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Class 11MathematicsRelations and FunctionsFunctions as a special kind of relationHard
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If (f:\mathbb{R}\to\mathbb{R}), (f(x)=x^2), why is (f^{-1}) as a relation not a function from (\mathbb{R}) to (\mathbb{R})?

Answer and explanation

Correct answer: Because (1) has preimages (1) and (-1)

In the inverse relation, (1) is related to both (1) and (-1). Two images for one input make the inverse relation not a function.

Related tags

Inverse RelationSquare FunctionHardFunctions As A Special Kind Of RelationRelations And FunctionsMathematicsClass 11 Mcq
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10 more MCQs
  1. 1When is a relation from (A) to (B) called a function?
  2. 2If \(A=\{1,2\}\) and \(B=\{3,4\}\), which of the following relations from \(A\) to \(B\) is a function?
  3. 3Why is R = {(1,2), (1,3), (2,4)} not a function from A = {1,2} to B = {2,3,4}?
  4. 4If (A={a,b,c}) and (R={(a,1),(b,1),(c,2)}), which statement about (R) is correct?
  5. 5Which of the following statements is always true for a function?
  6. 6If \(f:A\to B\) is a function, what is \(B\) called?
  7. 7If \(g(x)=2x\) and \(A=\{1,3,5\}\), what is the value of \(g(3)\)?
  8. 8Which mapping idea correctly identifies a function?
  9. 9If a relation from the set A={1,2,3} to the set B={a,b} is given by 1→a, 2→a, and 3→b, what type of relation is it?
  10. 10If A = {1, 2, 3} and R = {(1, 2), (2, 3)} is a relation, why is R not a function from A to A?
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