| Statement 1: \[x={{\left[ \frac{2}{3} \right]}^{4}}\div {{\left[ \frac{2}{3} \right]}^{2}},\] then value of \[{{x}^{2}}+2x+3\]is 0. |
| Statement 2: If \[{{\left[ -\frac{1}{2} \right]}^{4}}\times {{(-2)}^{8}}={{(-2)}^{4x}}\] then\[x=1\]. |
A) Only Statement-1
B) Only Statement-2
C) Both Statement-1 and Statement-2
D) Neither Statement-1 nor Statement-2
Correct Answer: B
Solution :
Statement-1: \[x={{\left( \frac{2}{3} \right)}^{4}}\div {{\left( \frac{2}{3} \right)}^{2}}\] \[={{\left( \frac{2}{3} \right)}^{4-2}}={{\left( \frac{2}{3} \right)}^{2}}=\frac{4}{9}\] Now,\[{{x}^{2}}+2x+3={{\left( \frac{4}{9} \right)}^{2}}+2\left( \frac{4}{9} \right)+3=4\frac{7}{81}\ne 0\] \[\therefore \] Statement-1 is false Statement-2: We have, \[{{\left( -\frac{1}{2} \right)}^{4}}\times {{\left( -2 \right)}^{8}}={{\left( -2 \right)}^{4x}}\] \[\Rightarrow \] \[{{(-1)}^{4}}\times {{(2)}^{-4}}\times {{(-1)}^{8}}\times {{(2)}^{8}}={{(-2)}^{4x}}\] \[\Rightarrow \] \[{{(-1)}^{12}}\times {{(2)}^{8-4}}={{(-2)}^{4x}}\Rightarrow {{(2)}^{4}}={{(-2)}^{4x}}\] \[\Rightarrow \] \[{{(-1)}^{4}}\times {{(2)}^{4}}={{(-2)}^{4x}}\Rightarrow {{(-2)}^{4}}={{(-2)}^{4x}}\] On comparing, we get \[4=4x\] \[\Rightarrow \] \[x=1\] \[\therefore \] Statement-2 is true.
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