JEE Main 2014 — Mathematics Algebra

$$f(1)=\alpha +\beta ,f(2)={\alpha }^{2}+{\beta }^{2},f(3)={\alpha }^{3}+{\beta }^{3},f(4)={\alpha }^{4}+{\beta }^{4}$$f(1)=α+β,f(2)=α2+β2,f(3)=α3+β3,f(4)=α4+β4

So, the given determinant can be written as

$| \begin{matrix} 1 + 1 + 1 & 1 + \alpha + \beta & 1 + {\alpha }^{2} + {\beta }^{2} \\ 1 + \alpha + \beta & 1 + {\alpha }^{2} + {\beta }^{2} & 1 + {\alpha }^{3} + {\beta }^{3} \\ 1 + {\alpha }^{2} + {\beta }^{2} & 1 + {\alpha }^{3} + {\beta }^{3} & 1 + {\alpha }^{4} + {\beta }^{4} \end{matrix} |$∣1+1+11+α+β1+α2+β2​1+α+β1+α2+β21+α3+β3​1+α2+β21+α3+β31+α4+β4​∣ $=|\begin{matrix}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {\alpha }^{2} & {\beta }^{2}\end{matrix}||\begin{matrix}1 & 1 & 1 \\ 1 & \alpha & {\alpha }^{2} \\ 1 & \beta & {\beta }^{2}\end{matrix}|$=∣111​1αα2​1ββ2​∣∣111​1αβ​1α2β2​∣

$$= | \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {\alpha }^{2} & {\beta }^{2} \end{matrix} | | \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {\alpha }^{2} & {\beta }^{2} \end{matrix} |$$=∣111​1αα2​1ββ2​∣∣111​1αα2​1ββ2​∣

$$= { | \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {\alpha }^{2} & {\beta }^{2} \end{matrix} | }^{2} = { [ ( \alpha {\beta }^{2} - {\alpha }^{2} \beta ) - ( {\beta }^{2} - \beta ) + ( {\alpha }^{2} - \alpha ) ] }^{2}$$=∣111​1αα2​1ββ2​∣2=[(αβ2−α2β)−(β2−β)+(α2−α)]2

$$= { [ \alpha \beta ( \beta - \alpha ) - ( \beta - \alpha ) ( \beta + \alpha ) + ( \beta - \alpha ) ] }^{2}$$=[αβ(β−α)−(β−α)(β+α)+(β−α)]2

$$={(\beta -a)}^{2}{[\alpha \beta -\beta -\alpha +1]}^{2}$$=(β−a)2[αβ−β−α+1]2

$$= { ( \beta - \alpha ) }^{2} { ( \alpha - 1 ) }^{2} { ( \beta - 1 ) }^{2}$$=(β−α)2(α−1)2(β−1)2

Hence, $K=1$K=1.