List of top Mathematics Questions asked in AP EAPCET

\( (1+\sqrt{3}i)^6 - (\sqrt{3}+i)^6 = \)

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 \( 3c+2 = \)

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

Evaluate the infinite series: \[ \left|\begin{array}{ccc} 2 & 1 & \frac{1}{3}
3 & 1 & 1
\end{array}\right| + \left|\begin{array}{ccc} 1 & \frac{1}{3} & \frac{1}{2}
3 & 1 & 1
\end{array}\right| + \left|\begin{array}{ccc} 1 & \frac{1}{4} & \frac{1}{9}
3 & 1 & 1
\end{array}\right| + \left|\begin{array}{ccc} 1 & \frac{1}{4} & \frac{1}{27}
3 & 1 & 1
\end{array}\right| + \cdots = ? \]

If \( A = \begin{bmatrix} 1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{bmatrix} \) and the rank of \( A \) is 2, then the value of \( x \) is equal to:

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 \( (k-1)n \).