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Question
Let
\[f : R \to R\] be given by \[f\left( x \right) = x^2 - 3\] Then, \[f^{- 1}\] is given by
Options
\[\sqrt{x + 3}\]
\[\sqrt{x} + 3\]
\[x + \sqrt{3}\]
None of these
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Solution
(d)
\[\text{Let} f^{- 1} \left( x \right) = y\]
\[f\left( y \right) = x\]
\[ y^2 - 3 = x\]
\[ y^2 = x + 3\]
\[y = \pm \sqrt{x + 3}\]
So, the answer is (d).
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ASSERTION (A): The relation f : {1, 2, 3, 4} `rightarrow` {x, y, z, p} defined by f = {(1, x), (2, y), (3, z)} is a bijective function.
REASON (R): The function f : {1, 2, 3} `rightarrow` {x, y, z, p} such that f = {(1, x), (2, y), (3, z)} is one-one.
