Let f-35- -35 bedefined by fx-3x-25x-3- Then

Let nbsp; y=f(x)=3x+25x 3 ∀ x ∈ ℝ {35} ⇒ 5xy 3y=3x+2 ⇒ x=3y+25y 3 ∴ f^ 1(x)=3x+25x 3=f(x) ⇒ f(f(x))=x or fof(x)=x Hence, option nbsp; (a) nbsp; is co ...

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Standard XI

Business Studies

Outsourcing and Its Types

Let f:ℝ-35→ ℝ...

Question

Let $f : R - {\frac{3}{5}} \to R - {\frac{3}{5}}$ be

defined by $f (x) = \frac{3 x + 2}{5 x - 3} .$ Then

f^{- 1} (x) = f (x)

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(f o f) x = - x

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f^{- 1} (x) = - f (x)

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f^{- 1} (x) = \frac{1}{19} f (x)

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Solution

The correct option is A $f^{- 1} (x) = f (x)$
Let $y = f (x) = \frac{3 x + 2}{5 x - 3} \forall x \in R - {\frac{3}{5}}$

$\Rightarrow 5 x y - 3 y = 3 x + 2$

$\Rightarrow x = \frac{3 y + 2}{5 y - 3}$

$∴ f^{- 1} (x) = \frac{3 x + 2}{5 x - 3} = f (x)$

$\Rightarrow f (f (x)) = x o r f o f (x) = x$

Hence, option (a) is correct.

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Similar questions

Q. Show that if

f : R - {\frac{7}{5}} \to R - {\frac{3}{5}}

is defined by

f (x) = \frac{3 x + 4}{5 x - 7}

and

g : R - {\frac{3}{5}} \to R - {\frac{7}{5}}

is defined by

g (x) = \frac{7 x + 4}{5 x - 3},

then

f o g = I_{A}

and

g o f = I_{B},

where

A = R - {\frac{3}{5}}, B = R - {\frac{7}{5}} I_{A} (x) = x, \forall x \in A, I_{B} (x) = x, \forall x \in B

are called identity functions on sets

A

and

B

, respectively.

Q. Let

f : R - {\frac{3}{5}} \to R

be defined by

f (x) = \frac{3 x + 2}{5 x - 3}

. Then,

Let $f : R - {\frac{- 4}{3}} \to R$ be a function defined as $f (x) = \frac{4 x}{3 x + 4}$ The inverse of $f$ is map $g :$ Range $f \to R - {\frac{- 4}{3}}$ given by

Q. Let

f : R \to R

be the function defined by

f (x) = x^{3} + 5.

Then

f^{- 1} (x)

If $f : R \to R$ be given by $f (x) = (3 - x^{3})^{\frac{1}{3}}$ , then $f$ of f $(x)$ is

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Outsourcing and Its Types

Standard XI Business Studies

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Let f:R−{35}→ R−{35} be defined by f(x)=3x+25x−3. Then

The correct option is A f−1(x)=f(x)Let y=f(x)=3x+25x−3 ∀ x ∈ R−{35} ⇒ 5xy−3y=3x+2 ⇒ x=3y+25y−3 ∴ f−1(x)=3x+25x−3=f(x) ⇒ f(f(x))=x or fof(x)=x Hence, option (a) is correct.

Let $f : R - {\frac{3}{5}} \to R - {\frac{3}{5}}$ be

defined by $f (x) = \frac{3 x + 2}{5 x - 3} .$ Then