A) \[A\to 3;B\to 2;C\to 1\]
B) \[A\to 2;B\to 1;C\to 3\]
C) \[A\to 1;B\to 2;C\to 3\]
D) None of these
Correct Answer: B
Solution :
(A) \[{{a}^{m}}.{{a}^{n}}={{a}^{mn}}\] \[\Rightarrow \] \[{{a}^{m+n}}={{a}^{mn}}\] \[\Rightarrow \] \[m+n=mn\] \[m(n-2)+n(m-2)-2mn-2(m+n)\] \[=2mn-2mn=0\] (B) \[{{2}^{x}}={{4}^{y}}={{8}^{z}}\Rightarrow {{2}^{x}}={{2}^{2y}}={{2}^{3z}}\] \[\Rightarrow \]\[x=2y=3z\] Consider \[\frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}=\frac{1}{2x}+\frac{1}{2x}+\frac{1}{2x}\] \[=\frac{3}{2x}\] \[\frac{3}{2x}=\frac{24}{7}\Rightarrow x=\frac{7}{16}\] \[\therefore \] \[z=\frac{7}{48}\] (C) \[\frac{{{3}^{12+n}}\times {{9}^{2n-7}}}{{{3}^{5n}}}=\frac{{{3}^{12}}{{.3}^{n}}\times {{3}^{4n}}{{.3}^{-14}}}{{{({{3}^{n}})}^{5}}}\] \[={{3}^{-2}}=\frac{1}{9}\]
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