A)
(i) (ii) (iii) \[\pm \sqrt{35}\] \[\pm 6\] 849
B)
(i) (ii) (iii) \[\pm \sqrt{23}\] \[\pm 5\] 833
C)
(i) (ii) (iii) \[\pm \sqrt{33}\] \[\pm 5\] 833
D)
(i) (ii) (iii) \[\pm \sqrt{29}\] \[\pm 3\] 849
Correct Answer: C
Solution :
(i)\[{{(x+y)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy\] \[\therefore {{(x+y)}^{2}}=29+2\times 2\] [Using \[{{x}^{2}}+{{y}^{2}}=29\] and\[xy=2\]]. \[{{(x+y)}^{2}}=29+4=33\] \[x+y=\pm \sqrt{33}\] (ii) \[{{(x-y)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy\] \[\therefore {{(x-y)}^{2}}=29-(2\times 2)\] \[\Rightarrow {{(x-y)}^{2}}=29-4=25\] \[\Rightarrow x-y=\pm \sqrt{25}\Rightarrow x-y=\pm 5\]. (iii) \[{{x}^{2}}+{{y}^{2}}=29\] ?(i) Squaring both sides, we have \[{{({{x}^{2}})}^{2}}+{{({{y}^{2}})}^{2}}+2{{x}^{2}}{{y}^{2}}={{(29)}^{2}}\] \[\Rightarrow {{x}^{4}}+{{y}^{4}}+2{{(2)}^{2}}={{(29)}^{2}}\] \[[\because xy=2]\] \[\Rightarrow {{x}^{4}}+{{y}^{4}}=833\]
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