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Provide solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise  13.1 question 10

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Answer: 28.21

Hint: Here we use this below formula

f(x+\Delta x)-f(x)=\Delta y

Given:f(2.01), f(x)=4 x^{2}+5 x+2

Solution: f 2.01=f(x+\Delta x)

=4(x+\Delta x)^{2}+5(x+\Delta x)+2

\begin{aligned} &\Delta y=f(x+\Delta x)-f(x) \\\\ &\Rightarrow f(x)+f(x) x \Delta x \\\\ &f(2.01)=\left(4 x^{2}+5 x+2\right)+(8 x+5) \Delta x \end{aligned}

\begin{aligned} &=\left(4(2)^{2}+5((2)+2)[8(2)+5](0.01)\right. \\\\ &=(16+10+2)+(16+5)(0.01) \\\\ &=28+0.21 \\\\ &=28.21 \end{aligned}

Hence, the approximate value of f(2.01) \text { is } 28.21

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