MHT CET · Maths · Application of Derivatives
The approximate value of \((66)^{\frac{1}{3}}\) is
- A \(4.0416\)
- B \(4.0447\)
- C \(4.0433\)
- D \(4.0481\)
Answer & Solution
Correct Answer
(A) \(4.0416\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{a}=64, \mathrm{~h}=2\) and let \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{\frac{1}{3}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{3} \mathrm{x}^{-\frac{2}{3}}=\frac{1}{3 \mathrm{x}^{\frac{2}{3}}}\)
\(\therefore \mathrm{f}(\mathrm{a})=64^{\frac{1}{3}}=4 \quad\) and \(\quad \mathrm{f}^{\prime}(\mathrm{a})=\frac{1}{3(64)^{\frac{2}{3}}}=\frac{1}{3(16)}=\frac{1}{48}\)
We have \(\mathrm{f}(\mathrm{a}+\mathrm{h}) \neq \mathrm{f}(\mathrm{a})+\mathrm{h} \cdot \mathrm{f}^{\prime}(\mathrm{a})\)
\(\therefore(66)^{\frac{1}{3}} \div 4+(2)\left(\frac{1}{48}\right) \div 4+\frac{1}{24} \div 4.0416\)
\(\therefore \mathrm{f}(\mathrm{a})=64^{\frac{1}{3}}=4 \quad\) and \(\quad \mathrm{f}^{\prime}(\mathrm{a})=\frac{1}{3(64)^{\frac{2}{3}}}=\frac{1}{3(16)}=\frac{1}{48}\)
We have \(\mathrm{f}(\mathrm{a}+\mathrm{h}) \neq \mathrm{f}(\mathrm{a})+\mathrm{h} \cdot \mathrm{f}^{\prime}(\mathrm{a})\)
\(\therefore(66)^{\frac{1}{3}} \div 4+(2)\left(\frac{1}{48}\right) \div 4+\frac{1}{24} \div 4.0416\)
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