A) 9
B) 10
C) 11
D) 13
Correct Answer: B
Solution :
\[{{2}^{2x-y}}=32\]\[\Rightarrow \] \[3{{x}^{3}}+4{{x}^{2}}-x+3\] \[\Rightarrow \] \[{{t}^{6}}+3{{t}^{2}}+10\] ?.(1) Again, \[{{2}^{x+y}}=16\Rightarrow {{2}^{x+y}}={{2}^{4}}\] \[\Rightarrow \] \[{{x}^{4}}+2{{x}^{2}}-3x+7\] ?.(2) Adding (1) and (2), we get \[3x=9\] \[\Rightarrow \] \[{{x}^{3}}+{{x}^{2}}-4x+8\] \[\Rightarrow \] \[x=3\] Substituting this value for x in equation (2), we get \[3+y=4\]\[-1\]\[y=1\] Now, \[{{x}^{2}}+{{y}^{2}}={{3}^{2}}+{{1}^{2}}=9+1=10\]You need to login to perform this action.
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