Q.Let $A = \mathbb{R} - \{3\}$, $B = \mathbb{R} - \{1\}$. Let $f : A \to B$ be defined by $f(x) = \dfrac{x - 2}{x - 3}$, $\forall\, x \in A$. Then show that $f$ is bijective.

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. …

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. …

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. …

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. …

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. $x\to-\infty$). …

Concept and Intuition

A quadratic $f(x)=Ax^2+Bx+C$ with $A>0$ is a downward-then-upward parabola with a single minimum at its vertex $x_0=-\dfrac{B}{2A}$. Restricted to $[x_0,\infty)$, it is strictly increasing, hence one-one (injective); its range there is exactly $[f(x_0),\infty)$, so it is automatically onto (surjective) that range. So for $f:[a,\infty)\to[b,\infty)$ to be a bijection, $a$ must equal the vertex $x_0$ (any larger $a$ would still be injective, but then $b$ must equal $f(a)$ for surjectivity to hold onto exactly $[b,\infty)$ — and the natural/expected choice, matching how these problems are always set, is that $a$ is exactly the vertex itself, the smallest valid starting point making it a bijection onto its natural range).

Step-by-Step Solution

1. Here $A=2, B=-3$. Vertex: $a = -\dfrac{B}{2A} = -\dfrac{-3}{4} = \dfrac34$.
2. Minimum value: $b = f(a) = 2\left(\dfrac34\right)^2 - 3\left(\dfrac34\right) + 5$. …

We are given a piecewise function:

and told that the codomain is $(-\infty, 5)$. We must decide whether $f$ is injective, surjective, both, or neither.

Intuition — Why a Bijection Proof Approach Works

The key is to find the actual range of $f$ over its entire domain. If the range equals the codomain, the function is surjective. If every output comes from exactly one input, it is injective. Because the two pieces are on disjoint intervals, we can analyze each separately and then combine. The danger is assuming that because the pieces are separate, the function is automatically one-to-one — but the same $y$-value could appear in both pieces.

Step-by-step reasoning

1. Simplify the first piece The expression $3\sin x - 4\cos x$ can be rewritten as a single sine wave. Recall: $R\sin(x - \phi) = R\sin x\cos\phi - R\cos x\sin\phi$. We want $R\cos\phi = 3$ and $R\sin\phi = 4$. Then $R = \sqrt{3^2 + 4^2} = 5$, and $\tan\phi = \frac{4}{3}$, so $\phi = \tan^{-1}\frac{4}{3}$. Hence:

2. Determine the range of the first piece on its interval The interval for $x$ is:

Let $u = x - \tan^{-1}\frac{4}{3}$. Then as $x$ runs over $I_1$, $u$ runs over $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

On this open interval, $\sin u$ is strictly increasing from $-1$ to $1$, but does not include the endpoints because the interval is open.

So $5\sin u$ takes all values in $(-5, 5)$.

Thus the range of the first piece is $(-5, 5)$.

3. Analyze the second piece

   For $x \in I_2 = \left(\frac{5\pi}{6}, \pi\right)$, we have $f(x) = \log(\sin x)$.

   On this interval, $\sin x$ is positive and decreasing: at $x = \frac{5\pi}{6}$, $\sin x = \frac{1}{2}$; as $x \to \pi^-$, $\sin x \to 0^+$.

   So $\sin x \in (0, \frac{1}{2}]$.

   Taking $\log$ (natural log, presumably), $\log(\sin x)$ ranges from $\log(0^+) = -\infty$ up to $\log(\frac{1}{2}) = -\log 2$.

   Hence the range of the second piece is $(-\infty, -\log 2]$.

4. Combine the ranges

   First piece: $(-5, 5)$

   Second piece: $(-\infty, -\log 2]$

   Note that $-\log 2 \approx -0.693$, which lies inside $(-5, 5)$. So the union of the two ranges is:

   $(-\infty, 5).$ …