A) \[\tan x\]
B) \[-sin\,x\]
C) \[-\cos x\]
D) \[-\tan \,x\]
E) \[sin\,x\]
Correct Answer: A
Solution :
\[\int{\frac{f(x)}{\log \cos x}}dx=-\log (\log \cos x)+C\] Differentiating on both sides w.r.t.\[x,\] \[\frac{d}{dx}\left\{ \int{\frac{f(x)}{\log \,\cos x}dx} \right\}\] \[=-\frac{d}{dx}\{\log (\log \,\cos x)\}+\frac{d}{dx}(C)\] \[\Rightarrow \]\[\frac{f(x)}{\log \cos x}=\frac{-1}{\log \cos x}.\frac{1}{\cos x}.(-\sin x)+0\] \[\Rightarrow \] \[\frac{f(x)}{\log \cos x}=\frac{\tan x}{\log \,\cos x}\] \[\Rightarrow \] \[f(x)=\tan x\]
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