42. Let f:R−{−43}→R be a function defined as f(x)=4x3x+4 . The inverse of f is map g Range f→R−{−43} g(y)=3y3−4y g(y)=4y4−3y g(y)=4y3−4y g(y)=3y4−3y

It is given that f:R−{−43}→R is defined as f(x)=4x3x−4 Let y be an arbitrary element of Range f. Then, there exists x∈R−{−43} such that y=f(x) ⇒y=4x3x+4 ⇒3xy+4y=4x let us define g: Range f→R−{−43}as,g(y)=4y4−3y ⇒x(4−3y)=4yNow,(gof)(x)=g(f(x))=g(4x3x+4)⇒x=4y4−3y=4(4x3x+4)4−3(4x3x+4)=16x12x+16−12x=16x16=xAnd,(fog)(y)=f(g(y))=f(4y4−3y)=4(4y4−3y)3(4y4−3y)+4=16y12y+16−12y=16y16=y∴gof=IR−(−43)and,fog=I_"Range,f" Thus, g is the inverse of f i.e., f−1=g. Hence, the inverse of f is the map g: Range f→R−{−43} which is given by g(y)=4y4−3y The correct answer is B.

42. 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 $f \to R - {- \frac{4}{3}}$

$g (y) = \frac{3 y}{3 - 4 y}$

$g (y) = \frac{4 y}{4 - 3 y}$

$g (y) = \frac{4 y}{3 - 4 y}$

$g (y) = \frac{3 y}{4 - 3 y}$

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Vishal Baghel | Contributor-Level 10

9 months ago

It is given that $f : R - {- \frac{4}{3}} \to R$ is defined as $f (x) = \frac{4 x}{3 x - 4}$

Let y be an arbitrary element of Range f.

Then, there exists $x \in R - {- \frac{4}{3}}$ such that $y = f (x)$

$\Rightarrow y = \frac{4 x}{3 x + 4}$

$\Rightarrow 3 x y + 4 y = 4 x$ let us define g: Range $f \to R - {- \frac{4}{3}} a s, g (y) = \frac{4 y}{4 - 3 y}$

$\begin{array}{l} \Rightarrow x (4 - 3 y) = 4 y \\ N o w, (g o f) (x) = g (f (x)) = g (\frac{4 x}{3 x + 4}) \\ \Rightarrow x = \frac{4 y}{4 - 3 y} \\ = \frac{4 (4 \frac{x}{3 x + 4})}{4 - 3 (\frac{4 x}{3 x + 4})} = \frac{16 x}{12 x + 16 - 12 x} = \frac{16 x}{16} = x \\ A n d, (f o g) (y) = f (g (y)) = f (\frac{4 y}{4 - 3 y}) \\ = \frac{4 (\frac{4 y}{4 - 3 y})}{3 (\frac{4 y}{4 - 3 y}) + 4} = \frac{16 y}{12 y + 16 - 12 y} = \frac{16 y}{16} = y \\ ∴ g o f = I_{R - (- \frac{4}{3})} a n d, f o g = I_" R a n g e, f " \end{array}$

Thus, g is the inverse of f i.e., $f^{- 1} = g .$

Hence, the inverse of f is the map g: Range $f \to R - {- \frac{4}{3}}$ which is given by

$g (y) = \frac{4 y}{4 - 3 y}$

The correct answer is B.

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