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 \[{{(0.007)}^{1/3}}=\frac{23}{120}\]

"> 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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The approximate value of \[{{(0.007)}^{1/3}}\]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 10 months ago

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[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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Posted on 10 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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