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

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Application of Derivatives

Let

f : R \to R

and

g : R \to R

be two functions defined by

f (x) = lo g_{e} (x^{2} + 1) - e^{- x} + 1

and

g (x) = \frac{1 - 2 e ^{2 x}}{e ^{x}}

. Then, for which of the following range of

α

, the inequality

f (g (\frac{( α - 1 ) ^{2}}{3})) > f (g (α - \frac{5}{3}))

holds?

Answer: (A)

First, analyze the monotonicity of both functions. For

f (x) = lo g_{e} (x^{2} + 1) - e^{- x} + 1

, compute

f^{'} (x) = \frac{2 x}{x ^{2} + 1} + e^{- x} > 0

for all

x \in R

, so

f

is strictly increasing.

For

g (x) = e^{- x} - 2 e^{x}

, compute

g^{'} (x) = - e^{- x} - 2 e^{x} < 0

for all

x \in R

, so

g

is strictly decreasing.

Since

f

is increasing,

f (g (\frac{( α - 1 ) ^{2}}{3})) > f (g (α - \frac{5}{3}))

implies

g (\frac{( α - 1 ) ^{2}}{3}) > g (α - \frac{5}{3})

.

Since

g

is decreasing, this means

\frac{( α - 1 ) ^{2}}{3} < α - \frac{5}{3}

.

Multiplying by 3:

(α - 1)^{2} < 3 α - 5

Expanding:

α^{2} - 2 α + 1 < 3 α - 5

α^{2} - 5 α + 6 < 0

(α - 2) (α - 3) < 0

Therefore

α \in (2, 3)

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option