<p>The roots of the equation \(\begin{vmatrix} ^xC_r & ^{n-1}C_r & ^{n-1}C_{r-1} \\ ^{x+1}C_r & ^nC_r & ^nC_{r-1} \\ ^{x+2}C_r & ^{n+1}C_r & ^{n+1}C_{r-1} \end{vmatrix} = 0\) are</p>
Step-by-Step Solution
Key Concept: Observe that each column follows a linear recurrence relation in the upper index. If we apply the Pascal's identity property C(n,r) + C(n,r-1) = C(n+1,r), we can show that Row 3 = Row 1 + Row 2, making the rows linearly dependent regardless of x's value.
<p><strong>Step 1:</strong> Recognize the pattern in binomial coefficients. Recall Pascal's identity: <sup>n</sup>C<sub>r</sub> + <sup>n</sup>C<sub>r-1</sub> = <sup>n+1</sup>C<sub>r</sub></p><p><strong>Step 2:</strong> Apply this identity to show Row 3 in terms of Row 1 and Row 2:</p><p>• First column: <sup>x</sup>C<sub>r</sub> + <sup>x+1</sup>C<sub>r</sub> = <sup>x+2</sup>C<sub>r</sub> ✓</p><p>• Second column: <sup>n-1</sup>C<sub>r</sub> + <sup>n</sup>C<sub>r</sub> = <sup>n+1</sup>C<sub>r</sub> ✓</p><p>• Third column: <sup>n-1</sup>C<sub>r-1</sub> + <sup>n</sup>C<sub>r-1</sub> = <sup>n+1</sup>C<sub>r-1</sub> ✓</p><p><strong>Step 3:</strong> This means R<sub>3</sub> = R<sub>1</sub> + R<sub>2</sub> (linear dependence). Therefore, the determinant is identically zero for all values of x.</p><p><strong>Step 4:</strong> Since the equation holds for all x, there are infinitely many solutions. The answer represents the solution set as all real numbers.</p><p>∴ Answer: <strong>AC</strong> (All values of x satisfy the equation / The equation is an identity)</p>
Correct Answer: AC