📝 Restricting domain to make function invertible (14 MCQs)

📖 Explanation: The function $f(x)=x^{2}$ fails the horizontal line test on $\mathbb{R}$ because each positive $y$ value corresponds to two $x$ values. Restricting the domain to $x\ge0$ eliminates the negative branch, making the function one‑to‑one and therefore invertible. Hence the appropriate restriction is $x\ge0$.

📖 Explanation: When the domain of $g(x)=x^{3}$ is limited to $[0,\infty)$, the function remains strictly increasing, so it passes the horizontal line test. However, because its outputs are only non‑negative, it does not cover the entire set of real numbers, meaning it is not onto $\mathbb{R}$. Therefore, the statement that it is one‑to‑one but not onto is correct.

📖 Explanation: The function $h(x)=\sqrt{x}$ is defined for $x\ge0$ and maps to non‑negative outputs. Its inverse must reverse this mapping, giving $h^{-1}(x)=x^{2}$. Since the original output range is $x\ge0$, the inverse’s domain must also be restricted to non‑negative values, making option B the correct description.

📖 Explanation: For the restriction $f_{1}(x)=x^{2}$ with $x\ge0$, the inverse function undoes the squaring by taking the positive square root, yielding $f_{1}^{-1}(x)=\sqrt{x}$. For the restriction $f_{2}(x)=x^{2}$ with $x\le0$, the inverse must return the negative root, giving $f_{2}^{-1}(x)=-\sqrt{x}$. This pair matches option A.

📖 Explanation: The interval $[- \\frac{\\pi}{2},\\frac{\\pi}{2}]$ is chosen because on this interval the sine function is strictly increasing and therefore one‑to‑one. This monotonic behavior ensures that every $y$ in the range $[-1,1]$ corresponds to exactly one $x$ in the interval, satisfying the horizontal line test and allowing an inverse to exist.

📖 Explanation: Both \\\arcsin x\ and \\\arccos x\ are defined only for inputs between \-1\ and \1\, so their domains are \[-1,1]\. However, their ranges differ: \\\arcsin x\ returns angles from \-\\frac{\\pi}{2}\ to \\\frac{\\pi}{2}\, while \\\arccos x\ returns angles from \0\ to \\\pi\. Option B correctly states these domain‑range pairs.

📖 Explanation: On the interval \[0,2]\ the function \p(x)=x^{3}-x\ is strictly increasing because its derivative \p'(x)=3x^{2}-1\ is positive for \x\\ge0\. Consequently, the function passes the horizontal line test on this interval, making it one‑to‑one and thus invertible. Option B accurately reflects this conclusion.

📖 Explanation: The tangent function is restricted to \(-\\frac{\\pi}{2},\\frac{\\pi}{2})\ to obtain a one‑to‑one branch. Its inverse, \\\arctan x\, therefore returns the angle that lies in the same interval. Hence the range of \\\arctan x\ is \(-\\frac{\\pi}{2},\\frac{\\pi}{2})\, which corresponds to option C.

📖 Explanation: When \f(x)=\\frac{1}{x}\ is limited to positive inputs, the output is also positive, preserving a one‑to‑one relationship. Solving \y=\\frac{1}{x}\ for \x\ yields \x=\\frac{1}{y}\, so the inverse function is \f^{-1}(x)=\\frac{1}{x}\ with domain \x>0\. This matches option A.

📖 Explanation: To guarantee an invertible restriction, one must select an interval where the original function is monotonic (either strictly increasing or decreasing). Monotonicity ensures that each output value is produced by exactly one input, satisfying the horizontal line test. Option B correctly describes this essential strategy.

📖 Explanation: A global inverse for \\\sin x\ does not exist because the sine function repeats its values periodically, violating the one‑to‑one requirement. By restricting the domain to \[-\\frac{\\pi}{2},\\frac{\\pi}{2}]\, the function becomes monotonic increasing, thereby satisfying the necessary condition for an inverse. Monotonicity is the key property, making option C correct.

📖 Explanation: On the interval \[-2,2]\ the cubic function \g(x)=x^{3}\ maps to the range \[-8,8]\. Solving \y=x^{3}\ for \x\ gives \x=\\sqrt[3]{y}\. Therefore the inverse is \g^{-1}(x)=\\sqrt[3]{x}\ with domain \[-8,8]\, which corresponds to option A.

📖 Explanation: For the restricted cosine function on \[0,\\pi]\, the inverse \\\arccos x\ is defined only for \x\ in \[-1,1]\ and yields values in \[0,\\pi]\. It is a decreasing function, and \\\arccos 0 = \\frac{\\pi}{2}\ are true statements. However, \\\arccos\ is not defined for all real numbers, making option D the false statement.

📖 Explanation: Choosing the principal value branch when defining an inverse trigonometric function selects a specific interval where the original function is one‑to‑one. This restriction ensures that the inverse is a genuine function, providing continuity across the selected interval and avoiding discontinuities that would arise from multiple possible angles. Option B captures this reasoning.