| Column -I | Column -II |
| (a) If \[\frac{3}{x+8}=\frac{4}{6-x},\]then \[x\]is ______. | (i) 3 |
| (b) if, \[\frac{{{2}^{x-1}}{{.4}^{2x+1}}}{{{8}^{x-1}}}=64,\] | (ii) \[-2\] |
| (c) if \[{{4}^{x}}-{{4}^{x}}^{-1}=24\] | (iii)\[-2\] |
| (d) \[{{4}^{x+1}}=256,\]then \[x\]is | (iv) 1 |
A) \[(a)\to (i);(b)\to (ii);(c)\to (iii);(d)\to (iv)\]
B) \[(a)\to (iii);(b)\to (iv);(c)\to (i);(d)\to (ii)\]
C) \[(a)\to (i);(b)\to (iv);(c)\to (ii);(d)\to (iii)\]
D) \[(a)\to (iii);(b)\to (iv);(c)\to (iii);(d)\to (i)\]
Correct Answer: D
Solution :
We have. \[\frac{3}{x+8}=\frac{4}{6-x}\] \[\Rightarrow \]\[18-3x=4x+32\Rightarrow 7x=-14\Rightarrow x=-2\] (b) We have, \[\frac{{{2}^{x-1}}{{.4}^{2x+1}}}{{{8}^{x-1}}}=64\] \[\Rightarrow \] \[\frac{{{2}^{x}}{{2}^{-1}}{{.4}^{2x}}.4}{{{8}^{x}}{{8}^{-1}}}=64\] \[\Rightarrow \]\[\frac{1}{{{4}^{x}}}{{.4}^{2x}}=64\Rightarrow {{4}^{x}}\Rightarrow \frac{64}{16}=4\] On comparing, we get \[x=1\] (c) we have, \[{{4}^{x-1}}=24\] \[\Rightarrow \]\[{{4}^{x}}\left( 1-\frac{1}{4} \right)=24\Rightarrow {{({{2}^{2}})}^{x}}=\frac{24\times 4}{3}=32\] \[\Rightarrow \]\[{{2}^{2x}}={{2}^{5}}\] On comparing. We get \[2x=5\Rightarrow x=\frac{5}{2}\] Now, \[{{(2x)}^{x}}={{(5)}^{5/2}}\] (d) We have, \[{{4}^{x+1}}=256\] \[\Rightarrow \] \[{{4}^{x+1}}={{(4)}^{4}}\] On comparing. We get \[x+1=4\Rightarrow x=3\]
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