A) \[2\]
B) \[\frac{7}{8}\]
C) \[\frac{5}{4}\]
D) \[2\]
Correct Answer: C
Solution :
Let \[y=\cos \theta -\cos 2\theta \] \[\Rightarrow \] \[y'=-\sin \theta +2\sin 2\theta \] \[\Rightarrow \] \[y''=-\cos \theta +4\cos \,2\theta \] For maxima or minima, put \[y'=0\] \[\Rightarrow \] \[2\sin \,2\theta =\sin \theta \] \[\Rightarrow \] \[4\,sin\,\theta \,\cos \,\theta =\sin \theta \] \[\Rightarrow \] \[\cos \theta =\frac{1}{4}\] and \[\sin \theta =0\] \[\Rightarrow \] \[\theta ={{\cos }^{-1}}\left( \frac{1}{4} \right),\theta ={{0}^{o}}\] At \[\theta ={{\cos }^{-1}}\left( \frac{1}{4} \right),y''<0,\] maximum \[\therefore \] Maximum value, \[{{y}_{\max }}={{(\cos \,\theta -2{{\cos }^{2}}\theta +1)}_{\theta ={{\cos }^{-1}}_{(1/4)}}}\] \[=\frac{1}{4}-2\left( \frac{1}{16} \right)+1\] \[=\frac{5}{4}-\frac{1}{8}\] \[=\frac{9}{8}\]
You need to login to perform this action.
You will be redirected in
3 sec
