Q.A function $f$ is defined by $f(x) = 2x - 5$. Write down the values of

Domain $A$. We need the radicand strictly positive: $|x| - [x] > 0$.

• For $x \ge 0$: $|x| = x$, so $|x| - [x] = x - [x] = \{x\}$ (the fractional part). This is $> 0$ exactly when $x$ is not an integer. So all non-negative integers ($0, 1, 2, \dots$) are excluded.
• For $x < 0$: $|x| = -x$ and $[x] \le -1$. Writing $x = -m - f$ with integer $m \ge 0$ and $f \in [0,1)$, we get $|x| - [x] = (m+f) - (-m-1) = 2m + f + 1 \ge 1 > 0$. So every negative real (including negative integers) is in the domain.

Hence $A = \mathbb{R} - (\mathbb{Z}^+ \cup \{0\})$ — the reals with the non-negative integers removed. …

Concept & Intuition

A bijection requires the function to be both injective (one‑one) and surjective (onto). For a function defined by a formula, we must restrict its domain $A$ and codomain $B$ so that these conditions hold. The expression inside the arcsine is $\sqrt{x^2-4x+5}$. The square root forces the radicand to be non‑negative, and the arcsine itself only accepts inputs in $[-1,1]$. Moreover, arcsine is one‑one only on $[-1,1]$ and its range is $[-\pi/2,\pi/2]$. We need to find the largest possible domain $A$ and codomain $B$ that make $f$ a bijection — but the problem gives $f:A\to B$ as a bijection, so $A$ and $B$ are already chosen to make that true. We must deduce what they must be.

Step‑by‑step reasoning

1. Simplify the expression under the square root

   $x^2-4x+5 = (x-2)^2 + 1$.

   So $\sqrt{x^2-4x+5} = \sqrt{(x-2)^2+1}$.

   The smallest value of $(x-2)^2$ is $0$ (at $x=2$), so the minimum of the square root is $\sqrt{0+1}=1$.

   As $x\to\pm\infty$, $(x-2)^2\to\infty$, so the square root $\to\infty$.

   Hence the range of $\sqrt{x^2-4x+5}$ is $[1,\infty)$.

2. Domain restriction from the arcsine

   $\sin^{-1}(t)$ is defined only for $t\in[-1,1]$.

   But our $t = \sqrt{x^2-4x+5}$ is always $\ge 1$.

   For $f(x)$ to be defined, we need $t\in[-1,1]\cap[1,\infty) = \{1\}$.

   So the only possible input to $\sin^{-1}$ is $t=1$.

   That forces $\sqrt{x^2-4x+5}=1 \implies x^2-4x+5=1 \implies x^2-4x+4=0 \implies (x-2)^2=0 \implies x=2$.

   Therefore the only real number that makes $f$ defined is $x=2$.

   So the domain $A$ must be $\{2\}$ (or a subset thereof, but to have a function we need at least this point).

3. Determine the codomain $B$ for bijection

   At $x=2$, $f(2)=\sin^{-1}(\sqrt{1}) = \sin^{-1}(1) = \frac{\pi}{2}$.

   So the only output is $\pi/2$. …

The greatest integer function $\lfloor t\rfloor$ jumps at every integer value of $t$ that $t$ actually crosses (not merely touches from one side). Since $x^2-3$ is continuous, $f$ can only be discontinuous where $x^2-3$ equals an integer.

1. Find the integers $x^2-3$ can equal for $x\in(-1,2)$.

   For $x\in(-1,2)$, $x^2\in[0,4)$, so $x^2-3\in[-3,1)$. The integers in this range are $n=-3,-2,-1,0$.

2. Solve $x^2-3=n$, i.e. $x=\pm\sqrt{n+3}$, and keep only roots inside $(-1,2)$.

   • $n=-3$: $x=0$.
   • $n=-2$: $x=\pm1$; only $x=1$ lies in $(-1,2)$ ($x=-1$ is the excluded endpoint).
   • $n=-1$: $x=\pm\sqrt2$; only $x=\sqrt2\approx1.414$ lies in $(-1,2)$.
   • $n=0$: $x=\pm\sqrt3$; only $x=\sqrt3\approx1.732$ lies in $(-1,2)$.

   Candidates: $x=0,1,\sqrt2,\sqrt3$.

3. Check $x=0$ separately.

   $x^2-3$ has its global minimum, $-3$, exactly at $x=0$ (since $x^2\ge0$). So for $x$ near $0$ on either side, $x^2-3$ is slightly greater than $-3$ (never less), meaning $\lfloor x^2-3\rfloor=-3$ on both sides, matching $f(0)=-3$. The function only touches the integer $-3$ here without crossing it, so $f$ is continuous at $x=0$ — it is not a discontinuity.

4. Check the remaining three points. …

Concept & Intuition

A piecewise function is like a chameleon: it changes its formula depending on where $x$ lives. To solve $f(x)=8$, we must treat each piece separately, solving the equation on that piece’s interval, and then discard any solution that doesn’t actually fall inside that interval. This is the classic “solve, then check the domain” approach — many students forget the second step and count extraneous solutions.

Step-by-step solution

1. First piece: $x < 2$ Here $f(x) = x^2 - 4x + 3$. Set equal to 8:

Factor: $(x-5)(x+1)=0$, so $x = 5$ or $x = -1$.

Check the domain condition: $x < 2$.

• $x=5$ is not less than 2 → discard.
• $x=-1$ is less than 2 → keep. So from this branch we get one valid solution: $x = -1$.

2. Second piece: $x \geq 2$ Here $f(x) = x - 3$. Set equal to 8:

Check the domain condition: $x \geq 2$.

• $11 \geq 2$ is true → keep. …

Concept and Intuition

For a function $f: A \to B$ to have an inverse function, it must be a bijection. This means $f$ must be both:

1. Injective (one-to-one): Every distinct element in the domain $A$ maps to a distinct element in the codomain $B$.
2. Surjective (onto): Every element in the codomain $B$ is the image of at least one element in the domain $A$. In other words, the codomain $B$ must be exactly equal to the range of the function $f$.

The problem defines $f(x)$ as an infinite series. Our first step is to find a closed-form expression for this series. Once we have $f(x)$ in a simpler form, we need to determine its range for the given domain. This range will then be the set $B$ that ensures $f$ is surjective, and thus bijective, allowing an inverse function to exist.

Step-by-step Derivation

1. Identify the function as an infinite geometric series:

   The given function is $f(x) = x + x^2 + x^3 + \ldots \infty$.

   This is an infinite geometric series where the first term is $a = x$ and the common ratio is $r = x$.

2. Determine the condition for convergence of the series:

   An infinite geometric series converges if and only if the absolute value of its common ratio is less than 1, i.e., $|r| < 1$.

   In this case, $|x| < 1$, which means $-1 < x < 1$. This condition perfectly matches the given domain of the function, $f : (-1, 1) \to B$.

3. Find the sum of the infinite geometric series:

   The sum $S$ of an infinite geometric series with first term $a$ and common ratio $r$ (where $|r|<1$) is given by:

   $S = \frac{a}{1-r}$

   Substituting $a=x$ and $r=x$ into the formula, we get the closed-form expression for $f(x)$:

4. Determine the range of $f(x)$ for $x \in (-1, 1)$: To find the range of $f(x) = \frac{x}{1-x}$ for $x \in (-1, 1)$, we can analyze its behavior. First, let's check if the function is monotonic (strictly increasing or strictly decreasing) over its domain. This will also confirm if it is injective. We can do this by finding its derivative:

Using the quotient rule, $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$:

For $x \in (-1, 1)$, the term $(1-x)$ is never zero, and $(1-x)^2$ is always positive. Therefore, $f'(x) = \frac{1}{(1-x)^2} > 0$ for all $x \in (-1, 1)$. …