A) \[\frac{2}{3}\]
B) \[\frac{8}{7}\]
C) \[\frac{3}{8}\]
D) \[\frac{9}{4}\]
Correct Answer: D
Solution :
\[f(x)=\cos x+\frac{1}{2}\cos 2x-\frac{1}{3}\cos 3x\] \[\Rightarrow \]\[f(x)=\sin x+\sin 2x-\sin 3x\] Put \[f(x)=0\] \[\Rightarrow \]\[2\sin \frac{3x}{2}\cos \frac{x}{2}=2\sin \frac{3x}{2}\cos \frac{3x}{2}\] \[\Rightarrow \]\[\sin \frac{3x}{2}=0,\cos \frac{3x}{2}=\cos \frac{x}{2}\] \[\Rightarrow \]\[x=\frac{2n\pi }{3},\frac{3x}{2}=2n\pi \pm \frac{x}{2}\] \[\Rightarrow \]\[x=0,\frac{2\pi }{3},x=0,\] At \[x=0,\] \[f(x)=1+\frac{1}{2}-\frac{1}{3}=\frac{7}{6}\] At \[x=\frac{2\pi }{3}\] \[f(x)=-\frac{1}{2}-\frac{1}{4}-\frac{1}{3}\] \[=-\frac{13}{12}\] \[\therefore \]Difference \[=\frac{7}{6}+\frac{13}{12}\] \[=\frac{27}{12}\] \[=\frac{9}{4}\]
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