A) \[2+\sqrt{2}\]
B) \[3\sqrt{2}\]
C) \[2\sqrt{3}\]
D) \[3+\sqrt{2}\]
E) 7
Correct Answer: B
Solution :
Let \[y=2\cos \theta +\frac{1}{\sin \theta }+\sqrt{2}\tan \theta \] \[\frac{dy}{d\theta }=-2\sin \theta -\cos ec\theta \cot \theta +\sqrt{2}{{\sec }^{2}}\theta \] \[=-2\sin \theta -\frac{\cos \theta }{\sin \theta }.\frac{1}{\sin \theta }+\sqrt{2}\frac{1}{{{\cos }^{2}}\theta }\] For extremum, put, \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[\theta =\frac{\pi }{4}\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-2\cos \theta -[-\cos e{{c}^{3}}\theta +{{\cot }^{3}}\cos ec\theta ]\] \[+\sqrt{2}.2\sec \theta \tan \theta \sec \theta \] \[=-2\cos \theta +\cos e{{c}^{3}}\theta -{{\cot }^{2}}\theta \cos ec\theta \] \[+2\sqrt{2}{{\sec }^{2}}\theta \tan \theta \] \[>0\] for\[\theta =\frac{\pi }{4}\] \[\therefore \]y is minimum for \[\theta =\frac{\pi }{4}.\] \[\Rightarrow \]\[\min (y)=2.\frac{1}{\sqrt{2}}+\sqrt{2}+\sqrt{2}(1)\] \[=2\sqrt{2}+\sqrt{2}=3\sqrt{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
