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Using the concept of differentials, $\sqrt{36.6}$ is approximately equal to: |
6.05 6.01 6.1 6.5 |
6.05 |
The correct answer is Option (1) → 6.05 Let $f(x)=\sqrt{x}$ $f(36)=\sqrt{36}=6$ $Δx=0.6$ so $f(36+0.6)=f(36)+\frac{dy}{dx}Δx$ $=6+\left(\frac{1}{2\sqrt{x}}\right)×0.6$ $[x=36]$ $=6+\frac{1}{12}×0.6=6.05$ |
