A) A prime number
B) An irrational number
C) Has non-zero fractional part
D) None of these
Correct Answer: B
Solution :
| [b] We have, |
| \[{{\left\{ {{3}^{{{\log }_{3}}\sqrt{{{25}^{x-1}}+7}}}+{{3}^{-1/8{{\log }_{3}}\left( {{5}^{x-1}}+1 \right)}} \right\}}^{10}}\] |
| \[={{\left[ \sqrt{{{25}^{x-1}}+7}+{{\left( {{5}^{x-1}}+1 \right)}^{-1/8}} \right]}^{10}}\] |
| (since \[{{a}^{{{\log }_{a}}N}}=N\]) |
| Here, \[{{T}_{9}}=180\] |
| \[\Rightarrow {{\,}^{10}}{{C}_{8}}{{\left\{ \sqrt{{{25}^{x-1}}+7} \right\}}^{10-8}}{{\left\{ {{\left( {{5}^{x-1}}+1 \right)}^{-1/8}} \right\}}^{8}}=180\] |
| \[\Rightarrow {{\,}^{10}}{{C}_{8}}\left( {{25}^{x-1}}+7 \right){{\left( {{5}^{x-1}}+1 \right)}^{-1}}=180\] |
| \[\Rightarrow 45\frac{\left( {{25}^{x-1}}+7 \right)}{{{5}^{x-1}}+1}=180\Rightarrow \frac{{{25}^{x-1}}+7}{{{5}^{x-1}}+1}=4\] |
| \[\Rightarrow \frac{{{y}^{2}}+7}{y+1}=4,\] where \[y={{5}^{x-1}}\Rightarrow {{y}^{2}}-4y+3=0\] |
| \[\Rightarrow y=3,1\Rightarrow {{5}^{x-1}}=3\] or \[{{5}^{x-1}}=1\] |
| \[\Rightarrow {{5}^{x}}=15\] or \[{{5}^{x}}=5\Rightarrow x={{\log }_{5}}15\] or \[x=1\] |
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