List of top Questions asked in AP EAPCET- 2025

If \(P\) is a variable point which is at a distance of 2 units from the line \(2x - 3y + 1 = 0\) and \(\sqrt{13}\) units from the point (5, 6), then the equation of the locus of \(P\) is
Domain of the real-valued function \(f(x) = \log(x^2 - 1) + x \, \coth^{-1}x\) is
If the transformed equation of the equation \[ 2x^2 + 3xy - 2y^2 - 17x + 6y + 8 = 0 \] after translating the coordinate axes to a new origin \((\alpha, \beta)\) is \[ aX^2 + 2h XY + bY^2 + c = 0, \] then find \(3\alpha + c\).
If the distance of a variable point \(P\) from a point \(A(2,-2)\) is twice the distance of \(P\) from the Y-axis, then the equation of locus of \(P\) is:
The image of a point \( (2, -1) \) with respect to the line \( x - y + 1 = 0 \) is
If A(1,0), B(0,-2), C(2,-1) are three fixed points, then the equation of the locus of a point P such that area of \( \triangle \text{PAB} \) is equal to area of \( \triangle \text{PAC} \) is
The coordinate axes are rotated about the origin in the counterclockwise direction through an angle \( 60^\circ \). If \( a \) and \( b \) are the intercepts made on the new axes by a straight line whose equation referred to the original axes is \( x + y = 1 \), then \( \dfrac{1}{a^2} + \dfrac{1}{b^2} = \, ? \)
\( \sum_{k=1}^{n} k(k+1)(k+2)...(k+r-1) = \)
If the values \( x = \alpha, y = \beta, z = \gamma \) satisfy all the 3 equations \(x+2y+3z=4\), \(3x+y+z=3\) and \(x+3y+3z=2\), then \(3\alpha + \gamma = \)
If $a$ is the determinant of the adjoint of the matrix $\begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 3 \end{bmatrix}$ and $b$ is the determinant of the inverse of the matrix $\begin{bmatrix} 2 & 1 & 3 \\ 1 & -4 & -1 \\ 2 & 1 & 4 \end{bmatrix}$, then $\frac{b+1}{18b} = $
For all \( n \in \mathbb{N} \), if \( 1^3 + 2^3 + 3^3 + \cdots + n^3>x \), then a value of \( x \) among the following is:
If \[ A = \begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 2\\ -1 & 1 & -2 \end{bmatrix}, \] then find $A + 2A^{-1}$.
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): \[ t_n = \dfrac{1}{n(n+2)} = \dfrac{1}{2} \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \]
\( \sum_{k=1}^{n} k(k+1)(k+2)...(k+r-1) = \)
Let \[ A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix} \text{be a singular matrix for} \]
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:
The remainder obtained when \( (2m + 1)^{2n} \), \( m, n \in \mathbb{N} \) is divided by 8 is
If the system of equations \( 2x + py + 6z = 8 \), \( x + 2y + qz = 5 \) and \( x + y + 3z = 4 \) has infinitely many solutions, then \( p = \)?
A value of \( \theta \) lying between \( 0 \) and \( \dfrac{\pi}{2} \) and satisfying \[ \begin{vmatrix} 1 + \sin^2 \theta & \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4\sin 4\theta \end{vmatrix} = 0 \] is:
If $A = \begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}$ and $\alpha, \beta, \gamma$ are the roots of the equation represented by $|A - xI| = 0$, then $\alpha^2 + \beta^2 + \gamma^2 =$
Let \[ A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix} \text{be a singular matrix for} \]
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): \[ t_n = \dfrac{1}{n(n+2)} = \dfrac{1}{2} \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \]
If the values \( x = \alpha, y = \beta, z = \gamma \) satisfy all the 3 equations \(x+2y+3z=4\), \(3x+y+z=3\) and \(x+3y+3z=2\), then \(3\alpha + \gamma = \)
Consider two systems of 3 linear equations in 3 unknowns $AX = B$ and $CX = D$. If $AX = B$ has the unique solution $X = D$ and $CX = D$ has the unique solution $X = B$, then the solution of $(A - C^{-1})X = 0$ is
If \( A = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 1 & 6 \end{pmatrix} \) and \( |adj(adj(A))|(adj A)^{-1} = kA \), then k =