Exams Joint Entrance Examination Mathematics Complex Numbers Question 115 Question 115 of 455 Hard MCQ
Let α \alpha α be a root of the equation ( a − c ) x 2 + ( b − a ) x + ( c − b ) = 0 (a-c)x^2+(b-a)x+(c-b)=0 ( a − c ) x 2 + ( b − a ) x + ( c − b ) = 0 where a , b , c a,b,c a , b , c are distinct real numbers such that the matrix ( α 2 α 1 1 1 1 a b c ) \begin{pmatrix}\alpha^2 & \alpha & 1\\ 1 & 1 & 1\\ a & b & c\end{pmatrix} α 2 1 a α 1 b 1 1 c is singular. Then the value of
\[\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}\]
is
Solutions · 1 Post your solution Official solution
\[
\begin{aligned}
& \Delta=0=\left|\[\begin{array}{|ccccccc|}\alpha^2 & \alpha & 1 \\ \\ 1 & 1 & 1 \\ \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\end{aligned} \]\right| \\\\
\& \Rightarrow \alpha\^2(\mathrm{c}-\mathrm{b})-\alpha(\mathrm{c}-\mathrm{a})+(\mathrm{b}-\mathrm{a})=0
\]
It is singular when α = 1 \alpha=1 α = 1
\[
( a − c ) 2 ( b − a ) ( c − b ) + ( b − a ) 2 ( a − c ) ( c − b ) + ( c − b ) 2 ( a − c ) ( b − a ) ( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 ( a − b ) ( b − c ) ( c − a ) = 3 ( a − b ) ( b − c ) ( c − a ) ( a − b ) ( b − c ) ( c − a ) = 3 \begin{aligned}
& \frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)} \\\\
& \frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} \\\\
& =3 \frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3
\end{aligned} ( b − a ) ( c − b ) ( a − c ) 2 + ( a − c ) ( c − b ) ( b − a ) 2 + ( a − c ) ( b − a ) ( c − b ) 2 ( a − b ) ( b − c ) ( c − a ) ( a − b ) 3 + ( b − c ) 3 + ( c − a ) 3 = 3 ( a − b ) ( b − c ) ( c − a ) ( a − b ) ( b − c ) ( c − a ) = 3
\] Answer the question to unlock