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:
- A$A-IV, B-V, C-III, D-I$
- B$A-V, B-IV, C-I, D-III$
- C$A-IV, B-V, C-II, D-I$
- D$A-III, B-V, C-IV, D-I$
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Similar Questions
If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} x+4 & \text{for } x < -4 \\ 3x+2 & \text{for } -4 \leq x < 4 \\ x-4 & \text{for } x \geq 4 \end{cases}$ then the correct matching of List-$I$ from List-$II$ is:
List-$I$ List-$II$ $(A)$ $f(-5) + f(-4)$ $(i)$ $14$ $(B)$ $f(|f(-8)|)$ $(ii)$ $4$ $(C)$ $f(f(-7) + f(3))$ $(iii)$ $-11$ $(D)$ $f(f(f(f(0)))) + 1$ $(iv)$ $-1$ $(v)$ $1$ $(vi)$ $0$
| List-$I$ | List-$II$ |
| $(A)$ $f(-5) + f(-4)$ | $(i)$ $14$ |
| $(B)$ $f(|f(-8)|)$ | $(ii)$ $4$ |
| $(C)$ $f(f(-7) + f(3))$ | $(iii)$ $-11$ |
| $(D)$ $f(f(f(f(0)))) + 1$ | $(iv)$ $-1$ |
| $(v)$ $1$ | |
| $(vi)$ $0$ |
DifficultAP EAMCET 2006
View Solution$f: X \rightarrow R$, where $X = \{x \mid 0 < x < 1\}$, is defined as $f(x) = \frac{2x-1}{1-|2x-1|}$. Then:
MediumWBJEE 2022
View SolutionLet $[\cdot]$ denote the greatest integer function. If $f(x) = [x]$ and $g(x) = 3[\frac{x}{3}]$, then the set of all real $x$ such that $f(x) = g(x)$ is
$A$ function $f: R - \{ 0 \} \rightarrow R$ is defined as $f(x) = \begin{cases} x^2 + 3x - 7, & x > 0 \\ h(x), & x < 0 \end{cases}$ If $f(x)$ is an odd function, then $h(x) =$
$f: N \rightarrow N, f(x) = x^3$ is . . . . . . .
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