JEE Main 2014Session, Shift 1MathematicsHardMatrices and Determinants245× since 2002If α,β≠0,\alpha ,\beta \ne 0,α,β=0, and f(n)=αn+βnf\left( n \right) = {\alpha ^n} + {\beta ^n}f(n)=αn+βn and ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣\left| {\matrix{ 3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr } } \right|31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4) =K(1−α)2(1−β)2(α−β)2,= K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2},=K(1−α)2(1−β)2(α−β)2, then KKK is equal to :AOption A. 111BOption B. −1-1−1COption C. αβ\alpha \betaαβDOption D. 1αβ{1 \over {\alpha \beta }}αβ1CheckSkip · show solution
