मराठी

Let a = { X : − 1 ≤ X ≤ 1 } and F : a → a Such that F ( X ) = X | X | (A) a Bijection (B) Injective but Not Surjective (C) Surjective but Not Injective (D) Neither Injective Nor Surjective

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प्रश्न

Let

\[A = \left\{ x : - 1 \leq x \leq 1 \right\} \text{and} f : A \to \text{A such that f}\left( x \right) = x|x|\]

 

पर्याय

  • a bijection

  • injective but not surjective

  • surjective but not injective

  • neither injective nor surjective

MCQ
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उत्तर

Injectivity:
Let x and y be any two elements in the domain A.
Case-1: Let x and y be two positive numbers, such that\[f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow x\left| x \right| = y\left| y \right|\] 
\[ \Rightarrow x\left( x \right) = y\left( y \right)\] 
\[ \Rightarrow x^2 = y^2 \] 
\[ \Rightarrow x = y\] 

Case-2: Let x and y be two negative numbers, such that
\[f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow x\left| x \right| = y\left| y \right|\] 
\[ \Rightarrow x\left( - x \right) = y\left( - y \right)\] 
\[ \Rightarrow - x^2 = - y^2 \] 
\[ \Rightarrow x^2 = y^2 \] 
\[ \Rightarrow x = y\] 
Case-3: Let be positive and y be negative.
\[\text{Then},x \neq y\] 
\[ \Rightarrow f\left( x \right) = x\left| x \right| \text{is positive and }f\left( y \right) = y\left| y \right| \text{is negative}\] 
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\] 
\[So, x \neq y\] 
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\]
So, f is one-one.
Surjectivity:
Let y be an element in the co-domain, such that y = f (x)
\[Case-1: \text{Let } y>0. \text{Then}, 0<y\leq1\] 
\[y = f\left( x \right) = x\left| x \right| > 0\] 
\[ \Rightarrow x > 0\] 
\[ \Rightarrow \left| x \right| = x\] 
\[ \Rightarrow f\left( x \right) = y\] 
\[ \Rightarrow x\left| x \right| = y\] 
\[ \Rightarrow x\left( x \right) = y\] 
\[ \Rightarrow x^2 = y\] 
\[ \Rightarrow x = \sqrt{y} \in A \left( \text{We do not get}\pm, \text{as } x > 0 \right)\] \[\] \[Case-2: Lety<0.\text{Then},-1\leq y<0\] 
\[y = f\left( x \right) = x\left| x \right| < 0\] 
\[ \Rightarrow x < 0\] 
\[ \Rightarrow \left| x \right| = - x\] 
\[ \Rightarrow f\left( x \right) = y\] 
\[ \Rightarrow x\left| x \right| = y\] 
\[ \Rightarrow x\left( - x \right) = y\] 
\[ \Rightarrow - x^2 = y\] 
\[ \Rightarrow x^2 = - y\] 
\[ \Rightarrow x = - \sqrt{- y} \in A \left( \text{We do not get}\pm, \text{ as } x>0 \right)\]
\[\Rightarrow\]is onto
\[\Rightarrow\] is a bijection.
So, the answer is (a).
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पाठ 2: Functions - Exercise 2.6 [पृष्ठ ७६]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 2 Functions
Exercise 2.6 | Q 16 | पृष्ठ ७६

व्हिडिओ ट्यूटोरियलVIEW ALL [5]

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