Piecewise Functions and Properties
DAILY_CHALLENGE
Grade None
Question:
Let $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$ be functions defined by
$$f(x)=\begin{cases}x|x|\sin\!\left(\dfrac{1}{x}\right), & x\neq0,\\ 0, & x=0,\end{cases}\quad\text{and}\quad g(x)=\begin{cases}1-2x, & 0\leq x\leq\dfrac{1}{2},\\ 0, & \text{otherwise.}\end{cases}$$
Let $a,b,c,d\in\mathbb{R}$. Define $h:\mathbb{R}\to\mathbb{R}$ by
$$h(x)=af(x)+b\!\left[g(x)+g\!\left(\tfrac{1}{2}-x\right)\right]+c(x-g(x))+dg(x),\quad x\in\mathbb{R}.$$
Match each entry in List-I to the correct entry in List-II.
**List-I**
(P) If $a=0,b=1,c=0,d=0$, then
(Q) If $a=1,b=0,c=0,d=0$, then
(R) If $a=0,b=0,c=1,d=0$, then
(S) If $a=0,b=0,c=0,d=1$, then
**List-II**
(1) $h$ is one-one.
(2) $h$ is onto.
(3) $h$ is differentiable on $\mathbb{R}$.
(4) the range of $h$ is $[0,1]$.
(5) the range of $h$ is $\{0,1\}$.
(P)→(4) (Q)→(3) (R)→(1) (S)→(2)
(P)→(5) (Q)→(2) (R)→(4) (S)→(3)
(P)→(5) (Q)→(3) (R)→(2) (S)→(4)
(P)→(4) (Q)→(2) (R)→(1) (S)→(3)
Step-by-Step Solution
Key Concept: Analyse each case by computing h explicitly on different intervals; identify range and injectivity/differentiability
(P) $h=g(x)+g(\frac{1}{2}-x)$. For $x\in[0,\frac{1}{2}]$: $g(x)=1-2x$, $g(\frac{1}{2}-x)=1-2(\frac{1}{2}-x)=2x$. Sum $=1$. For $x<0$ or $x>\frac{1}{2}$: both $g$ terms are 0. So $h=1$ on $[0,\frac{1}{2}]$, $h=0$ elsewhere. Range $=\{0,1\}$→(5).
(Q) $h=f(x)=x|x|\sin(1/x)$. At $x=0$: $f'(0)=\lim_{x\to0}|x|\sin(1/x)=0$. For $x\neq0$: $f'(x)=2|x|\sin(1/x)-\cos(1/x)$, which exists. So $h$ is differentiable on $\mathbb{R}$→(3). (Range is all of $\mathbb{R}$, not (2).)
(R) $h=x-g(x)$. For $x\in[0,\frac{1}{2}]$: $h=x-(1-2x)=3x-1\in[-1,\frac{1}{2}]$. For $x<0$ or $x>\frac{1}{2}$: $h=x$. Combined range $=\mathbb{R}$, so $h$ is onto→(2). ($h(-1)=h(0)=-1$ shows not one-one.)
(S) $h=g(x)$. Range of $g=[0,1]$→(4).
Answer: (P)→(5),(Q)→(3),(R)→(2),(S)→(4) → C.
Correct Answer: C