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 \[a={{(1)}^{2}}+{{(-6)}^{2}}\] \[=1+36=37\]You need to login to perform this action.
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