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

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Answer: 0.57560442

Hint: Here, we use f(x+\Delta y)-f(x)

Given: \cos \left(\frac{11 \pi}{36}\right)

Solution:  Let  x=\frac{12 \pi}{36}=\frac{\pi}{3}

    \begin{aligned} &\text { So, } x+\Delta x=\frac{\pi}{36} \\\\ &\Rightarrow \Delta x=\frac{-\pi}{36}=\frac{-22}{7}=-0.0873 \end{aligned}

Differentiating f(x)

        \frac{d f}{d x}=\frac{d}{d x}(\cos x)

we know

        \begin{aligned} &\frac{d}{d x}(\cos x)=-\sin x \\\\ &\frac{d f}{d x}=-\sin x \end{aligned}

when, x=\frac{\pi}{3}

we have

        \frac{d f}{d x}=-\sin \left(\frac{\pi}{3}\right)

        \begin{aligned} &\Rightarrow\left(\frac{d f}{d x}\right)_{x=\frac{\pi}{3}}=-0.86603 \\\\ &\Rightarrow f(x+\Delta x)-f(x) \\\\ &\Delta y=\left(\frac{d y}{d x}\right) \Delta x \end{aligned}

\begin{aligned} &\text { Here, } \frac{d f}{d x}=-0.86603 \\\\ &\Delta x=0.0873 \\\\ &\Rightarrow \Delta f=0.07560442 \end{aligned}

now, we have

        \begin{aligned} &f\left(\frac{11 \pi}{36}\right)=f\left(\frac{\pi}{3}\right)+\Delta f \\\\ &f\left(\frac{11 \pi}{36}\right)=\cos \left(\frac{\pi}{3}\right)+0.07560442 \end{aligned}

        \begin{aligned} &f\left(\frac{11 \pi}{36}\right)=0.5+0.07560442 \\\\ &f\left(\frac{11 \pi}{36}\right)=0.57560442 \end{aligned}



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