📝 Irrational exponents and real powers (12 MCQs)

📖 Explanation: An irrational exponent is a number that cannot be expressed as a fraction of two integers, i.e., it is not rational. Because it cannot be written in the form $p/q$ with integers $p,q$, its decimal expansion is non‑terminating and non‑repeating, which is the defining property of irrational numbers.

📖 Explanation: Since the exponent $\sqrt{2}>0$, the derivative f&#039;(x)=\sqrt{2}\,x^{\sqrt{2}-1} is negative when $0<x<1$ (because the power $\sqrt{2}-1$ is positive but $x^{\text{positive}}$ is less than 1) and positive when $x>1$. Thus the function falls on $(0,1)$ and rises after $x=1$.

📖 Explanation: Because $\sqrt{2}\approx1.414<2$, raising a number greater than 1 to a larger exponent yields a larger value, so for $x>1$ we have $x^{\sqrt{2}}<x^{2}$. Conversely, for numbers between 0 and 1, a larger exponent makes the result smaller, reversing the inequality. At $x=1$ the two functions coincide.

📖 Explanation: The cube root $\sqrt[3]{x}$ is $x^{1/3}$. Raising this to the power $\sqrt{2}$ multiplies the exponents: $(x^{1/3})^{\sqrt{2}}=x^{\sqrt{2}/3}$. This follows directly from the law $(a^{m})^{n}=a^{mn}$.

📖 Explanation: Substituting the first equation into the second gives $(a^{\sqrt{2}})^{\sqrt{2}}=a$, which simplifies to $a^{2}=a$. Solving $a^{2}-a=0$ yields $a=0$ or $a=1$. Since $a>0$, the admissible solution is $a=1$.

📖 Explanation: The radicand must be non‑negative: $x^{\sqrt{2}}-1\ge0\Rightarrow x^{\sqrt{2}}\ge1$. Because the base $x$ is required to be positive for the irrational exponent to be real, this inequality reduces to $x\ge1$. Hence the domain is the interval $[1,\infty)$.

📖 Explanation: For any positive exponent, the function $x^{c}$ approaches 0 as $x$ approaches 0 from the right. Since $\sqrt{2}>0$, the limit $\lim_{x\to0^{+}}x^{\sqrt{2}}=0$. This follows from the continuity of the power function on $(0,\infty)$.

📖 Explanation: When the base is negative and the exponent is irrational, the result cannot be expressed as a real number because the exponent cannot be written as a fraction with an odd denominator that would allow a real root. Consequently, $x^{\sqrt{2}}$ is undefined in the real number system for every negative $x$.

📖 Explanation: Here the numerator 2 is even and the denominator 3 is odd. According to the rule, we replace the original function with $g(x)=|x|^{p/q}$; thus $g(x)=|x|^{2/3}$ correctly produces real values for both positive and negative inputs.

📖 Explanation: The expression $x^{\sqrt{2}}$ is only defined for $x>0$ when we restrict ourselves to real numbers, because a negative or zero base with an irrational exponent leads to an undefined or complex value. Therefore the function does not exist at $x=0$, and continuity cannot be assessed there.

📖 Explanation: Power functions with positive exponents dominate logarithmic functions for large arguments. Since $\sqrt{2}>0$, $x^{\sqrt{2}}$ increases without bound much more rapidly than $\ln x$, which grows only logarithmically. Hence $f(x)$ outpaces $g(x)$ as $x$ becomes large.

📖 Explanation: Rewrite $16$ as $2^{4}$. Then $(2^{4})^{\sqrt{2}/2}=2^{4\cdot(\sqrt{2}/2)}=2^{2\sqrt{2}}$. This simplification uses the exponent rule $(a^{m})^{n}=a^{mn}$, yielding the exact expression $2^{2\sqrt{2}}$.