Question
If $$f\left( x \right) = \cos \left\{ {\frac{\pi }{2}\left[ x \right] - {x^3}} \right\},\,1 < x < 2,$$ and $$\left[ x \right]$$ the greatest integer $$ \leqslant x,$$ then $$f'\left( {\root 3 \of {\frac{\pi }{2}} } \right)$$ is equal to :
A.
0
B.
$$3{\left( {\frac{\pi }{2}} \right)^{\frac{2}{3}}}$$
C.
$$ - 3{\left( {\frac{\pi }{2}} \right)^{\frac{3}{2}}}$$
D.
none of these
Answer :
0
Solution :
$$\eqalign{
& {\text{Around}}\,{\text{ }}x = \root 3 \of {\frac{\pi }{2}} ,\,\left[ x \right] = 1. \cr
& {\text{So, }}\,f\left( x \right) = \cos \left\{ {\frac{\pi }{2} - {x^3}} \right\} = \sin \,{x^3}\,{\text{around }}x = \root 3 \of {\frac{\pi }{2}} \cr
& \therefore f'\left( x \right) = 3{x^2}\cos \,{x^3} \cr
& \therefore f'\left( {\root 3 \of {\frac{\pi }{2}} } \right) = 3.{\left( {\frac{\pi }{2}} \right)^{\frac{2}{3}}}.\cos \frac{\pi }{2} = 0 \cr} $$