A) \[0.2\]
B) \[0.2083\]
C) \[0.0032\]
D) \[0.0083\]
Correct Answer: B
Solution :
Use the relation; \[f(x+\Delta x)=f(x)+\Delta xf'(x);\] to determine the approximate value. Consider, \[f(x)={{(x)}^{1/3}}\,\,\,\Rightarrow \,\,\,\,f'(x)=\frac{1}{3}{{x}^{-2/3}}\] Let \[x=\frac{8}{1000}\] and \[\Delta x=\frac{1}{1000}\] Now, \[f'(x+\Delta x)=f(x)+\Delta xf'(x)\] \[\Rightarrow \] \[{{(x+\Delta x)}^{1/3}}={{x}^{1/3}}+\frac{1}{3{{x}^{2/3}}}\times \Delta x\] \[\Rightarrow \]\[{{\left( \frac{8}{1000}+\frac{1}{1000} \right)}^{1/3}}={{\left( \frac{8}{1000} \right)}^{1/3}}+\frac{\frac{1}{1000}}{3{{\left( \frac{8}{1000} \right)}^{1/3}}}\] \[\Rightarrow \]\[{{\left( \frac{9}{1000} \right)}^{1/3}}=\frac{2}{10}+\frac{1}{1000\times 3\times \frac{2}{10}}\] \[=\frac{1}{5}+\frac{1}{600}=\frac{121}{600}\] \[\therefore \] \[{{(0.009)}^{1/3}}=0.20166\]
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