| (i) If \[{{m}^{2}}={{27}^{2/3}}\times {{16}^{-3/2}}\]then m = P . |
| (ii) If \[ab=1\]then \[\frac{1}{1+{{a}^{-1}}}+\frac{1}{1+{{b}^{-1}}}=\] Q . |
| (iii) If x = \[({{8}^{2/3}}\cdot {{32}^{-2/5}})\]then \[{{x}^{-5}}=\] R |
A)
P Q R 5/4 0 10/7
B)
P Q R 1 1 5/16
C)
P Q R 3/8 1 1
D)
P Q R 7/8 0 7/8
Correct Answer: C
Solution :
(i) We have, \[{{m}^{2}}={{27}^{2/3}}\times {{16}^{-3/2}}\] \[\Rightarrow {{m}^{2}}={{({{3}^{3}})}^{2/3}}\times {{({{4}^{2}})}^{-3/2}}\Rightarrow {{m}^{2}}={{3}^{2}}\times {{4}^{-3}}\] \[\Rightarrow {{m}^{2}}=\frac{{{3}^{2}}}{{{4}^{3}}}=\frac{9}{64}\Rightarrow m=\frac{3}{8}\] (ii) We have, \[\frac{1}{1+{{a}^{-1}}}+\frac{1}{1+{{b}^{-1}}}\] \[=\frac{1}{1+\frac{1}{a}}+\frac{1}{1+\frac{1}{b}}=\frac{1}{\frac{a+1}{a}}+\frac{1}{\frac{b+1}{b}}\] \[=\frac{a}{a+1}+\frac{b}{b+1}=\frac{a(b+1)+b(a+1)}{(a+1)(b+1)}\] \[=\frac{ab+a+ab+b}{(a+1)(b+1)}=\frac{1+a+1+b}{ab+a+b+1}\] \[(\because ab=1)\] \[=\frac{1+a+1+b}{1+a+b+1}=1\] (iii) We have, \[x={{8}^{2/3}}\cdot {{32}^{-2/5}}\] \[={{({{2}^{3}})}^{2/3}}\cdot {{({{2}^{5}})}^{-2/5}}={{2}^{2}}\cdot {{2}^{-2}}=1\] \[\therefore \] \[{{x}^{-5}}={{(1)}^{-5}}=1\]
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