A) 3/2
B) 1
C) 1/2
D) 11/4
Correct Answer: C
Solution :
Given equation is \[{{x}^{2}}+(2+\lambda )x-\frac{1}{2}(1+\lambda )=0\] So, \[\alpha +\beta =-(2+\lambda )=0\] and \[\alpha \beta =-\left( \frac{1+\lambda }{2} \right)\] Now, \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\left[ -(2+\lambda ) \right]}^{2}}+2.\frac{(1+\lambda )}{2}\] Þ \[{{\alpha }^{2}}+{{\beta }^{2}}={{\lambda }^{2}}+4+4\lambda +1+\lambda ={{\lambda }^{2}}+5\lambda +5\] which is minimum for \[\lambda =1/2\].
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