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Need solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise 13.1 question 11

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Answer: 34.495

Hint: Here we use this below formula

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

Given:  x^{3}-7 x^{2}+15

Solution:  x=5 \Delta x=0.001

⇒ then we have,

\begin{aligned} &f(5.001)=f(x+\Delta x)=(x+\Delta x)^{2}-7(x+\Delta x)^{2} \\ &=f 15 \end{aligned}

Now,   \Delta y=f(x+\Delta x)-f(x)

\begin{aligned} &\therefore f(x+\Delta x)=f(x)+\Delta y \\\\ &=f(x)+f(x) \Delta x \\\\ &\Rightarrow f(5.001)=\left(x^{3}-7 x^{2}+15\right)+\left(3 x^{2}-14 x\right) \Delta x \\\\ &=\left[(5)^{3}-7(5)^{2}+15\right]+\left[3(5)^{2}-14(5)\right](0.001) \end{aligned}

\begin{aligned} &=-35+(s)(0.001) \\\\ &=-35+0.005 \\\\ &=-34.995 \end{aligned}

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