A) \[\frac{5}{12}\]
B) \[\frac{1}{4}\]
C) \[\frac{-1}{12}\]
D) \[\frac{1}{12}\]
Correct Answer: D
Solution :
Given, \[{{f}_{k}}(x)=\frac{1}{k}(si{{n}^{k}}x+co{{s}^{k}}x)\] Now.\[{{f}_{4}}(k)-{{f}_{6}}(k)\] \[=\frac{1}{4}(si{{n}^{4}}x+co{{s}^{4}}x)-\frac{1}{6}(si{{n}^{6}}x+co{{s}^{6}}x)\] \[=\frac{1}{4}\left( 1-\frac{1}{2}{{\sin }^{2}}2x \right)-\frac{1}{6}\left( 1-\frac{3}{4}{{\sin }^{2}}2x \right)\] \[=\frac{1}{4}-\frac{1}{4}{{\sin }^{2}}2x-\frac{1}{6}+\frac{1}{8}{{\sin }^{2}}2x=\frac{1}{12}\]
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