A) \( \Large \frac{3}{2} \) |
B) 1 |
C) \( \Large \frac{1}{2} \) |
D) \( \Large \frac{11}{4} \) |
C) \( \Large \frac{1}{2} \) |
Given equation is \( \Large x^{2}+ \left(2+\lambda\right)x-\frac{1}{2} \left(1+\lambda \right)=0 \).
Let \( \Large \alpha \ and \ \beta \) are the roots of given equations
=> \( \Large \alpha + \beta =- \left(2+ \lambda \right) \) and \( \Large \alpha \beta = - \left(\frac{1+\lambda}{2}\right) \)
Now, \( \Large \alpha ^{2}+ \beta ^{2}= \left( \alpha + \beta \right)^{2}-2 \alpha \beta \)
=> \( \Large \alpha ^{2}+ \beta ^{2}= \left[ - \left(2+\lambda\right)^{2}+2\frac{ \left(1+\lambda\right) }{2} \right] \)
=>\( \Large \alpha ^{2}+ \beta ^{2}= \lambda^{2}+4+4\lambda+1+\lambda=\lambda^{2}+5\lambda+5 \)
Now we take the option simultaneously
=> It is minimum for \( \Large \lambda = \frac{1}{2} \).