If x ≠ y ≠ z & x, y, z are in GP and D = 0, then y is equal to -
Step-by-Step Solution
Key Concept: The determinant D is a Vandermonde-like determinant. For x, y, z in GP, let x=a/r, y=a, z=ar. The determinant D = xyz(x-y)(y-z)(z-x)(x+y+z). Setting D=0 with x, y, z distinct implies x+y+z=0, which is not possible for positive real numbers in GP. However, the determinant is actually D = xyz(x-y)(y-z)(z-x)(x+y+z) is incorrect. The determinant is D = (x-y)(y-z)(z-x)(xy+yz+zx). For D=0, xy+yz+zx=0. For GP, a^2/r + a^2 + a^2r = 0, which implies 1/r + 1 + r = 0. This leads to r^2+r+1=0, which has no real roots. Re-evaluating the determinant: D = xyz(x-y)(y-z)(z-x)(x+y+z) is not correct. The determinant is D = (x-y)(y-z)(z-x)(xy+yz+zx). Wait, the determinant is |x x^3 x^4-1; y y^3 y^4-1; z z^3 z^4-1|. This is |x x^3 x^4; y y^3 y^4; z z^3 z^4| - |x x^3 1; y y^3 1; z z^3 1|. This simplifies to xyz(x-y)(y-z)(z-x)(x+y+z) - (x-y)(y-z)(z-x)(xy+yz+zx). For D=0, xyz(x+y+z) = xy+yz+zx. For GP, y^2=xz, so y^3=xyz. y^3(x+y+z) = xy+y^2+yz = y(x+y+z). Since x,y,z are distinct, x+y+z is not 0. Thus y^2=1, so y=1.
The determinant D = |x x^3 x^4-1; y y^3 y^4-1; z z^3 z^4-1| = |x x^3 x^4; y y^3 y^4; z z^3 z^4| - |x x^3 1; y y^3 1; z z^3 1|. This simplifies to xyz(x-y)(y-z)(z-x)(x+y+z) - (x-y)(y-z)(z-x)(xy+yz+zx). Given D=0 and x,y,z distinct, we have xyz(x+y+z) = xy+yz+zx. Since x,y,z are in GP, let x=a/r, y=a, z=ar. Then y^2=xz. Substituting, a^3(a/r + a + ar) = a^2/r + a^2 + a^2r. a^3(a/r + a + ar) = a^2(1/r + 1 + r). a^3 = a^2, so a=1 (since x,y,z are in R+). Thus y=1.
Correct Answer: B