Q57 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics

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

  • A
    $\dfrac{x + 5}{2x - 1}$, $x \neq \dfrac{1}{2}$
  • Correct answer
    B
    $\dfrac{5x + 1}{2 - x}$, $x \neq 2$
  • C
    $\dfrac{5x - 1}{2 - x}$, $x \neq 2$
  • D
    $\dfrac{x - 5}{2x + 1}$, $x \neq -\dfrac{1}{2}$

Solution

Let $y = \dfrac{2x - 1}{x + 5}$

$y(x + 5) = 2x - 1$

$xy + 5y = 2x - 1$

$x(y - 2) = -5y - 1$

$x = \dfrac{5y + 1}{2 - y}$

Therefore, $f^{-1}(x) = \dfrac{5x + 1}{2 - x}$, $x \neq 2$

Answer: $\dfrac{5x + 1}{2 - x}$, $x \neq 2$

Hence, the correct answer is option: 'B'.

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