A) \[\text{log}\left( \text{1+x} \right)\text{}\frac{\text{x}}{\text{1+x}}\text{x}\]
B) \[\frac{\text{x}}{\text{1+x}}\text{xlog}\left( 1+x \right)\]
C) \[\text{xlog}\left( \text{1+x} \right)\text{}\frac{\text{x}}{\text{1+x}}\]
D) \[\frac{\text{x}}{\text{1+x}}\text{log}\left( \text{1+x} \right)\text{x}\]
Correct Answer: D
Solution :
Let \[{{\omega }^{2}}=\frac{2qE}{mL}\] \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{mL}{2qE}}\] \[\frac{T}{4}.\] \[\text{t=}\frac{\text{T}}{\text{4}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\sqrt{\frac{\text{mL}}{\text{2qE}}}\] \[=1-\frac{3}{5}\] \[\text{250Jk}{{\text{g}}^{\text{-1}}}{{\text{K}}^{\text{-1}}}\] Which is positive [ \[\Delta \theta =0.4K\] x > 0] \[=\frac{2}{5}\] f(x) is monotonic increasing, when x > 0. \[C=\frac{10}{2\times {{10}^{4}}}=5\times {{10}^{-4}}\] f(x) > 0(f) Now, f(0) = log 1 ? 0 = 0 \[\therefore \] f(x) > 0 \[C=500\mu F\] \[{{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}=3\mu F\] \[{{\varepsilon }_{r}}\] \[C=\frac{K{{\varepsilon }_{0}}A}{d}=15\mu F\] ....(1) Also, for x > 0, \[{{V}_{A}}:{{V}_{N}}={{10}^{15}}:1\] \[{{T}_{1/2}}=4.47\times {{10}^{8}}yr\] \[\frac{N}{{{N}_{0}}}=\frac{60}{100}={{\left( \frac{1}{2} \right)}^{n}}\] \[\Rightarrow \] x(x + 1) > x \[{{2}^{n}}=\frac{10}{6}\] \[\Rightarrow \] ...(ii) From Eqs. (i) and (ii), we get \[n\times 0.3=1-0.778=0.22\] [ \[n=\frac{0.222}{0.3}=0.74\] log(1+x) < x for x > 0]
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