A) 3
B) 4
C) 5
D) none of these
Correct Answer: B
Solution :
Let\[f(\theta )=12\sin \theta -9{{\sin }^{2}}\theta \] On differentiating w.r.t. \[\theta ,\] we get \[f(\theta )=12\cos \theta =18\sin \theta \cos \theta \] \[=12\cos \theta -9\sin 2\theta \] For maximum or minimum, put \[f(\theta )=0.\] \[12\cos \theta -18\sin \theta \cos \theta =0\] \[\Rightarrow \]\[6\cos \theta (2-3sin\theta )=0\] \[\Rightarrow \]\[\cos \theta =0\]or \[\sin \theta =\frac{2}{3}\] \[\Rightarrow \]\[\theta =\frac{\pi }{2}\]or \[\theta ={{\sin }^{-1}}\frac{2}{3}\] \[f(\theta )=-12\sin \theta -18\cos 2\theta \] At \[\theta =\frac{\pi }{2}\] \[f(\theta )=-12-18(-1)\] \[=6>0,\]minimum At \[\sin \theta =\frac{2}{3}\] \[f(\theta )=-12\left( \frac{2}{3} \right)-18\left( 1-2{{\left( \frac{2}{3} \right)}^{2}} \right)\] \[=-8-18\left( \frac{9-8}{9} \right)\] \[=-10<0,\]maximum \[\therefore \] The maximum value of \[f(\theta )\]at \[\theta ={{\sin }^{-1}}\frac{2}{3}\]is \[f(\theta )=12\left( \frac{2}{3} \right)-9{{\left( \frac{2}{3} \right)}^{2}}\] \[=8-4=4\] Alternate Solution: Let \[f(\theta )=12\sin \theta -9{{\sin }^{2}}\theta \] \[\Rightarrow \] \[f(\theta )=-(9si{{n}^{2}}\theta -12\sin \theta +4)+4\] \[=4-{{(3sin\theta -2)}^{2}}\le 4\] \[\therefore \] Maximum value of \[12\sin \theta -9{{\sin }^{2}}\theta \] is 4. Note: The extreme value of a function is sequentialy maxima and minima.
You need to login to perform this action.
You will be redirected in
3 sec
