(D) Given $g(x)=1, \forall x \in R$ and the signum function $f(x)= \begin{cases}-1, & x 0\end{cases}$.For $x < 0$,$f(x)=-1$,

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$g(x)= \begin{cases}f(x)+2, & x 0\end{cases}$

(D) Given $g(x)=1, \forall x \in R$ and the signum function $f(x)= \begin{cases}-1, & x 0\end{cases}$.For $x For $x=0$,$f(x)=0$,so $g(x)=1 = 1+0 = 1+f(x)$.For $x>0$,$f(x)=1$,so $g(x)=1 = 0+1 = 0+f(x) = f(x)$.Thus,$g(x)= \begin{cases}f(x)+2, & x 0\end{cases}$.

Class 12 Mathematics Relations and Functions Type of Functions based on Mapping

Easy

If $f(x)$ is the signum function,then in terms of $f(x)$,the constant function $g(x)=1, \forall x \in R$ is

AP EAMCET 2023 All AP EAMCET

A
$g(x)= \begin{cases}2-f(x), & x < 0 \\ f(x), & x \geq 0\end{cases}$
B
$g(x)= \begin{cases}f(x)+f(-x), & x < 0 \\ f(x) f(-x), & x \geq 0\end{cases}$
C
$g(x)= \begin{cases}1+f(x), & x>0 \\ 1-f(x), & x \leq 0\end{cases}$
D
$g(x)= \begin{cases}f(x)+2, & x < 0 \\ 1+f(x), & x=0 \\ f(x), & x>0\end{cases}$

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Class 12

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Relations and Functions

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Type of Functions based on Mapping

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AP EAMCET 2023 Paper All AP EAMCET years →

Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be defined by $f(x) = x^3 + 2$. Then,$f$ is . . . . . . .

EasyGUJCET 2024

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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-$II$

$(A)$ $f(-5) + f(-4)$ $(i)$ $14$

$(B)$ $f(|f(-8)|)$ $(ii)$ $4$

$(C)$ $f(f(-7) + f(3))$ $(iii)$ $-11$

$(D)$ $f(f(f(f(0)))) + 1$ $(iv)$ $-1$

$(v)$ $1$

$(vi)$ $0$

DifficultAP EAMCET 2006

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Consider the identity function $I_{N}: N \rightarrow N$ defined as $I_{N}(x) = x$ for all $x \in N$. Show that although $I_{N}$ is onto,the function $I_{N} + I_{N}: N \rightarrow N$ defined as $(I_{N} + I_{N})(x) = I_{N}(x) + I_{N}(x) = x + x = 2x$ is not onto.

Easy

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The function $f: C \rightarrow C$ defined by $f(x) = \frac{ax + b}{cx + d}$ for $x \in C$,where $ad - bc \neq 0$,reduces to a constant function if:

MediumTS EAMCET 2005

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Which of the following functions is injective but not surjective?

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List-$I$	List-$II$
$(A)$ $f(-5) + f(-4)$	$(i)$ $14$
$(B)$ $f(\|f(-8)\|)$	$(ii)$ $4$
$(C)$ $f(f(-7) + f(3))$	$(iii)$ $-11$
$(D)$ $f(f(f(f(0)))) + 1$	$(iv)$ $-1$
	$(v)$ $1$
	$(vi)$ $0$

If $f(x)$ is the signum function,then in terms of $f(x)$,the constant function $g(x)=1, \forall x \in R$ is

Similar Questions

Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be defined by $f(x) = x^3 + 2$. Then,$f$ is . . . . . . .

Consider the identity function $I_{N}: N \rightarrow N$ defined as $I_{N}(x) = x$ for all $x \in N$. Show that although $I_{N}$ is onto,the function $I_{N} + I_{N}: N \rightarrow N$ defined as $(I_{N} + I_{N})(x) = I_{N}(x) + I_{N}(x) = x + x = 2x$ is not onto.

The function $f: C \rightarrow C$ defined by $f(x) = \frac{ax + b}{cx + d}$ for $x \in C$,where $ad - bc \neq 0$,reduces to a constant function if:

Which of the following functions is injective but not surjective?

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