A) \[f(\theta )-f''(\theta )+C\]
B) \[f(\theta )+f''(\theta )+C\]
C) \[f'(\theta )+f''(\theta )+C\]
D) \[f'(\theta )-f''(\theta )+C\]
Correct Answer: B
Solution :
Given, \[u=-f''(\theta ).\sin \theta +f'(\theta )\cos \theta \] and \[v=f''(\theta )\cos \theta +f'(\theta )\sin \theta \] On differentiating w.r.t.\[\theta \]respectively, we get \[\frac{du}{d\theta }=-f''\,'(\theta ).\sin \theta -f''(\theta )\cos \theta \] \[+f''\,'(\theta ).\cos \theta -f''(\theta )\sin \theta \] \[=-f'\,'\,'(\theta )\sin \theta -f'(\theta )\sin \theta \] and \[\frac{dv}{d\theta }=f'\,'\,'(\theta ).\cos \theta -f'\,'(\theta ).\sin \theta \] \[+f'\,'(\theta )\sin \theta +f'\,'(\theta ).\cos \theta \] \[=f'\,'\,'(\theta ).\cos \theta +f'(\theta )\cos \theta \] \[\therefore \]\[{{\left( \frac{du}{d\theta } \right)}^{2}}+{{\left( \frac{dv}{d\theta } \right)}^{2}}={{[f'\,'\,'(\theta )]}^{2}}\] \[+{{[f'(\theta )]}^{2}}+2f'(\theta ).f'\,'\,'(\theta )\] \[={{[f'\,'\,'(\theta )+f'(\theta )]}^{2}}\] Now, \[{{\int{\left[ {{\left( \frac{du}{d\theta } \right)}^{2}}+{{\left( \frac{dv}{d\theta } \right)}^{2}} \right]}}^{1/2}}d\theta \] \[=\int{[f'\,'\,'(\theta )+f'(\theta )]}\,d\theta \] \[=f'\,'(\theta )+f(\theta )+C\]
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