Question

Use differential to approximate $\sqrt{101}$.

• A. 10
• B. 10.05
• C. 11
• D. Non of these

Answer 1

The correct option is B 10.05

Let $f\left(x\right)=y=\sqrt{x}$
Then, $y+\Delta y=\sqrt{x+\Delta x}$
$\therefore \Delta y=\sqrt{x+\Delta x}-\sqrt{x}$ .....(i)
$\Delta y=f^{\prime }\left(x\right).\Delta x$
$\Rightarrow \sqrt{x+\Delta x}-\sqrt{x}=f^{\prime }\left(x\right).\Delta x$
$\Rightarrow \sqrt{x+\Delta x}-\sqrt{x}=\frac{1}{2\sqrt{x}}.\Delta x$
Put $x=100$, $\Delta x=1$
$\sqrt{101}=10+\frac{1}{20}\times 1=10+0.05=10.05$