Let $x$ denote the total number of one-one functions from a set $A$ with $3$ elements to a set $B$ with $5$ elements,and $y$ denote the total number of one-one functions from the set $A$ to the set $A \times B$. Then ...... .
- A$y=273x$
- B$2y=91x$
- C$y=91x$
- D$2y=273x$
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Let $X$ be a set with exactly $5$ elements and $Y$ be a set with exactly $7$ elements. If $\alpha$ is the number of one-one functions from $X$ to $Y$ and $\beta$ is the number of onto functions from $Y$ to $X$,then the value of $\frac{1}{5!}(\beta-\alpha)$ is.
DifficultJEE MAIN 2018
View SolutionMatch the following:
$(A)$ $f: R \rightarrow R$ is such that $f(x)=px+q$ $(p \neq 0)$,$\forall x \in R$ $I.$ $f$ is neither one-one nor onto $(B)$ $f: R \rightarrow R^{+} \cup\{0\}$ is such that $f(x)=x^2$,$\forall x \in R$ $II.$ $f$ is both one-one and onto $(C)$ $f: N \rightarrow N$ is such that $f(n)=n^2+2n+3$,$\forall n \in N$ $III.$ $f$ is one-one but not onto $(D)$ $f: R \rightarrow R$ is such that $f(x)=2(\cos ^2 5x+\sin ^2 5x)$ $\forall x \in R$ $IV.$ $f$ is onto but not one-one $V.$ $f$ is a constant function and also a bijection
| $(A)$ $f: R \rightarrow R$ is such that $f(x)=px+q$ $(p \neq 0)$,$\forall x \in R$ | $I.$ $f$ is neither one-one nor onto |
| $(B)$ $f: R \rightarrow R^{+} \cup\{0\}$ is such that $f(x)=x^2$,$\forall x \in R$ | $II.$ $f$ is both one-one and onto |
| $(C)$ $f: N \rightarrow N$ is such that $f(n)=n^2+2n+3$,$\forall n \in N$ | $III.$ $f$ is one-one but not onto |
| $(D)$ $f: R \rightarrow R$ is such that $f(x)=2(\cos ^2 5x+\sin ^2 5x)$ $\forall x \in R$ | $IV.$ $f$ is onto but not one-one |
| $V.$ $f$ is a constant function and also a bijection |
Let $f(x) = \begin{cases} -a & \text{if } -a \leq x \leq 0 \\ x+a & \text{if } 0 < x \leq a \end{cases}$ where $a > 0$ and $g(x) = \frac{f(|x|) - |f(x)|}{2}$. Then the function $g: [-a, a] \rightarrow [-a, a]$ is
DifficultJEE MAIN 2024
View SolutionIf $f: R \rightarrow R$ is defined by $f(x) = x^2 + 3x + 4$,then the function $f$ is . . . . . . .
$f:[-2,2] \rightarrow[-2,2]$ 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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