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><mo>=</mo><mn>0</mn></math> :-
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 2 log beta = log alpha + log gamma. Applying R1 -> 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. => a + c = 2b. Also alpha, beta, gamma are in G.P. => beta^2 = alpha * gamma => 2 log 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), x+b+x+c-2(x+b), ...) = (0, 0, 0). Since one column is zero, Delta = 0 for all x. Thus, it is an identity.
Correct Answer: 1, 2, 3