Let \(\alpha\) be a root of the equation \((\mathrm{a}-\mathrm{c}) \mathrm{x}^{2}+(\mathrm{b}-\mathrm{a}) \mathrm{x}+(\mathrm{c}-\mathrm{b})=0\) where \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are distinct real numbers such that the matrix \(\left[\begin{array}{ccc}\alpha^{2} & \alpha & 1 \\ 1 & 1 & 1 \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right]\) is singular. Then the value of \(\frac{(a-c)^{2}}{(b-a)(c-b)}+\frac{(b-a)^{2}}{(a-c)(c-b)}+\frac{(c-b)^{2}}{(a-c)(b-a)}\) is

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Determinants

Let

α

be a root of the equation

(a - c) x^{2} + (b - a) x + (c - b) = 0

where

a, b, c

are distinct real numbers such that the matrix

α^{2} 1 a α 1 b 11 c

is singular. Then the value of

\frac{( a - c ) ^{2}}{( b - a ) ( c - b )} + \frac{( b - a ) ^{2}}{( a - c ) ( c - b )} + \frac{( c - b ) ^{2}}{( a - c ) ( b - a )}

Answer: (B)

Because

(a - c) + (b - a) + (c - b) = 0

, the value

1

is a root of

(a - c) x^{2} + (b - a) x + (c - b) = 0

, and it also makes the given matrix singular, so such an

α

exists. Putting

u = a - c, v = b - a, w = c - b

gives

u + v + w = 0

, and the expression becomes

\frac{u ^{2}}{v w} + \frac{v ^{2}}{w u} + \frac{w ^{2}}{uv} = \frac{u ^{3} + v ^{3} + w ^{3}}{uv w}

. Since

u + v + w = 0

makes

u^{3} + v^{3} + w^{3} = 3 uv w

, the value is

3.

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option