Let S (x) = \int {dx}/{e^x + 8e^{-x} + 4e^{-3x}} , R (x) = \int {dx}/{e^{3x}+8e^x+4e^{-x} and M (x) = S (x) - 2R (x). If M (x) = 1/2 \tan^{-1} (f (x)) + c where c is an arbitrary constant then f(\log_e^2) =

JEE Mains 2020 Model Paper 2 New Pattern

Question : 6 of 75

Marks: +1, -0

Let S (x) = ∫

ex+8e−x+4e−3x

, R (x) = ∫

e3x+8ex+4e−x

and M (x) = S (x) - 2R (x). If M (x) =

tan−1 (f (x)) + c where c is an arbitrary constant then f(loge2) =

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