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Provide solution for RD Sharma maths class 12 chapter 13 Differentials Errors and Approximations exercise  very short answer question 6

Answers (1)

Answer: F(2.1)=9.2

Hint:  Here we use this below formula,

                F(x+\Delta x)=F(x)+F^{\prime}(x) \Delta x

Given:

                F(x)=x^{4}-10

Solution:

We have,

            F(x)=x^{4}-10

So,       F^{\prime}(x)=4 x^{3}

and   x=2, \Delta x=0.1

The let’s put the value in formula

F(2+0.1)=F(2)+F^{\prime}(2) \Delta x

\begin{aligned} &=2^{4}-10+4(2)^{3}(0.1) \\\\ &=16-10+(32)(0.1) \\\\ &=16-10+3.2 \\\\ &=9.2 \end{aligned}

\mathrm{So}_{,} \; F(2.1) \text { is } 9.2

 

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