(B) Step $1$: Let $y = f(x) = \frac{2x - 1}{x + 5}$.Step $2$: Solve for $x$ in terms of $y$: $y(x + 5) = 2x - 1$.Step $3$: Expand and rearrange: $yx +

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$\frac{5x + 1}{2 - x}, x \neq 2$

(B) Step $1$: Let $y = f(x) = \frac{2x - 1}{x + 5}$.Step $2$: Solve for $x$ in terms of $y$: $y(x + 5) = 2x - 1$.Step $3$: Expand and rearrange: $yx + 5y = 2x - 1 \implies yx - 2x = -5y - 1$.Step $4$: Factor out $x$: $x(y - 2) = -(5y + 1) \implies x = \frac{-(5y + 1)}{y - 2} = \frac{5y + 1}{2 - y}$.Step $5$: Replace $y$ with $x$ to get $f^{-1}(x) = \frac{5x + 1}{2 - x}, x \neq 2$.

Class 12 Mathematics Relation and Function Inverse Function

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If $f(x) = \frac{2x - 1}{x + 5}, x \neq -5$ then $f^{-1}(x)$ is equal to

MHT CET 2026 All MHT CET

A
$\frac{x + 5}{2x - 1}, x \neq \frac{1}{2}$
B
$\frac{5x + 1}{2 - x}, x \neq 2$
C
$\frac{5x - 1}{2 - x}, x \neq 2$
D
$\frac{x - 5}{2x + 1}, x \neq -\frac{1}{2}$

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If $f(x) = \frac{2x - 1}{x + 5}, x \neq -5$ then $f^{-1}(x)$ is equal to

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