A)
(i) (ii) 9/2048 7/2048
B)
(i) (ii) 3/1024 5/2048
C)
(i) (ii) -3/1024 -10/2048
D)
(i) (ii) -9/2048 -15/1024
Correct Answer: D
Solution :
We have, \[a={{2}^{-2}}-{{2}^{-3}}\] \[=\frac{1}{{{2}^{2}}}-\frac{1}{{{2}^{3}}}=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}\] Also, \[b={{2}^{-3}}-{{2}^{-4}}\] \[=\frac{1}{{{2}^{3}}}-\frac{1}{{{2}^{4}}}=\frac{1}{8}-\frac{1}{16}=\frac{1}{16}\] And \[c={{2}^{-4}}-{{2}^{-2}}\] \[=\frac{1}{{{2}^{4}}}-\frac{1}{{{2}^{2}}}=\frac{1}{16}-\frac{1}{4}=\frac{-3}{16}\] (i) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( \frac{1}{8} \right)}^{3}}+{{\left( \frac{1}{16} \right)}^{3}}+{{\left( \frac{-3}{16} \right)}^{3}}\] \[=\frac{1}{512}+\frac{1}{4096}-\frac{27}{4096}=\frac{8+1-27}{4096}\] \[=\frac{-18}{4096}=\frac{-9}{2048}\] (ii) \[10abc=10\left( \frac{1}{8} \right)\left( \frac{1}{16} \right)\left( \frac{-3}{16} \right)=-\frac{15}{1024}\]
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