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