Exams › JEE Main › Maths

The number of values of k, for which the system of equations (k + 1)x + 8y = 4k kx + (k + 3)y = 3k − 1 has no solution, is

1. infinite
2. 1
3. 2
4. 3

Correct answer: 2

Solution

The system of equations has no solution when the lines represented by the equations are parallel, which occurs when the ratios of the coefficients of x and y are equal but the ratio of the constants is different. By analyzing the coefficients and constants, we find two specific values of k that lead to this condition, confirming that there are exactly 2 values of k for which the system has no solution.

Related JEE Main Maths questions

• In a triangle $ABC$, if the determinant $\begin{vmatrix}\sin A & \sin^2 A\\ \sin B & \sin^2 B\\ \sin C & \sin^2 C\end{vmatrix}=0$, then the triangle must be
• If $[\,\cdot\,]$ represents the greatest integer less than or equal to a real number, and $-1\le x<0$, $0\le y<1$, $1\le z<2$, then the determinant \[ \begin{vmatrix} [x] & [y] & [z] \\ [x+1] & [y+1] & [z] \\ [x] & [y] & [z]+1 \end{vmatrix} \] is equal to:
• Let $x$, $y$, and $z$ be complex numbers. If $\Delta$ denotes the determinant of the matrix \[ \begin{bmatrix} 0 & -y & -z\\ y & 0 & -x\\ z & x & 0 \end{bmatrix}, \] then $\Delta$ is
• For the linear system \[ \begin{aligned} x_1+2x_2+3x_3&=6\\ x_1+3x_2+5x_3&=9\\ 2x_1+5x_2+ax_3&=b \end{aligned} \] if it is consistent and admits infinitely many solutions, then which statement must be true?
• For which value(s) of $x$ does the determinant of the matrix \[ \begin{vmatrix} 1 & x-3 & (x-3)^2\\ 1 & x-4 & (x-4)^2\\ 1 & x-5 & (x-5)^2 \end{vmatrix} \] become zero?
• Let $\Delta_r$ denote the determinant \[ \Delta_r=\begin{vmatrix} \binom{m}{r} & 1 & 1\\ m^2-1 & 2^m & m+1\\ \sin^2(m^2) & \sin^2(m) & \sin^2(m+1) \end{vmatrix}. \] Then the value of $\sum_{r=0}^{m} \Delta_r$ is

⚔️ Practice JEE Main Maths free + battle 1v1 →