Let f : ( 0 , 1 ) be a function defined by f ( x ) = 1 1 - e - x and g ( x ) = ( f ( - x ) - f ( x ) ) . Consider the following two statements: (I) g is an increasing function in (0, 1) (II) g is one-one in (0, 1) Then,

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Solution

Q.
Let $f : (0, 1) \to ℝ$ be a function defined by $f (x) = \frac{1}{1 - e^{- x}} and g (x) = (f (- x) - f (x)) .$ Consider the following two statements:

(I) $g$ is an increasing function in (0, 1)

(II) $g$ is one-one in (0, 1)

Then, [2023]

1 Only (I) is true

2 Both (I) and (II) are true

3 Neither (I) nor (II) is true

4 Only (II) is true

Ans.
(2)

$f (x) = \frac{e^{x}}{e^{x} - 1}$

$f (- x) = \frac{e^{- x}}{e^{- x} - 1} = \frac{1}{1 - e^{x}} = \frac{- 1}{e^{x} - 1};$ $g (x) = - \frac{(e^{x} + 1)}{e^{x} - 1}$

$g' (x) = - [\frac{(e^{x} - 1) e^{x} - (e^{x} + 1) e^{x}}{{(e^{x} - 1)}^{2}}] = \frac{2 e^{x}}{{(e^{x} - 1)}^{2}} > 0 \forall x \in (0, 1)$

$Hence, g is an increasing function in (0, 1) and is also one-one.$

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