Question

The inverse of the funtion f : R → (-1, 3) is given by
f (x) = eˣ - e⁻ˣ/eˣ + e⁻ˣ + 2
(a) log (x - 1/x + 1)⁻² (b) log ( x - 2/x - 1)¹/²
(c) log (x/2 - x)¹/² (d) log (x - 1/3 - x)¹/²

Answer 1

Answer:

$\large\boxed{\sf{(d)\:log{( \dfrac{x - 1}{3 - x}) }^{ \frac{1}{2} }}}$

Step-by-step explanation:

Given a function such that f : R → (-1, 3) is given by,

$f(x) = \dfrac{ {e}^{x} - {e}^{ - x} }{ {e}^{x} + {e}^{ - x} } + 2$

Now, to find the inverse of f(x),

Let's assume f(x) = y

Therefore, we will get,

$= > y = \dfrac{ {e}^{x} - \frac{1}{ {e}^{x} } }{ {e}^{x} + \frac{1}{ {e}^{x} } } + 2 \\ \\ = > y = \dfrac{ {e}^{2x} - 1 }{ {e}^{2x} + 1} + 2 \\ \\ = > y = \dfrac{ {e}^{2x} - 1 + 2 {e}^{2x} + 2}{ {e}^{2x} + 1} \\ \\ = > y = \frac{3 {e}^{2x} + 1} { {e}^{2x} + 1 } \\ \\ = > y( {e}^{2x} + 1) = 3 {e}^{2x} + 1 \\ \\ = > y {e}^{2x} + y = 3 {e}^{2x} + 1 \\ \\ = > y {e}^{2x} - 3 {e}^{2x} = 1 - y \\ \\ = > {e}^{2x} (y - 3) = 1 - y \\ \\ = > {e}^{2x} = \dfrac{1 - y}{y - 3} \\ \\ = > {e}^{2x} = \dfrac{y - 1}{3 - y} \\ \\ = > 2x = log( \dfrac{y - 1}{3 - y} ) \\ \\ = > x = \dfrac{1}{2} log( \dfrac{y - 1}{ 3 - y} ) \\ \\ = > x = log {( \dfrac{y - 1}{3 - y} )}^{ \frac{1}{2} }$

Now, replacing x with y and y with x, we get,

$= > y = log{( \dfrac{x - 1}{3 - x}) }^{ \frac{1}{2} }$

Hence, the correct option is $\bold{(d)\:log{( \dfrac{x - 1}{3 - x}) }^{ \frac{1}{2} } }$