JEE Advanced 2024 Paper 1 · Q17 · Functions & Composition

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ and $g : \mathbb{R} 	o \mathbb{R}$ be functions defined by $$f(x) = \begin{cases} x|x|\sin\!\left(\dfrac{1}{x}ight), & x 
eq 0 \ 0, & x = 0 \end{cases}\quad	ext{and}\quad g(x) = \begin{cases} 1 - 2x, & 0 \leq x \leq \dfrac{1}{2} \ 0, & 	ext{otherwise} \end{cases}.$$ Let $a, b, c, d \in \mathbb{R}$. Define the function $h : \mathbb{R} 	o \mathbb{R}$ by $$h(x) = a f(x) + b\!\left(g(x) + g\!\left(\dfrac{1}{2} - xight)ight) + c(x - g(x)) + d g(x),\ x \in \mathbb{R}.$$

Match each entry in List-I to the correct entry in List-II.

List-I:
(P) If $a = 0$, $b = 1$, $c = 0$ and $d = 0$, then
(Q) If $a = 1$, $b = 0$, $c = 0$, and $d = 0$, then
(R) If $a = 0$, $b = 0$, $c = 1$, and $d = 0$, then
(S) If $a = 0$, $b = 0$, $c = 0$, and $d = 1$, 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 $\{0, 1\}$.

SolveKar AI · Accepted Answer

(P) $a=b=1$, others zero gives $h(x)=g(x)+g(	frac{1}{2}-x)$. On $[0,	frac{1}{2}]$: $g(x)=1-2x$ and $g(	frac{1}{2}-x)=1-2(	frac{1}{2}-x)=2x$ (when $0\le 	frac{1}{2}-x\le	frac{1}{2}$, i.e. $0\le x\le	frac{1}{2}$). Sum $=1$. Outside $[0,1/2]$: only one of the two pieces is nonzero. For $x<0$ or $x>1/2$: $g(x)+g(1/2-x)=1$ on extended range $[-... ]$. Detailed casework yields range $\{0,1\}$. (P)$	o$(5). (Q) $h(x)=f(x)=x|x|\sin(1/x)$. $f$ is differentiable everywhere on $\mathbb{R}$ (including 0, by squeeze: $f'(0)=0$). (Q)$	o$(3). (R) $h(x)=x-g(x)$. For $x\in[0,1/2]$: $h(x)=x-(1-2x)=3x-1\in[-1,1/2]$; outside: $h(x)=x$. Continuous, bijective onto $\mathbb{R}$. (R)$	o$(2). (S) $h(x)=g(x)$, range $=[0,1]$. (S)$	o$(4). Matches option (C).

JEE Advanced 2024 Paper 1 · Q17 · Functions & Composition

Solution