Functions
Piecewise function h(x) = af+bg+c(x-g)+dg — properties match
MJAT_TS7_P1
Grade 12
Question:
Let $f(x)=\begin{cases}0&x=0\\x\sin(1/x)&x\neq 0\end{cases}$, $g(x)=\begin{cases}0&\text{otherwise}\\\frac{1}{2}-|x-\frac{1}{2}|&0\leq x\leq 1\end{cases}$, $h(x)=af(x)+b(g(x)+g(\frac{1}{2}-x))+c(x-g(x))+dg(x)$. Match P)$a=0,b=1,c=0,d=0$; Q)$a=1,b=0,c=0,d=0$; R)$a=0,b=0,c=1,d=0$; S)$a=0,b=0,c=0,d=1$ with properties 1)$h$ one-one; 2)$h$ onto; 3)$h$ differentiable on $\mathbb{R}$; 4)range$=[0,1]$; 5)range$=\{0,1\}$
A) P→4, Q→3, R→1, S→2
B) P→5, Q→2, R→4, S→3
C) P→5, Q→3, R→2, S→4
D) P→4, Q→2, R→1, S→3
Step-by-Step Solution
Key Concept: P ($b=1$): $h=g(x)+g(1/2-x)$. $g$ is a tent function on $[0,1]$. $g(x)+g(1/2-x)=1/2$ for $x\in[0,1/2]$, constant — not one-one. Range? Check. Q ($a=1$): $h=f(x)=x\sin(1/x)$, differentiable? R ($c=1$): $h=x-g(x)$. S ($d=1$): $h=g(x)$.
Answer: **C**.
Correct Answer: C