If f(x)=cosπ2x-x3, 1<x<2, and x= the greatest integer ≤x, then f'3π2 is equal to :

f(x)=cos π2[x]-x3 =cos π2(1)-x3 1<x<2 ⇒ [x]=1 f(x)=sin x3 f1(x)=cos x3 (3x)2 f1π23=cos π2·3·π123 =0

If f(x)=cosπ2x-x3, 1<x<2, and x= the greatest integer ≤x, then f'3π2 is equal to :

f(x)=cos π2[x]-x3 =cos π2(1)-x3 1<x<2 ⇒ [x]=1 f(x)=sin x3 f1(x)=cos x3 (3x)2 f1π23=cos π2·3·π123 =0

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$If f (x) = \cos \{\frac{π}{2} [x] - x^{3}\}, 1 < x < 2, and [x] = the greatest integer \leq x, then f' (3 \sqrt{\frac{π}{2}}) is equal to :$

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$3 {(\frac{π}{2})}^{2 / 3}$

$- 3 {(\frac{π}{2})}^{3 / 2}$

answer is A.

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Detailed Solution

$f (x) = \cos \{\frac{π}{2} [x] - x^{3}\} = \cos \{\frac{π}{2} (1) - x^{3}\} (1 < x < 2 \Rightarrow [x] = 1) f (x) = \sin x^{3} f^{1} (x) = \cos x^{3} {(3 x)}^{2} f^{1} (\sqrt[3]{\frac{π}{2}}) = \cos \frac{π}{2} \cdot 3 \cdot {(\frac{π}{1})}^{\frac{2}{3}} = 0$