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प्रश्न
Let
\[A = \left\{ x \in R : x \leq 1 \right\} and f : A \to A\] be defined as
\[f\left( x \right) = x \left( 2 - x \right)\] Then,
\[f^{- 1} \left( x \right)\] is
विकल्प
\[1 + \sqrt{1 - x}\]
\[1 - \sqrt{1 - x}\]
\[\sqrt{1 - x}\]
\[1 \pm \sqrt{1 - x}\]
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उत्तर
LaTeX
\[\text{Let y be the element in the codomain R such that}\]
\[ f^{- 1} \left( x \right) = y . . . \left( 1 \right)\]
\[ \Rightarrow f\left( y \right) = x and y \leq1 \]
\[ \Rightarrow y\left( 2 - y \right) = x\]
\[ \Rightarrow 2y - y^2 = x\]
\[ \Rightarrow y^2 - 2y + x = 0\]
\[ \Rightarrow y^2 - 2y = - x\]
\[ \Rightarrow y^2 - 2y + 1 = 1 - x\]
\[ \Rightarrow \left( y - 1 \right)^2 = 1 - x\]
\[ \Rightarrow y - 1 = \pm \sqrt{1 - x}\]
\[ \Rightarrow y = 1 \pm \sqrt{1 - x}\]
\[ \Rightarrow y = 1 - \sqrt{1 - x} \left ( \because y \leq1 \right)\]
The correct answer is (b).
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