If f: [R{37}→[R−{37} is given by f(x)=3x+57x−3, then the statement which is not true, is

f(x)=3x+57x−3Let f−1(x)=y⇒ x=f(y)⇒ x=3y+57y−3⇒ 7xy−3x=3y+5⇒ 7xy−3y=3x+5⇒ y(7x−3)=3x+5⇒ y=3x+57x−3∴ f−1(x)=f(x)Let fof(x)=f(f(x))=f(3x+57x−3)=3(3x+57x−3)+57(3x+57x−3)−3⇒9x+15+35x−1521x+35−21x+9⇒44x44=xLet fofof(x)=f(fof(x))=f(x)

If f: [R{37}→[R−{37} is given by f(x)=3x+57x−3, then the statement which is not true, is

f(x)=3x+57x−3Let f−1(x)=y⇒ x=f(y)⇒ x=3y+57y−3⇒ 7xy−3x=3y+5⇒ 7xy−3y=3x+5⇒ y(7x−3)=3x+5⇒ y=3x+57x−3∴ f−1(x)=f(x)Let fof(x)=f(f(x))=f(3x+57x−3)=3(3x+57x−3)+57(3x+57x−3)−3⇒9x+15+35x−1521x+35−21x+9⇒44x44=xLet fofof(x)=f(fof(x))=f(x)

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$I f f : [R {\frac{3}{7}} \to [R - {\frac{3}{7}} i s g i v e n b y f (x) = \frac{3 x + 5}{7 x - 3}, t h e n t h e s t a t e m e n t w h i c h i s n o t t r u e, i s$

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

$(f o f) (x) = x$

$(f o f o f) (x) = x$

$(f o f o f o f) (x) = x$

answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{3 x + 5}{7 x - 3} \\ L e t f^{- 1} (x) = y \Rightarrow x = f (y) \\ \Rightarrow x = \frac{3 y + 5}{7 y - 3} \\ \Rightarrow 7 x y - 3 x = 3 y + 5 \\ \Rightarrow 7 x y - 3 y = 3 x + 5 \\ \Rightarrow y (7 x - 3) = 3 x + 5 \\ \Rightarrow y = \frac{3 x + 5}{7 x - 3} \\ ∴ f^{- 1} (x) = f (x) \\ L e t f o f (x) = f (f (x)) = f (\frac{3 x + 5}{7 x - 3}) \\ = \frac{3 (\frac{3 x + 5}{7 x - 3}) + 5}{7 (\frac{3 x + 5}{7 x - 3}) - 3} \Rightarrow \frac{9 x + 15 + 35 x - 15}{21 x + 35 - 21 x + 9} \Rightarrow \frac{44 x}{44} = x \\ L e t f o f o f (x) = f (f o f (x)) = f (x) \end{array}$