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  • Please solve RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (xii) maths textbook solution

Please solve RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (xii) maths textbook solution

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Answer: 0.1996

Hint: Here we use  f(x+\Delta x)-f(x)

Given: \frac{1}{\sqrt{25.1}}

Solution:  let us assume,

f(x)=\frac{1}{\sqrt{25.1}}

⇒ Also let  x=25 \text { so } \Delta x=0.1

⇒ Differentiating f(x) with respect to x

        \begin{aligned} &\frac{d f}{d x}=\frac{1}{d x}\left(\frac{1}{\sqrt{x}}\right) \Rightarrow \frac{d f}{d x}=\frac{d}{d x} \frac{1}{x^{\frac{1}{2}}} \\\\ &\Rightarrow \frac{d f}{d x}=\frac{-1}{2 x^{\frac{3}{2}}} \end{aligned}

⇒ When x = 25 we have

        \left(\frac{d f}{d x}\right)_{x=25}=\frac{-1}{2(25)^{\frac{3}{2}}}

        \begin{aligned} &\left(\frac{d f}{d x}\right)_{x=25}=\frac{-1}{2(125)}=\frac{-1}{250} \\\\ &\frac{d f}{d x}=\frac{-1}{250}=-0.004 \end{aligned}

        \begin{aligned} &\Rightarrow \Delta y=\left(\frac{d y}{d x}\right) \Delta x \\\\ &\Delta f=-0.0004 \\\\ &\Rightarrow f(25.1)=\frac{1}{\sqrt{25}}=-0.004 \end{aligned}

        \begin{aligned} &f(25.1)=\frac{1}{5}=0.0004 \\\\ &f(25.1)=0.2-0.0004 \\\\ &f(25.1)=0.1996 \\\\ &\frac{1}{\sqrt{25.1}}=0.1996 \end{aligned}

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