$Let f (x) = \{\begin{array}{l} x^{3} - 1, x < 2 \\ x^{2} + 3, x \geq 2 \end{array} . Then$

Moderate

A

$f^{- 1} (x) = \{\begin{cases} (x + 1)^{1 / 3}, x < 2 \\ (x - 3)^{1 / 2}, x \geq 2 \end{cases}$
B

$f^{- 1} (x) = \{\begin{cases} (x + 1)^{1 / 3}, x < 7 \\ (x - 3)^{1 / 2}, x \geq 7 \end{cases}$
C

$f^{- 1} (x) = \{\begin{cases} (x + 1)^{1 / 3}, x < 1 \\ (x - 3)^{1 / 2}, x \geq 7 \end{cases}$
D

$f^{- 1} (x) does not exist$

Solution

$f (x) = \{\begin{cases} x^{3} - 1, x < 2 \\ x^{2} + 3, x \geq 2 \end{cases}$
$\begin{array}{l} For f (x) = x^{3} - 1, x < 2, f^{- 1} (x) = (x + 1)^{1 / 3}, x < 7 \\ (as x < 2 \Rightarrow x^{3} < 8 \Rightarrow x^{3} - 1 < 7) \\ For f (x) = x^{2} + 3, x \geq 2, f^{- 1} (x) = (x - 1)^{1 / 2}, x \geq 7 \\ (as x \geq 2 \Rightarrow x^{2} \geq 4 \Rightarrow x^{2} + 3 \geq 7) \end{array}$

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