A) 1
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
We have, \[\sqrt{3{{x}^{2}}+6x+12}+\sqrt{5{{x}^{2}}+10x+9}\] \[=4-2x-{{x}^{2}}\] |
\[3{{x}^{2}}+6x+12=3{{\left( x-1 \right)}^{2}}+9\] |
\[\therefore \,\,\,\,\,\,\,\,\,\,\,\sqrt{3{{x}^{2}}+6x+12}\ge 3\] |
\[5{{x}^{2}}+10x+9=5{{\left( x+1 \right)}^{2}}+4\] |
\[\sqrt{5{{x}^{2}}+10x+9}\ge 2\] |
\[\therefore \,\,\,\,\,\,\,\,\,\,\,LHS\ge 3+2=5\] |
\[RHS\,\,\,\,\,\,\,4-2x-{{x}^{2}}=5-{{\left( x+1 \right)}^{2}}\] |
\[RHS\,\le 5\] |
\[\therefore \] Only equality is possible. |
\[\therefore \] Equality is possible if \[x=-1\] |
\[\therefore \] \[x=-1\]is the only solutions. |
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