Aspire Faculty ID #18002 · Topic: JEE Main 2026 (23 January Morning Shift) · Just now
JEE Main 2026 (23 January Morning Shift)

Among the statements: I : \( \begin{vmatrix} 1 & \cos\alpha & \cos\beta \\ \cos\alpha & 1 & \cos\gamma \\ \cos\beta & \cos\gamma & 1 \end{vmatrix} = \begin{vmatrix} 0 & \cos\alpha & \cos\beta \\ \cos\alpha & 0 & \cos\gamma \\ \cos\beta & \cos\gamma & 0 \end{vmatrix} \), then \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = \frac{3}{2} \), and --- II : \( \begin{vmatrix} x^2 + x & x + 1 & x - 2 \\ 2x^2 + 3x - 1 & 3x & 3x - 3 \\ x^2 + 2x + 3 & 2x - 1 & 2x - 1 \end{vmatrix} = px + q \), then \( p^2 = 196q^2 \)

Solution

Let \( \cos\alpha = x,\; \cos\beta = y,\; \cos\gamma = z \) \( \begin{vmatrix} 0 & x & y \\ x & 0 & z \\ y & z & 0 \end{vmatrix} = \begin{vmatrix} 1 & x & y \\ x & 1 & z \\ y & z & 1 \end{vmatrix} \) Expanding both sides, we get \( x^2 + y^2 + z^2 = 1 \) i.e. \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \) Statement 1 is false --- Now, \( \begin{vmatrix} x^2 + x & x + 1 & x - 2 \\ 2x^2 + 3x - 1 & 3x & 3x - 3 \\ x^2 + 2x + 3 & 2x - 1 & 2x - 1 \end{vmatrix} = px + q \) --- Put \( x = 0 \) \( q = \begin{vmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 3 & -1 & -1 \end{vmatrix} \) \( q = -12 \) --- Put \( x = 1 \) \( p + q = \begin{vmatrix} 2 & 2 & -1 \\ 4 & 3 & 3 \\ 6 & 1 & 1 \end{vmatrix} = 42 \) \( p = 54 \) --- \( p^2 = 54^2,\quad 196q^2 = 196 \cdot (-12)^2 \) \( p^2 \ne 196q^2 \)

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