A) 2
B) 1
C) 3
D) 0
Correct Answer: B
Solution :
| Let \[\alpha ,\beta \]be the roots of the equation \[{{x}^{2}}-\left( a-2 \right)x-a+1=0,\] then \[\alpha +\beta =a-2,\alpha \beta =-a+1\] |
| \[\therefore \,\,z={{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \]\[={{\left( a-2 \right)}^{2}}+2\left( a-1 \right)={{a}^{2}}-2a+2\] |
| \[\frac{dz}{da}=2a-2=0\Rightarrow a=1\] |
| \[\frac{{{d}^{2}}z}{d{{a}^{2}}}=2>0,\], so z has minima at\[a=1\]. |
| So \[{{\alpha }^{2}}+{{\beta }^{2}}\] has least value for\[a=1\]. This is because we have only one stationary value at which we have minima. |
| Hence\[a=1\]. |
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