question_answer

Let\[P(x)={{x}^{2}}-2({{a}^{2}}+a+1)x+{{a}^{2}}+5a+2\]. If the minimum value of \[P(x)\] for \[x\le 0\] is 8, then the sum of the squares of all possible values of a is

A) \[13\]                    

B)        \[17\]                    

C) \[37\]             

D)        \[49\]

Correct Answer: C

Solution :

   [c] Abscissa of vertex of \[P(x):\,\,\,x=({{a}^{2}}+a+1)>0\] Hence, \[P(x)\] is decreasing in \[(-\infty ,0)\] Therefore, minimum value occurs at \[x=0\]. \[\therefore \,\,\,\,\,\,f(0)=8\] \[\Rightarrow \,\,\,\,\,\,\,{{a}^{2}}+5a+2=8\] \[\Rightarrow \,\,\,\,\,\,\,{{a}^{2}}+5a-6=0\] \[\Rightarrow \,\,\,\,\,\,\,(a+6)(a-1)=0\] \[\therefore \,\,\,\,\,\,a=1\] or \[-6\] \[\therefore \]   Sum of squares of values of \[a={{(1)}^{2}}+{{(-6)}^{2}}\] \[=1+36=37\]