A) \[\frac{23}{120}\]
B) \[\frac{27}{120}\]
C) \[\frac{19}{120}\]
D) \[\frac{17}{120}\]
Correct Answer: A
Solution :
[a] Let \[f(x)={{x}^{1/3}}\Rightarrow f'(x)=\frac{1}{3}{{x}^{-2/3}}\] Now \[f(x+\Delta x)-f(x)=f'(x)\cdot \Delta x=\frac{\Delta x}{3({{x}^{2/3}})}\] We may write, \[0.007=0.008-0.001\], taking. \[x=0.008\] and \[dx=-0.001.\] We have \[f(0.007)-f(0.008)=-\frac{0.001}{3{{(0.008)}^{2/3}}}\] \[\Rightarrow f(0.007)-{{(0.008)}^{1/3}}=-\frac{0.001}{3{{(0.2)}^{2}}}\] \[\Rightarrow f(0.007)=0.2-\frac{0.001}{3(0.04)}=0.2-\frac{1}{120}=\frac{23}{120}\] Hence \[{{(0.007)}^{1/3}}=\frac{23}{120}\]
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