A) \[248\sqrt{2}\]
B) \[280\sqrt{2}\]
C) \[-32\sqrt{2}\]
D) \[-280\sqrt{2}\]
Correct Answer: D
Solution :
\[{{x}^{2}}-4\sqrt{2}kx+2{{e}^{4\ln k}}-1=0\] or\[{{x}^{2}}-4\sqrt{2}kx+2{{k}^{4}}-1=0\] \[\alpha +\beta =4\sqrt{2}k\]and\[\alpha .\beta =2{{k}^{4}}-1\] Squaring both sides, we get \[{{(\alpha +\beta )}^{2}}={{(4\sqrt{2}k)}^{2}}\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta =32{{k}^{2}}\] \[66+2\alpha \beta =32{{k}^{2}}\] \[66+2(2{{k}^{4}}-1)=32{{k}^{2}}\] \[66+4{{k}^{4}}-2=32{{k}^{2}}\Rightarrow 4{{k}^{4}}-32{{k}^{2}}+64=0\] or\[{{k}^{4}}-8{{k}^{2}}+16=0\Rightarrow {{({{k}^{2}})}^{2}}-8{{k}^{2}}+16=0\] \[\Rightarrow \]\[({{k}^{2}}-4)({{k}^{2}}-4)=0\Rightarrow {{k}^{2}}=4,{{k}^{2}}=4\] \[\Rightarrow \]\[k=\pm 2\] Now, \[{{\alpha }^{3}}+{{\beta }^{3}}=(4\sqrt{2}k)[66-(2{{k}^{4}}-1)]\] Putting k = ? 2, (k = +2 cannot be taken because it does not satisfy the above equation) \[\therefore \]\[{{\alpha }^{3}}+{{\beta }^{3}}=(4\sqrt{2}(-2))[66-2{{(-2)}^{4}}-1]\] \[{{\alpha }^{3}}+{{\beta }^{3}}=(-8\sqrt{2}(66-32+1)=(-8\sqrt{2})(35)\] \[\therefore \]\[{{\alpha }^{3}}+{{\beta }^{3}}=-280\sqrt{2}\]
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