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Please solve RD Sharma class 12 chapter 13 Differentials Errors and Approximations exercise multiple choice question 13 maths textbook solution

Answers (1)

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Answer: (a) 0.32

Hint: Here we use the basic concept of x=x+\Delta x

Given: y=x^{4}-10

Solution:

Let  x=1.99

So  

      \begin{aligned} &x+\Delta x=2 \\ &\Delta x=2-1.99=0.01 \end{aligned}

Let

        \begin{gathered} d x=\Delta x=0.01 \\ y=x^{4}-10 \end{gathered}

Let’s differentiate,

        \begin{aligned} &y=x^{4}-10 \\\\ &\frac{d y}{d x}=4 x^{3} \end{aligned}

        \left(\frac{d y}{d x}\right)_{x=2}=4(2)^{3}=32

\therefore \quad \frac{d y}{d x} \times d x=d y

So,        d y=32(0.01)=0.32(\text { approx })

So, approximate change in y =0.32

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