Let f- - be defined by fx-ex-e-x2-The inverse of the given function is-

Let nbsp; y=f(x)=(e^x e^ x)/2 ⇒ 2y=e^x e^ x ⇒ e^2x 2ye^x 1=0 ⇒ e^x=2y±√(4y^2+4)2=y±√(y^2+1) ⇒ e^x=y+ √(y^2+1) (∵ nbsp;y √(y^2+1) lt;0 for all y but e^x i ...

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Standard XII

Mathematics

Higher Order Equations

Let f:ℝ→ ℝ be...

Question

Let $f : R \to R$ be defined by $f (x) = \frac{(e^{x} - e^{- x})}{2}$ .
The inverse of the given function is:

f^{- 1} (x) = {log}_{e} (x + \sqrt{x^{2} + 1})

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f^{- 1} (x) = {log}_{e} (x - \sqrt{x^{2} + 1})

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f^{- 1} (x) = {log}_{e} (x + \sqrt{x^{2} - 1})

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f^{- 1} (x) = {log}_{e} (x - \sqrt{x^{2} - 1})

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Solution

The correct option is A $f^{- 1} (x) = {log}_{e} (x + \sqrt{x^{2} + 1})$
Let $y = f (x) = (e^{x} - e^{- x}) / 2$
$\Rightarrow 2 y = e^{x} - e^{- x}$
$\Rightarrow e^{2 x} - 2 y e^{x} - 1 = 0$
$\Rightarrow e^{x} = \frac{2 y \pm \sqrt{4 y^{2} + 4}}{2} = y \pm \sqrt{y^{2} + 1}$
$\Rightarrow e^{x} = y + \sqrt{y^{2} + 1}$
$(∵ y - \sqrt{y^{2} + 1} < 0$ for all $y$ but $e^{x}$ is always positive $)$
$\Rightarrow x = {log}_{e} (y + \sqrt{y^{2} + 1})$
$∴ f^{- 1} (x) = {log}_{e} (x + \sqrt{x^{2} + 1})$

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Higher Order Equations

Standard XII Mathematics

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Let f:R→R be defined by f(x)=(ex−e−x)2. The inverse of the given function is:

The correct option is A f−1(x)=loge(x+√x2+1)Let y=f(x)=(ex−e−x)/2 ⇒2y=ex−e−x ⇒e2x−2yex−1=0 ⇒ex=2y±√4y2+42=y±√y2+1 ⇒ex=y+√y2+1 (∵y−√y2+1<0 for all y but ex is always positive) ⇒x=loge(y+√y2+1) ∴f−1(x)=loge(x+√x2+1)

Let $f : R \to R$ be defined by $f (x) = \frac{(e^{x} - e^{- x})}{2}$ .
The inverse of the given function is: