Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ is defined by,

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}\mathrm{x}+4& \mathrm{for}\mathrm{x}<-4\\ 3\mathrm{x}+2& \mathrm{for}-4\le \mathrm{x}\le 4\\ \mathrm{x}-4& \mathrm{for}\mathrm{x}\ge 4\end{array}\right.$ then the correct matching of List I from List II is

List I	List II
A. $\mathrm{f}(-5)+\mathrm{f}(-4)$	1. 14
B. $\mathrm{f}\left(\left|\mathrm{f}(-8)\right|\right)$	2. 4
C. $\mathrm{f}\left[\mathrm{f}(-7)+\mathrm{f}\left(3\right)\right]$	3. -11
D. $\mathrm{f}\left[\mathrm{f}\left\{\mathrm{f}\left(0\right)\right\}\right]+1$	4. -1
	5. 1
	6. 0

• A. $A=3,B=6,C=2,D=5$
• B. $A=3,B=4,C=2,D=5$
• C. $A=4,B=3,C=2,D=1$
• D. $A=3,B=6,C=5,D=2$

Answer 1

The correct option is A

$A=3,B=6,C=2,D=5$

Explanation for the correct answer:

Step 1: Solving, A. $\mathrm{f}(-5)+\mathrm{f}\left(4\right)$,

$\begin{array}{rcl}\therefore \mathrm{f}(-5)& =& -5+4\left[\because \mathrm{for}\mathrm{x}\ge 4,\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-4\right]\\ & =& -1\end{array}$

$\begin{array}{rcl}\therefore \mathrm{f}(-4)& =& 3(-4)+2\left[\because \mathrm{for}\mathrm{x}=-4,\mathrm{f}\left(\mathrm{x}\right)=3\mathrm{x}+2\right]\\ & =& -12+2\\ & =& -10\end{array}$

$$\begin{gathered} \therefore \mathrm{f}(-5)+\mathrm{f}(-4)=-1-10 \\ =-11 \end{gathered}$$

Step 2: Solving, B.$\mathrm{f}(\left|\mathrm{f}(-8)\right|$,

$\begin{array}{rcl}\therefore \mathrm{f}(\left|\mathrm{f}(-8)\right|& =& \mathrm{f}(-8+4)\\ & =& \mathrm{f}\left(4\right)\end{array}$

$\begin{array}{rcl}\therefore \mathrm{f}\left(4\right)& =& 4-4\left[\because \mathrm{for}\mathrm{x}=4,\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-4\right]\\ & =& 0\end{array}$

Step 3: Solving, C.$\mathrm{f}\left[\mathrm{f}(-7)+\mathrm{f}\left(3\right)\right]$

$\begin{array}{rcl}\therefore \mathrm{f}\left(\mathrm{f}\right(-7\left)\right)+\mathrm{f}\left(3\right)& =& \mathrm{f}(-7+4)+3\left(3\right)+2\\ & =& 3(-3)+2+3\left(3\right)+2\\ & =& -9+2+9+2\\ & =& 4\end{array}$

Step 4: Solving, D. $\mathrm{f}\left(\mathrm{f}\right(\mathrm{f}\left(\mathrm{f}\right(0\left)\right)\left)\right)+1$

$\begin{array}{rcl}\therefore \mathrm{f}\left(\mathrm{f}\right(\mathrm{f}\left(\mathrm{f}\right(0\left)\right)\left)\right)& =& \mathrm{f}\left(\mathrm{f}\right(\mathrm{f}\left(2\right)\left)\right)+1\\ & =& \mathrm{f}\left(\mathrm{f}\right(8\left)\right)+1\\ & =& \mathrm{f}\left(4\right)+1\\ & =& 0+1\\ & =& 1\end{array}$

Therefore, $\mathrm{A}=\mathrm{option}3\mathrm{in}\mathrm{List}\mathrm{II},\mathrm{B}=\mathrm{option}6\mathrm{in}\mathrm{List}\mathrm{II},\mathrm{C}=\mathrm{option}2\mathrm{in}\mathrm{List}\mathrm{II},\mathrm{D}=\mathrm{option}5\mathrm{in}\mathrm{List}\mathrm{II}$

Hence, the correct answer is an option (A).