📝 Domain and range of a function (18 MCQs)

📖 Explanation: The domain consists of every x‑value for which the function produces a real output. It is the collection of admissible inputs, i.e., all possible x‑values that can be substituted into the rule without causing undefined operations such as division by zero or taking square roots of negatives.

📖 Explanation: The vertical line test checks each vertical line x = c for multiple intersections. If any vertical line meets the curve at two or more points, the relation cannot assign a unique y‑value to that x, violating the definition of a function. Hence the test requires at most one intersection for each vertical line.

📖 Explanation: Two distinct points sharing the same x‑coordinate mean the relation assigns two different y‑values to a single input. This directly violates the definition of a function, which requires exactly one output for each input. Therefore the relation fails the vertical line test and cannot be a function.

📖 Explanation: The vertical line test permits at most one intersection per vertical line. Touching the line at a single point still satisfies the “at most once” condition, so the relation meets the criteria for a function across its entire domain, including $x = 2$.

📖 Explanation: The denominator $x-3$ cannot be zero because division by zero is undefined. Setting $x-3=0$ yields $x=3$. Consequently every real number except 3 is permissible, giving the domain $\mathbb{R}\setminus\{3\}$.

📖 Explanation: Passing the vertical line test guarantees $f$ is a function, but failing the horizontal line test means $f$ is not one‑to‑one. Without the one‑to‑one property, an inverse relation cannot assign a unique x‑value to each y, so $f^{-1}$ is not a function.

📖 Explanation: The failure at $x = -1$ indicates the function is not one‑to‑one over the entire domain. By removing or restricting the problematic x‑value (or the corresponding y‑values), the remaining portion can become one‑to‑one, allowing an inverse function to exist on that restricted domain.

📖 Explanation: The square‑root function demands the radicand be non‑negative, giving $x-2\ge0$ or $x\ge2$. The rational function cannot have a zero denominator, so $x-2\neq0$ or $x\neq2$. Thus $f$ allows $x=2$ while $g$ excludes it, matching option B.

📖 Explanation: The expression $-(x-3)^2$ is always non‑positive, attaining its maximum value 0 when $x=3$. Adding 4 shifts this maximum to 4. Therefore the function never exceeds 4 and can take any value less than or equal to 4, giving the range $(-\infty,4]$.

📖 Explanation: Squaring eliminates the distinction between positive and negative roots. Starting from $x = \sqrt{9}=3$ and squaring yields $x^2=9$, which also admits $x=-3$ as a solution. The squaring step therefore creates an extra solution that was not present in the original equation, leading to extraneous roots.

📖 Explanation: Adding a constant shifts every output upward by that constant. Since the smallest value of $f(x)$ is 0, the smallest value of $g(x)=x^2+5$ becomes $0+5=5$. All larger values are also increased by 5, so the range becomes $[5,\infty)$.

📖 Explanation: The polynomial $y = x^3 - x$ gives a single y‑value for each x, so no vertical line can intersect it more than once. The circle equation, the sideways parabola, and the expression involving $\sqrt{-x}$ each produce at least one vertical line intersecting the curve twice, violating the test.

📖 Explanation: The equation describes a circle centered at the origin. A vertical line with $|x|<3$ meets the circle at two points, providing two different y‑values for the same x. This violates the vertical line test, so the relation cannot be expressed as a single‑valued function of x.

📖 Explanation: The radicand $1 - x^2$ must be non‑negative, giving $-1\le x\le 1$. The square root then yields values from 0 up to the square root of 1, i.e., 1. Hence the domain is $[-1,1]$ and the range is $[0,1]$.

📖 Explanation: For any non‑zero real number $y$, we can solve $y = 1/x$ to obtain $x = 1/y$, which is defined as long as $y\neq0$. However, there is no real $x$ that yields $f(x)=0$ because $1/x$ is never zero. Thus the range excludes only 0.

📖 Explanation: Both pieces require positive arguments: $\sqrt{x}$ allows $x=0$ but $\ln(x)$ does not. Since the function must be defined for every x in its domain, the stricter condition $x>0$ governs, excluding zero and negative numbers.

📖 Explanation: A function is one‑to‑one when no two distinct x‑values share the same y‑value. This is equivalent to the graph passing the vertical line test without any vertical line intersecting more than once. When the function is one‑to‑one, each y corresponds to exactly one x, allowing the inverse relation to assign a unique input to each output, thus forming a proper function.

📖 Explanation: Two distinct points sharing the same x‑coordinate mean the relation assigns two different y‑values to that x. This violates the definition of a function for any domain that includes $x=4$. Therefore the relation cannot be expressed as a function on any set containing that x‑value.