Question

The approximate value of $\sqrt[5]{33}$ correct to $4$decimal

• A. $2.0000$
• B. $2.1001$
• C. $2.0125$
• D. $2.0500$

Answer 1

The correct option is C

$2.0125$

Find the approximate value $\sqrt[5]{33}$ :

Let

$$\begin{gathered} f\left(x\right)=x^{\frac{1}{5}} \\ ={(32+1)}^{\frac{1}{5}}\left[\because x=33\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{1}{5}{x}^{\frac{1}{5}-1} \\ =\frac{1}{5}{x}^{\frac{-4}{5}} \\ =\frac{1}{5x^{\frac{4}{5}}} \end{gathered}$$

We know that

$f(a+h)=f\left(a\right)+h{f}^{\text{'}}\left(a\right)$

Here, $a=32,h=1$

$$\begin{gathered} {(32+1)}^{\frac{1}{5}}={32}^{\frac{1}{5}}+\frac{1}{5\times {32}^{{\displaystyle \frac{4}{5}}}}\left[\because f\text{'}\left(x\right)=\frac{1}{5x^{\frac{4}{5}}}\right] \\ =2+\frac{1}{80} \\ =2.0125 \end{gathered}$$

Hence, option (C) is the correct answer.