If $f(x) = \begin{cases} [x] & \text{if } -3 < x \leq -1 \\ |x| & \text{if } -1 < x < 1 \\ |[x]| & \text{if } 1 \leq x < 3 \end{cases}$,then the set $\{x : f(x) \geq 0\}$ is equal to
- A$(-1, 3)$
- B$[-1, 3)$
- C$(-1, 3]$
- D$[0, 3)$
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Similar Questions
If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} 2x & : x > 3 \\ x^2 & : 1 < x \leq 3 \\ 3x & : x \leq 1 \end{cases}$,then the value of $f(-1) + f(2) + f(4)$ is:
Let $A = \{x, y, z, u\}$ and $B = \{a, b\}$. $A$ function $f: A \rightarrow B$ is selected randomly. The probability that the function is an onto function is
Let $R$ be the set of all real numbers. Let $f: R \rightarrow R$ be a function defined by $f(x) = \begin{cases} 2x-5 & x < -3 \\ x+2 & -3 \leq x < 5 \\ 3x+1 & x \geq 5 \end{cases}$
Match the following:
List-$I$ List-$II$ $(A) f(-5)+f(0)+f(-1)$ $(I) 16$ $(B) f(f(5)+10f(-3))$ $(II) 40$ $(C) f(f(-4))$ $(III) -31$ $(D) f(f(f(1)))$ $(IV) -12$ $(V) 19$
The correct match is:
Match the following:
| List-$I$ | List-$II$ |
| $(A) f(-5)+f(0)+f(-1)$ | $(I) 16$ |
| $(B) f(f(5)+10f(-3))$ | $(II) 40$ |
| $(C) f(f(-4))$ | $(III) -31$ |
| $(D) f(f(f(1)))$ | $(IV) -12$ |
| $(V) 19$ |
The correct match is:
DifficultTS EAMCET 2022
View SolutionLet $A$ and $B$ be sets. Show that $f: A \times B \rightarrow B \times A$ defined by $f(a, b) = (b, a)$ is a bijective function.
Easy
View SolutionThe function $f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right],$ defined as $f(x) = \frac{x}{1 + x^2}$ is
DifficultJEE MAIN 2017
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