Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Let}f:[-1,\infty )\to (0,\infty )\text{be defined by}f\left(x\right)=e^{x^2+|x|}\text{. Then}f\text{is}& \text{(P)}& \text{one-one}\\ \text{(B)}& \text{Let}f:(1,\infty )\to [3,\infty )\text{be defined by}f\left(x\right)=\sqrt{10-2x+x^2}.\text{Then}f\text{is}& \text{(Q)}& \text{into}\\ \text{(C)}& \text{Let}f:R\to Z\text{be defined by}f\left(x\right)={\tan }^5\pi [x^2+2x+3]\text{where [. ] denotes}& \text{(R)}& \text{many-one}\\ & \text{the greatest integer function. Then}f\text{is}& & \\ \text{(D)}& \text{Let}f:[3,4]\to [4,6]\text{be defined by}f\left(x\right)=|x-1|+|x-2|+|x-3|+|x-4|.& \text{(S)}& \text{onto}\\ & \text{Then}f\text{is}& & \\ & & \text{(T)}& \text{periodic}\end{array}$

Which of the following is a CORRECT combination?

• A. $\text{(A)}\to \text{(Q), (S)};\text{(B)}\to \text{(S), (T)}$
• B. $\text{(A)}\to \text{(Q), (R)};\text{(B)}\to \text{(S), (T)}$
• C. $\text{(A)}\to \text{(Q), (S)};\text{(B)}\to \text{(P), (Q)}$
• D. $\text{(A)}\to \text{(Q), (R)};\text{(B)}\to \text{(P), (Q)}$

Answer 1

The correct option is D $\text{(A)}\to \text{(Q), (R)};\text{(B)}\to \text{(P), (Q)}$

$\left(\text{A}\right)$
We have $f\left(x\right)=e^{x^2+|x|},x\in [-1,\infty )$
Clearly, $f(-1)=e^2=f\left(1\right)$
Also, $0\le x^2+|x|<\infty \forall x\in [-1,\infty )$
$\Rightarrow 1\le e^{x^2+|x|}<\infty$
Hence, range of $f$ is $[1,\infty )$
$\therefore f\left(x\right)$ is many-one, into.
$\text{(A)}\to \text{(Q), (R)}$


$\text{(B)}$
We have, $f\left(x\right)=\sqrt{10-2x+x^2}$
$f\left(x\right)=\sqrt{(x-1{)}^2+9}$
Clearly, range of $f$ is $(3,\infty )$ and also $f\left(x\right)$ is one-one.
$\therefore f\left(x\right)$ is one-one, into.
$\text{(B)}\to \text{(P), (Q)}$