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-$II$ |
| $(A)$ $f(-5) + f(-4)$ | $(i)$ $14$ |
| $(B)$ $f(|f(-8)|)$ | $(ii)$ $4$ |
| $(C)$ $f(f(-7) + f(3))$ | $(iii)$ $-11$ |
| $(D)$ $f(f(f(f(0)))) + 1$ | $(iv)$ $-1$ |
| $(v)$ $1$ | |
| $(vi)$ $0$ |
- A
- B
- C
- D
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| $A$. $f$ is one-one and onto,if | $1$. $A = R^{+}, B = R$ |
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DifficultTS EAMCET 2010
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