Q.Show that the function $f: \mathbf{R} \to \{x \in \mathbf{R} : -1 < x < 1\}$ defined by $f(x) = \frac{x}{1+|x|}$, $x \in \mathbf{R}$ is one one and onto function.

Concept and Intuition

To test a piecewise function for injectivity/surjectivity, examine each piece's range separately, then check for both (i) values in the codomain that are missed entirely, and (ii) values that are produced by both pieces (a collision across the pieces, not just within one).

Step-by-Step Solution

1. Piece 1 ($x<0$): $f(x)=a^x$ with $a>1$ is strictly increasing in $x$. As $x\to-\infty$, $a^x\to 0^+$; as $x\to 0^-$, $a^x\to 1^-$. So this piece is a bijection from $(-\infty,0)$ onto the open interval $(0,1)$ — it never actually equals $0$ or $1$.
2. Piece 2 ($x\ge0$): $f(x)=b^x$ with $0<b<1$ is strictly decreasing in $x$. At $x=0$, $f=1$; as $x\to\infty$, $f\to 0^+$. So this piece bijects $[0,\infty)$ onto $(0,1]$ — it attains $1$ (at $x=0$) but never actually reaches $0$.
3. Onto check: is every value in the codomain $[0,1]$ achieved? The value $0$ is approached but never attained by either piece, so $f$ is not onto.
4. One-one check: consider any value $v\in(0,1)$ (strictly between 0 and 1). Piece 1 gives exactly one $x_1<0$ with $a^{x_1}=v$ (since it's a bijection onto $(0,1)$). Piece 2 gives exactly one $x_2>0$ with $b^{x_2}=v$ (since it's a bijection from $(0,\infty)$ onto $(0,1)$, excluding the point $x=0\mapsto1$). So every $v\in(0,1)$ has two distinct pre-images, $x_1$ and $x_2$ — $f$ is not one-one.
5. Hence $f$ is neither one-one nor onto.

Common Mistakes

• Checking injectivity only within each piece and forgetting to check for collisions between the two pieces.
• Assuming the codomain restriction to $[0,1]$ automatically makes the map onto, without checking whether every value is actually attained.

ANSWER: D

Why a Bijection Proof Works

A bijection means every element of $A$ maps to a unique element of $B$ (injective), and every element of $B$ is hit (surjective). The function $f(x) = |x| - [x]$ involves the absolute value and the greatest integer (floor) function. The floor $[x]$ is the greatest integer $\le x$. For $x \ge 0$, $|x| = x$, so $f(x) = x - [x]$, which is the fractional part of $x$, always in $[0,1)$. For $x < 0$, $|x| = -x$, and $[x]$ is negative or zero, so the expression behaves differently. The trick is to see that $f$ is periodic in a certain sense, and we must pick intervals where it is strictly monotonic and covers a range exactly once.

Step-by-Step Reasoning

1. Understand $f$ for $x \ge 0$

   For $x \ge 0$, $|x| = x$, so $f(x) = x - [x]$. This is the fractional part $\{x\}$, which lies in $[0,1)$. It is periodic with period 1, and on each interval $[n, n+1)$ (for integer $n \ge 0$), $f(x) = x - n$, which increases linearly from $0$ to $1$ (excluding 1). So on $[0, \infty)$, $f$ is not injective because many $x$ give the same fractional part (e.g., $f(0.5)=0.5$, $f(1.5)=0.5$). It is surjective onto $[0,1)$.

2. Understand $f$ for $x < 0$

   For $x < 0$, $|x| = -x$, so $f(x) = -x - [x]$. Write $x = -n + r$ where $n$ is a positive integer and $r \in [0,1)$ (so $x$ is in $(-n, -n+1]$). Then $[x] = -n$ (since $x$ is between $-n$ and $-n+1$, the floor is $-n$). Then $f(x) = -(-n + r) - (-n) = n - r + n = 2n - r$. So on $(-n, -n+1]$, $f(x) = 2n - r$, which decreases from $2n$ (when $r=0$, i.e., $x = -n$) to $2n-1$ (when $r \to 1^-$, i.e., $x \to -n+1^-$). Note: at $x = -n$, $f(-n) = 2n$; at $x = -n+1$ (if included), $f = 2n-1$. So the range on each negative interval is $(2n-1, 2n]$ (or $[2n-1, 2n]$ depending on endpoints). These intervals are disjoint and cover positive numbers > 0.

3. Combine the two cases

   • For $x \ge 0$, $f(x) \in [0,1)$.
   • For $x < 0$, $f(x) \in (1, \infty)$ (since $2n-1 \ge 1$ for $n \ge 1$, and $2n$ grows without bound). So $f$ maps nonnegative numbers to $[0,1)$ and negative numbers to $(1, \infty)$. The value $1$ is never attained? Check: $f(x)=1$ would require $x - [x] = 1$ (impossible, fractional part < 1) or $2n - r = 1$ => $r = 2n-1$, but $r \in [0,1)$, so only $n=1$ gives $r=1$, but $r=1$ is not in $[0,1)$ (open at 1). So $1$ is not in the range. Also $0$ is attained at $x=0$ and at $x$ any integer $\ge 0$? Actually $f(0)=0$, $f(1)=0$, so not injective on $[0,\infty)$.

4. Condition for bijection

   For $f$ to be a bijection, the domain $A$ must be chosen so that $f$ is one-to-one on $A$ and onto $B$. Since $f$ is periodic on $[0,\infty)$, we must restrict $A$ to an interval of length less than 1 (or exactly 1 but half-open) to avoid repeats. Similarly, on the negative side, each interval $(-n, -n+1]$ gives a distinct range interval $(2n-1, 2n]$, so we can pick one such interval. The codomain $B$ must match the image of $A$.

5. Test the options

   • (A) $A = [-\infty, 0]$ (likely $(-\infty, 0]$) and $B = [0, \infty)$. On $(-\infty,0]$, $f$ takes values in $(1,\infty)$ plus $f(0)=0$, so $0$ is hit, but $[0,1)$ is not covered (except 0). Not onto $[0,\infty)$. Also not injective on $(-\infty,0]$ because multiple $x$ give same $f$? Actually on $(-\infty,0]$, each negative interval gives a distinct range, but $f$ is decreasing on each, so injective if we take a single interval. But $(-\infty,0]$ includes many intervals, so not injective. So no.
   • (B) $A = [-3, -2)$ and $B = (5, 6]$. For $x \in [-3,-2)$, we have $n=3$ (since $x$ in $(-3,-2]$? Actually $[-3,-2)$ includes $-3$ but not $-2$. At $x=-3$, $f(-3)=6$; as $x \to -2^-$, $f(x) \to 5$. So the image is $(5,6]$ (including 6, excluding 5). That matches $B = (5,6]$. And on this interval, $f$ is strictly decreasing, so injective. So this is a bijection! But wait: the question asks for $A$ and $B$ respectively. Option (B) gives $A=[-3,-2)$ and $B=(5,6]$, which works. However, we must check if any other option also works.
   • (C) $A = [1,2)$ and $B = [3,4)$. For $x \in [1,2)$, $f(x) = x - 1$ (since $[x]=1$), so $f(x) \in [0,1)$. That gives $B=[0,1)$, not $[3,4)$. So this is false unless we misinterpret? Wait, $f(x)=|x|-[x]$, for $x \ge 0$ it's fractional part, so $[1,2)$ maps to $[0,1)$. But option says $B=[3,4)$. That doesn't match. So (C) is wrong.
   • (D) $A = [0, \infty)$ and $B = [0,1)$. As argued, $f$ on $[0,\infty)$ is not injective (many $x$ give same fractional part), so not a bijection.

   So only (B) seems to work. But let's double-check (C) — maybe they intend $A$ to be negative? No, $[1,2)$ is positive. So (C) is out.

   However, wait: The problem says "A and B are respectively" meaning we need the pair. Option (B) gives a valid bijection. But is there any trick? For $A=[-3,-2)$, $f$ maps to $(5,6]$, which matches $B$. So (B) is correct.

   But let's verify the endpoints: $x=-3$ gives $f(-3)=|-3|-[-3]=3-(-3)=6$; $x$ approaching $-2$ from below gives $f(x)= -x - [x]$, with $[x]=-3$ for $x \in (-3,-2)$, so $f(x) = -x - (-3) = -x+3$, which as $x\to -2^-$ gives $2+3=5$, but not including 5. So image is $(5,6]$, exactly $B$. So yes.

   Thus the correct option is (B).

Concept and Intuition

$f(x)=\dfrac{4-x^2}{4+x^2}$ depends on $x$ only through $x^2$, so $f(-x)=f(x)$ — it is an even function and therefore NOT one-one over all of $\mathbb{R}$ (e.g. $f(1)=f(-1)$). For $f:A\to B$ to be stated as a bijection, its domain $A$ must be cut down to a set where distinct inputs never share the same $x^2$ value — i.e., restricted to only non-negative or only non-positive $x$. Since we are told $-4\in A$, the restriction must be to the non-positive branch, $A=(-\infty,0]$, and $B$ (as codomain of a bijection) must equal the exact range of $f$ on that branch.

Step-by-Step Solution

1. Rewrite $f(x) = \dfrac{4-x^2}{4+x^2} = \dfrac{-(4+x^2)+8}{4+x^2} = -1+\dfrac{8}{4+x^2}$.
2. On $A=(-\infty,0]$, let $u=x^2\in[0,\infty)$ as $x$ ranges over $A$ (bijective correspondence between $x\le 0$ and $u\ge 0$, since squaring is injective on non-positive reals). Then $f=-1+\dfrac{8}{4+u}$.
3. As $u$ increases from $0$ to $\infty$, $\dfrac{8}{4+u}$ decreases from $2$ to $0^+$, so $f$ decreases from $1$ (at $u=0$, i.e. $x=0$) to values approaching $-1$ (but never reaching it, as $u\to\infty$, i.e. $x\to-\infty$).
4. So as $x$ ranges over $(-\infty,0]$, $f(x)$ ranges over $(-1,1]$: this is exactly $B$ (since $f:A\to B$ is onto by definition of bijection).
5. Compute $A\cap B = (-\infty,0]\cap(-1,1]$: this keeps every value that is both $\le 0$ and in $(-1,1]$, giving $(-1,0]$.

Common Mistakes

• Taking $A=[0,\infty)$ by default (the "natural" injective branch) without checking that $-4\in A$ forces the other branch, $(-\infty,0]$.
• Getting the open/closed endpoints of the range wrong — the limit $-1$ is approached as $x\to-\infty$ but never attained (open), while $f(0)=1$ is attained exactly (closed).

ANSWER: D

Concept and Intuition

$f(x)=x^2-3x+2$ is a parabola opening upward with vertex at $x=\frac32$ (from $x=-b/2a=3/2$), where $f\left(\frac32\right)=\frac94-\frac92+2=-\frac14$. A parabola is never one-to-one on all of $\mathbb R$, so "bijection" forces us to pick a domain on which $f$ is strictly monotonic, and then the codomain $B$ must equal exactly the set of values $f$ takes there (otherwise it isn't onto).

Step-by-Step Solution

1. Vertex: $x=\frac32$, $f_{\min\text{ on right branch}}=-\frac14$.
2. Test option (B): $A=(-\infty,\frac32]$. On this ray $f$ is strictly decreasing (left of the vertex), so it's injective.
3. As $x\to-\infty$, $f(x)\to\infty$; at $x=\frac32$, $f=-\frac14$. So the range on $A$ is exactly $[-\frac14,\infty)$ — matching $B$ in option (B). Hence $f:A\to B$ is a bijection.
4. Check the rejects: (A) $A=(-\infty,0]$ gives range $[f(0),\infty)=[2,\infty)$, not $(-\infty,-\frac14]$. (C) $A=[\frac32,\infty)$ gives range $[-\frac14,\infty)$, but $B$ is stated as $(-\infty,-\frac14]$ — wrong. (D) the full line is not injective at all.

Common Mistakes

• Forgetting that a bijection needs the codomain to be the exact range, not just an over-estimate.
• Picking the wrong side of the vertex (left branch decreasing vs. right branch increasing) and mismatching it with the stated range.

ANSWER: B

Concept and Intuition

A Möbius-type map $f(x)=\dfrac{ax+b}{cx+d}$ is undefined exactly where the denominator is zero, and it never attains the value that would require dividing by zero when solving $y=f(x)$ for $x$ — that value is precisely $a/c$ (the horizontal asymptote). Removing both these single points from domain and codomain makes the map a genuine bijection.

Step-by-Step Solution

1. $f(x)=\dfrac{x+3}{x-2}$ is undefined at $x=2$, so the domain is $\mathbb{R}-\{2\}$, giving $l=2$.
2. Rewrite: $f(x) = \dfrac{(x-2)+5}{x-2} = 1+\dfrac{5}{x-2}$. This shows $f(x)=1$ would need $\dfrac{5}{x-2}=0$, impossible, so $y=1$ is never attained.
3. For $f$ to be a bijection onto $\mathbb{R}-\{m\}$, this excluded value must be exactly $m$: $m=1$.
4. $3l+2m = 3(2)+2(1) = 6+2=8$.

Common Mistakes

• Swapping which constant is $l$ (domain gap) and which is $m$ (codomain gap).
• Trying to solve $y=\frac{x+3}{x-2}$ for $x$ incorrectly and missing that $y=1$ is the singular value.

ANSWER: C