$A$ 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 $h(x) =$
- A$x^2 + 3x + 7$
- B$x^2 + 3x - 7$
- C$-x^2 + 3x + 7$
- D$-x^2 - 3x + 7$
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Similar Questions
Prove that the Greatest Integer Function $f: R \rightarrow R$,given by $f(x)=[x]$,is neither one-one nor onto,where $[x]$ denotes the greatest integer less than or equal to $x$.
Medium
View SolutionIf $f(x) = \begin{cases} [x] & \text{if } -3 < x \leq -1 \\ |x| & \text{if } -1 < x < 1 \\ |[x]| & \text{if } 1 \leq x < 3 \end{cases}$,then the set $\{x : f(x) \geq 0\}$ is equal to
$f(x)=\frac{x}{e^x-1}+\frac{x}{2}+2 \cos ^3 \frac{x}{2}$ on $R-\{0\}$ is
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$f:[-2,2] \rightarrow[-2,2]$ and $g:[-2,2] \rightarrow[0,4]$ are two functions defined as $f(x)=\begin{cases} -2, & -2 \leq x \leq 0 \\ x^2-2, & 0 \leq x \leq 2 \end{cases}$ and $g(x)=|f(x)|+f(|x|)$, then
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