A) \[\sqrt{1+4{{x}^{2}}}\]
B) \[\sqrt{1+2{{x}^{2}}}\]
C) \[\sqrt{1+{{x}^{2}}}\]
D) \[\sqrt{1+\frac{{{x}^{2}}}{2}}\]
Correct Answer: C
Solution :
Given, \[y=\frac{{{x}^{2}}}{2}\] |
\[\therefore \]\[\frac{dy}{dx}=\frac{2x}{2}=x\] |
Or \[\tan \theta =x\] |
Now \[\frac{\sin 90{}^\circ }{\sin \theta }=\mu \] |
Or \[\mu =\frac{1}{\left( \frac{1}{\sqrt{1+{{x}^{2}}}} \right)}=\sqrt{1+{{x}^{2}}}\] |
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