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Let $\alpha ,\beta $ be the roots of ${x^2} + (3 - \lambda )x - \lambda = 0.$ The value of $\lambda $ for which ${\alpha ^2} + {\beta ^2}$ is minimum, is
$0$
$1$
$2$
$3$
Solution
(c) $\alpha + \beta = \lambda – 3\,\,\,\,{\rm{and }}\,\,\, \alpha \beta = – \lambda $
${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} – 2\alpha \beta $ = ${(\lambda – 3)^2} + 2\lambda = {\lambda ^2} – 4\lambda + 9$
from options,
for $\lambda = 0,\,{({\alpha ^2} + {\beta ^2})_{\lambda = 0}} = 9$
for $\lambda = 1,\,{({\alpha ^2} + {\beta ^2})_{\lambda = 1}} = 1 – 4 + 9 = 6$
for $\lambda = 2,\,{({\alpha ^2} + {\beta ^2})_{\lambda = 2}} = 4 – 8 + 9 = 5$
for $\lambda = 3,\,{({\alpha ^2} + {\beta ^2})_{\lambda = 3}} = 9 – 12 + 9 = 6$
${\alpha ^2} + {\beta ^2}$ is minimum for $\lambda = 2$.
