Let f:R o R and g:R o R be two functions defined by f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1 and g(x) = {{1 - 2{e^{2x}}} \over {{e^x}}}. Then, for which of the following range of \alpha, the inequality f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} ight)} ight) f\left( {g\left( {\al

Question

Let f:R 	o R and g:R 	o R be two functions defined by f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1 and g(x) = {{1 - 2{e^{2x}}} \over {{e^x}}}. Then, for which of the following range of \alpha, the inequality f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} ight)} ight) > f\left( {g\left( {\alpha -{5 \over 3}} ight)} ight) holds ?

Accepted Answer

Answer: A. (2, 3) f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1 f'(x) = {{2x} \over {{x^2} + 1}} + {e^{ - x}} = {2 \over {x + {1 \over x}}} + {e^{ - x}} > 0\,\,\forall x \in R g(x) = {e^{ - x}} - 2{e^x} g'(x) - - {e^{ - x}} - 2{e^x} \Rightarrow f(x) is increasing and g(x) is decreasing function. f ( {g ( {{{{{(\alpha - 1)}^2}} \over 3}} )} ) > f ( {g ( {\alpha - {5 \over 3}} )} ) \Rightarrow {{{{(\alpha - 1)}^2}} \over 3} = {\alpha ^2} - 5\alpha + 6 = (\alpha - 2)(\alpha - 3) = \alpha \in (2,\,3)

Answer

Explanation

More JEE Math questions