If a, b, c are in A.P. and α, β, γ are positive real numbers in G.P., then the equation <br> <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|"><mtable><mtr><mtd><mi>x</mi><mo>+</mo><mi>a</mi></mtd><mtd><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mi>log</mi><mi>α</mi></mtd><mtd><mi>k</mi></mtd></mtr><mtr><mtd><mi>x</mi><mo>+</mo><mi>b</mi></mtd><mtd><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mi>log</mi><mi>β</mi></mtd><mtd><mi>k</mi></mtd></mtr><mtr><mtd><mi>x</mi><mo>+</mo><mi>c</mi></mtd><mtd><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mi>log</mi><mi>γ</mi></mtd><mtd><mi>k</mi></mtd></mtr></mtable></mfenced></math> = 0 :-
Step-by-Step Solution
Key Concept: Since a, b, c are in A.P., 2b = a + c. Since \alpha, \beta, \gamma are in G.P., \beta^2 = \alpha\gamma, so 2log\beta = log\alpha + log\gamma. Applying row operations R1 + R3 - 2R2 makes the first column and second column zero, implying the determinant is zero for all x.
Given a, b, c are in A.P. => 2b = a + c. Also \alpha, \beta, \gamma are in G.P. => \beta^2 = \alpha\gamma => 2log\beta = log\alpha + log\gamma. Let the determinant be \Delta. Applying R1 -> R1 + R3 - 2R2: The first column becomes (x+a+x+c-2(x+b)) = (2x+a+c-2x-2b) = 0. The second column becomes (x^2+log\alpha+x^2+log\gamma-2(x^2+log\beta)) = (2x^2+log\alpha+log\gamma-2x^2-2log\beta) = 0. The third column becomes (k+k-2k) = 0. Since all elements of the first row are zero, the determinant \Delta = 0 for all x. Thus, it is an identity.
Correct Answer: C