List of top Mathematics Questions

Consider the following statements
Statement-I: A function \( f: A \rightarrow B \) is said to be one-one if and only if \[ f(x) = f(y) \Rightarrow x = y \]
Statement-II: A relation \( f: A \rightarrow B \) is said to be a function if \[ x = y \Rightarrow f(x) \neq f(y) \]
Then which one of the following is true?

If \( t_n = \dfrac{1}{n(n+2)} \), \( n \in \mathbb{N} \), then which one of the following is true?
Assertion (A): \[ t_1 + t_2 + \cdots + t_{2003} = \dfrac{2003}{3005} \]
Reason (R): \[ t_n = \dfrac{1}{n(n+2)} = \dfrac{1}{2} \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \]

The range of the real valued function \( f(x) = \cos^{-1 \left( \dfrac{3}{\sqrt{9x^2 - 12x + 22}} \right) \) is}

If \( k \in N \) then \( \lim_{n\to\infty} \left[ \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \dots + \frac{1}{kn} \right] = \) (Note: The last term should be \( \frac{1}{n+ (k-1)n} = \frac{1}{kn} \) or sum up to \(n+(k-1)n\). The given form \(1/kn\) as the endpoint of the sum means sum from \(r=1\) to \((k-1)n\). The sum is usually \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). If the last term is \( \frac{1}{kn} \), it means \( n+r = kn \implies r = (k-1)n \). So it's \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \).) Let's assume the sum goes up to \( \frac{1}{n+(k-1)n} = \frac{1}{kn} \). So the sum is \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). No, this seems to be \( \frac{1}{n+1} + \dots + \frac{1}{n+(kn-n)} \). The sum should be written as \( \sum_{i=1}^{(k-1)n} \frac{1}{n+i} \). The dots imply the denominator goes up. The last term is \( \frac{1}{kn} \). This means the sum is actually \( \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{n+(k-1)n} \). The number of terms is \( (k-1)n \).