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If $f : R - \left\{\frac{3}{5}\right\} \to R$ be defined by $f(x) = \frac{3x + 2}{5x - 3}$, then |
$f^{-1}(x) = f(x)$ $f^{-1}(x) = -f(x)$ $(f \circ f)x = -x$ $f^{-1}(x) = \frac{1}{19} f(x)$ |
$f^{-1}(x) = f(x)$ |
The correct answer is Option (1) → $f^{-1}(x) = f(x)$ ## Given that, $f(x) = \frac{3x + 2}{5x - 3}$ Let $y = \frac{3x + 2}{5x - 3}$ $3x + 2 = 5xy - 3y \Rightarrow x(3 - 5y) = -3y - 2$ $x = \frac{3y + 2}{5y - 3} \Rightarrow f^{-1}(x) = \frac{3x + 2}{5x - 3}$ $∴f^{-1}(x) = f(x)$ |
