MCQ
${{(0.2)}^{{{\log }_{\sqrt{5}}}\left( \frac{\text{1}}{\text{4}}\,+\,\frac{\text{1}}{\text{8}}\,+\,\frac{\text{1}}{\text{16}}\,+\,.....\,\infty \right)}}$ નું મૂલ્ય:
- A$1$
- B$2$
- C$1/2$
- ✓$4$
$=\,\frac{\frac{1}{4}}{1-\frac{1}{2}}\,=\,\frac{1}{2}\,=\,{{2}^{-1}}$
$\therefore \,{{(0.2)}^{{{\log }_{\sqrt{5}}}\,\left( \frac{1}{4}\,+\,\frac{1}{8}\,+\,\frac{1}{16}\,+\,...\,\infty \right)}}$
$\,=\,{{\left( \frac{1}{5} \right)}^{{{\log }_{\sqrt{5}}}2^{-1}}}$
$=\,{{\left( {{5}^{-1}} \right)}^{\frac{-1}{1/2}{{\log }_{5}}2}}$
$=\,\,{{5}^{2{{\log }_{5}}2}}\,\,\,=\,{{5}^{{{\log }_{5}}{{(2)}^{2}}}}\,=\,{{2}^{2}}\,\,=\,4$
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કારણ ${\rm{(R)}}$ બિંદુઓ ${\rm{ (}}{{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{)}}$ એઅતિવલય ${\rm{ }}\,\,\frac{{{x^2}}}{{{a^2}}}\, - \,\,\frac{{{y^2}}}{{{b^2}}}\, = \,\,1$ ની અંદર આવેલું , તો $\frac{{x_{^1}^2}}{{{a^2}}}\, - \,\,\frac{{y_1^2}}{{{b^2}}}\, - \,\,1\,\, < \,\,0$
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