Let f- R-35 - R-35 be defined by fx-3 x-25 x-3- ThenA- f -1 x - f x B- f -1 x - f x C- fof x-xD- f-1x-119 fx

The correct option is A f 1(x)=f(x) Let nbsp;y nbsp;= nbsp;fxy=3x+25x 3⇒5xy 3y=3x+2⇒5xy 3x=3y+2⇒x5y 3=3y+2⇒x=3y+25y 3⇒f 1y=3y+25y 3 nbsp; nbsp; nbsp ...

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Byju's Answer

Standard X

Mathematics

Componendo and Dividendo

Let f: R-35 →...

Question

Let $f : R - (\frac{3}{5}} \to R - (\frac{3}{5}} b e d e f i n e d b y f (x) = \frac{3 x + 2}{5 x - 3} .$
Then

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

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

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f o f (x) = - x

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

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Solution

The correct option is A $f^{- 1} (x) = f (x)$
$Let y = f (x) y = \frac{3 x + 2}{5 x - 3} \Rightarrow 5 xy - 3 y = 3 x + 2 \Rightarrow 5 xy - 3 x = 3 y + 2 \Rightarrow x (5 y - 3) = 3 y + 2 \Rightarrow x = \frac{3 y + 2}{5 y - 3} \Rightarrow f^{- 1} (y) = \frac{3 y + 2}{5 y - 3} [y = f (x) \Rightarrow x = f^{- 1} (y)] \Rightarrow f^{- 1} (x) = \frac{3 x + 2}{5 x - 3} = f (x)$

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Similar questions

Q. Mark the correct alternative in the following question:

Let f : R

-

\{\frac{3}{5}\}

\to

R be defined by f(x) =

\frac{3 x + 2}{5 x - 3}

. Then,

(a) f ^{$-$ l}(x) = f(x) (b) f ^{$-$ 1}(x) =

-

f(x) (c) fof(x) =

-

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

\frac{1}{19}

f(x)

Q. If

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

and

f (x) = \frac{3 x + 1}{5 x - 3}

, then

Q. Let

f : R - (\frac{3}{5}} \to R - (\frac{3}{5}} b e d e f i n e d b y f (x) = \frac{3 x + 2}{5 x - 3} .

Then

Q. Let

f : R \to R

be defined by

f (x) = 3 x - 4

then

f^{- 1} (x)

Q. Let

f : R \to R

be defined by

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

then find

f^{- 1} (x)

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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)y=3x+25x−3⇒5xy−3y=3x+2⇒5xy−3x=3y+2⇒x(5y−3)=3y+2⇒x=3y+25y−3⇒f−1(y)=3y+25y−3 [ y=f(x)⇒x=f−1(y)]⇒f−1(x)=3x+25x−3=f(x)

Let $f : R - (\frac{3}{5}} \to R - (\frac{3}{5}} b e d e f i n e d b y f (x) = \frac{3 x + 2}{5 x - 3} .$
Then