Question

Answer the following by appropriately matching the lists based on the information given in Column I and Column II
​​​​​​$\begin{array}{cc}Column1& Column2\\ \text{a.}f:R\to [\frac{3\pi }{4},\pi )\text{and}f\left(x\right)={\cot }^{-1}(2x-x^2-2),& \\ \text{then}f\text{is}& \text{p. one-one}\\ \text{b.}f:R\to R\text{and}f\left(x\right)=e^{px}\sin qx& \\ \text{where}p,q\in R^{+},\text{then}f\text{is}& \text{q. into}\\ \text{c.}f:R^{+}\to [4,\infty )\text{and}f\left(x\right)=4+3x^2,& \\ \text{then}f\text{is}& \text{r. many-one}\\ \text{d.}f:R\to R\text{and}f\left(f\right(x\left)\right)=x,\forall x\in R& \\ \text{then}f\text{is}& \text{s. onto}\end{array}$

• A. $a-p,r;b-q,r;c-p,s;d-q,s$
• B. $a-s,r;b-r,s;c-p,q;d-p,s$
• C. $a-q,r;b-p,s;c-p,r;d-q,s$
• D. $a-p,q;b-q,s;c-r,s;d-p,r$

Answer 1

The correct option is B $a-s,r;b-r,s;c-p,q;d-p,s$

$a.$
$f\left(x\right)={\cot }^{-1}(2x-x^2-2)={\cot }^{-1}(-1-(x-1{)}^2)\le -1$
$\therefore f\left(0\right)=f\left(2\right)$
Hence, $f\left(x\right)$ is many-one. Thus,
${\cot }^{-1}(2x-x^2-2)\in [\frac{3\pi }{4},\pi )$
Hence, $f\left(x\right)$ is onto.
Also, $f\left(3\right)=f(-1)$. Hence, function is many-one.

$\text{b.}$

Clearly, from the graph, $f\left(x\right)$ is many-one and onto.

$\text{c.}$

Clearly, from the graph, $f\left(x\right)$ is one-one and into.

$\text{d.}$
Let $X=\{x_1,x_2,x_3,\dots ,x_n\}$
Let $f\left(x_1\right)=x_2$
or, $f\left(f\right(x_1\left)\right)=f\left(x_2\right)=x_1$
Thus, $f\left(x\right)$ is one-one and onto.