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Q. The value of $\sqrt{24 . 99}$ is

KCETKCET 2019Application of Derivatives

Solution:

$\sqrt{24.99}$
$\Delta y=f \left(x+\Delta x\right)-f\left(x\right)$
Take $ f\left(x\right)=\sqrt{x}$
$Let \,x = 25$
$\Delta x=-0.01$
$\therefore \, f\left(24.99\right)=\sqrt{25}+\Delta y$
$\approx5-0.001$
$\approx4.999$
$\Delta y \approx\frac{d y}{d x} \times\Delta x$
$\approx\frac{1}{2\sqrt{x}}\times-0.01$
$\approx-\frac{0.01}{10}$