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Need solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (xxvi)

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Answer: 7.036

Hint: Here we use this below formula

        \Delta y=f(x+\Delta x)-f(x)

Given: \sqrt{49.5}

Solution: \text { let } f(x)=\sqrt{x} \quad \text { where } x=49

\text { Let } \Delta x=0.5

f(x+\Delta x)=\sqrt{x+\Delta x}=\sqrt{49.5}

Now by definition approximately we can write

\begin{aligned} &f{}'(x)=\frac{f(x+\Delta x)-f(x)}{\Delta x} \\\\ &f(x)=\sqrt{x}=\sqrt{49}=7 \\\\ &\Delta x=0.5 \\\\ &\Rightarrow f(x)=\frac{1}{2 \sqrt{x}}=\frac{1}{2 \sqrt{49}}=\frac{1}{14} \end{aligned}

putting values in (i) we get,

\begin{aligned} &\frac{1}{14}=\frac{\sqrt{495}-7}{0.5} \\\\ &\sqrt{49.5}=\frac{0.5}{14}+7=\frac{05+98}{14}=\frac{485}{14} \\\\ &=7.036 \end{aligned}

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