 - 2 + \dots \) | (II) oscillatory |
(B) \( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \) | (IV) conditionally convergent |
(C) \( \sum_{n=0}^{\infty} \left( (n^3+1)^{1/3} - n \right) \) | (I) convergent |
(D) \( \sum_{n=1}^{\infty} \frac{1}{n \left( 1 + \frac{1}{n} \right)} \) | (III) divergent |
Choose the correct answer from the options given below:
Match List-I with List-II and choose the correct option:
LIST-I (Set) | LIST-II (Supremum/Infimum) |
---|---|
(A) \( S = \{2, 3, 5, 10\} \) | (III) Sup S = 10, Inf S = 2 |
(B) \( S = (1, 2] \cup [3, 8) \) | (IV) Sup S = 8, Inf S = 1 |
(C) \( S = \{2, 2^2, 2^3, \dots, 2^n, \dots\} \) | (II) Sup S = 5, Inf S = -5 |
(D) \( S = \{x \in \mathbb{Z} : x^2 \le 25\} \) | (I) Inf S = 2 |
Choose the correct answer from the options given below:
Which of the following are correct?
A. A set \( S = \{(x, y) \mid xy \leq 1 : x, y \in \mathbb{R}\} \) is a convex set.
B. A set \( S = \{(x, y) \mid x^2 + 4y^2 \leq 12 : x, y \in \mathbb{R}\} \) is a convex set.
C. A set \( S = \{(x, y) \mid y^2 - 4x \leq 0 : x, y \in \mathbb{R}\} \) is a convex set.
D. A set \( S = \{(x, y) \mid x^2 + 4y^2 \geq 12 : x, y \in \mathbb{R}\} \) is a convex set.
If p is a prime number and a group G is of the order p2, then G is: