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Subjects
Class 12MathematicsRelations and FunctionsOne-one functionHard
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For the function (f:\mathbb{R}\to\mathbb{R}), (f(x)=\begin{cases}x+3,&x<0\x^2+3,&x\geq0\end{cases}). Choose the correct option.

Answer and explanation

Correct answer: It is one-one

Step 1: For (x<0), the range of (x+3) is less than (3). Step 2: For (x\geq0), the range of (x^2+3) is at least (3). Step 3: The two ranges do not overlap and both pieces are one-one on their domains, so the whole function is one-one.

Related tags

One-One-FunctionPiecewise-FunctionRange-ComparisonInjectiveOne One FunctionRelations And FunctionsMathematicsClass 12 Mcq
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10 more MCQs
  1. 1Which statement is true for a one-one function?
  2. 2If (f(2)=5) and (f(3)=5), what can be said about (f)?
  3. 3If (f:\mathbb{R}\to\mathbb{R}) and (f(x)=2x+1), what type of function is (f)?
  4. 4If (f:\mathbb{R}\to\mathbb{R}) and (f(x)=7), what type of function is (f)?
  5. 5If (f:\mathbb{R}\to[0,\infty)) and (f(x)=|x|), what type of function is (f)?
  6. 6If (f:\mathbb{R}\to(0,\infty)) and (f(x)=e^x), what type of function is (f)?
  7. 7If (f:(0,\infty)\to\mathbb{R}) and (f(x)=\log x), which statement about (f) is correct?
  8. 8If (f:\mathbb{R}\to\mathbb{R}) and (f(x)=3x-5), what follows from (f(a)=f(b))?
  9. 9Which example proves that (f(x)=x^2+1) is not one-one on real numbers?
  10. 10If (f:\mathbb{N}\to\mathbb{N}) and (f(n)=n+4), what type of function is (f)?
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