List of top Mathematics Questions asked in AP EAPCET

A unit vector that is perpendicular to the vector \( 2\vec{i} - \vec{j} + 2\vec{k} \) and coplanar with the vectors \( \vec{i} + \vec{j} - \vec{k} \) and \( 2\vec{i} - 2\vec{j} - \vec{k} \) is

If the magnitudes of \( \vec{a} \), \( \vec{b} \), and \( \vec{a} + \vec{b} \) are respectively \( 3 \), \( 4 \), and \( 5 \), then the magnitude of \( \vec{a} - \vec{b} \) is:

Sech\(^{-1}(\sin \alpha)\) is equal to:

Out of 8 students in a classroom, 4 of them are chosen and they are arranged around a table. If the remaining 4 are arranged in a row, then the total number of arrangements that can be made with those 8 students is

If \( \alpha, \beta \) are the roots of the equation \( x^2 + bx + c = 0 \) satisfying the conditions \( \alpha+\beta=5 \) and \( \alpha^3+\beta^3=60 \), then \( 3c+2 = \)

If the sum of two roots of the equation \( x^4 + 2x^3 - 7x^2 - 8x + 12 = 0 \) is zero, then the sum of the squares of the other two roots is

For any two non-zero complex numbers \(z_1\) and \(z_2\), if \(|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2\), then

Let \[ A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix} \text{be a singular matrix for} \]

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 \( k \in N \) 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 \( (k-1)n \).