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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