Question
Let $$f\left( x \right) = 2\,{\sin ^3}x - 3\,{\sin ^2}x + 12\,\sin \,x + 5,\,0 \leqslant x \leqslant \frac{\pi }{2}.$$ Then $$f\left( x \right)$$ is :
A.
decreasing in $$\left[ {0,\,\frac{\pi }{2}} \right]$$
B.
increasing in $$\left[ {0,\,\frac{\pi }{2}} \right]$$
C.
increasing in $$\left[ {0,\,\frac{\pi }{4}} \right]$$ decreasing in $$\left[ {\frac{\pi }{4},\,\frac{\pi }{2}} \right]$$
D.
none of these
Answer :
increasing in $$\left[ {0,\,\frac{\pi }{2}} \right]$$
Solution :
$$\eqalign{
& f'\left( x \right) = 6\,{\sin ^2}x\,\cos \,x - 6\,\sin \,x\,\cos \,x + 12\,\cos \,x \cr
& = 6\,\cos \,x\left\{ {{{\sin }^2}x - \sin \,x + 2} \right\} \cr
& = 6\,\cos \,x\left\{ {{{\left( {\sin \,x - \frac{1}{2}} \right)}^2} + \frac{7}{4}} \right\} \cr
& \therefore \,{\text{in}}\left[ {0,\,\frac{\pi }{2}} \right],\,f'\left( x \right) \geqslant 0. \cr
& {\text{So, }}f\left( x \right){\text{ is increasing in }}\left[ {0,\,\frac{\pi }{2}} \right] \cr} $$