Kerala KEAM 5 June 2024 Solved Paper

The given expression is a determinant: $$\begin{vmatrix} x & 2 & -11 & x & 53 & 2 & x \end{vmatrix} = 0.$$We will calculate the determinant of the 3x3 matrix. Using cofactor expansion along the first row: $$Determinant = x \begin{vmatrix} x & 52 & x \end{vmatrix} - 2 \begin{vmatrix} 1 & 53 & x \end{vmatrix} + (-1) \begin{vmatrix} 1 & x3 & 2 \end{vmatrix}.$$Step 1: Now, calculate each of the 2x2 determinants: $$\begin{vmatrix} x & 52 & x \end{vmatrix} = x \cdot x - 5 \cdot 2 = x^2 - 10,$$ $$\begin{vmatrix} 1 & 53 & x \end{vmatrix} = 1 \cdot x - 5 \cdot 3 = x - 15,$$ $$\begin{vmatrix} 1 & x3 & 2 \end{vmatrix} = 1 \cdot 2 - x \cdot 3 = 2 - 3x.$$Step 2: Substitute these values into the cofactor expansion: $$Determinant = x(x^2 - 10) - 2(x - 15) - (2 - 3x).$$Simplifying the expression: $$Determinant = x^3 - 10x - 2x + 30 - 2 + 3x,$$ $$Determinant = x^3 - 9x + 28.$$Step 3: Set the determinant equal to zero: $$x^3 - 9x + 28 = 0.$$Step 4: To find the roots of the cubic equation, we can use the Rational Root Theorem. The possible rational roots are the factors of 28 divided by the factors of 1, i.e., $\pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28$. Testing $x = -4$: $$(-4)^3 - 9(-4) + 28 = -64 + 36 + 28 = 0.$$Thus, $x = -4$ is a solution.Thus, the correct answer is option (E). Quick Tip: When calculating the determinant of a 3x3 matrix, use cofactor expansion. You can then simplify the resulting expression to find the value of $x$.