मराठी

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 (A) \[1 + \Sqrt{1

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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

पर्याय

  • \[1 + \sqrt{1 - x}\]

  • \[1 - \sqrt{1 - x}\]

  • \[\sqrt{1 - x}\]

  • \[1 \pm \sqrt{1 - x}\]

MCQ
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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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पाठ 2: Functions - Exercise 2.6 [पृष्ठ ७८]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 2 Functions
Exercise 2.6 | Q 35 | पृष्ठ ७८

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