📝 Algebraic functions definition and examples (12 MCQs)

📖 Explanation: An algebraic function is built from polynomials using a finite number of algebraic operations such as addition, subtraction, multiplication, division, and root extraction. The expression $f(x)=\sqrt{x^{2}-4}$ uses only polynomial terms and a square‑root, which is an allowed root extraction, so it qualifies. The other choices involve transcendental functions (sine, exponential, logarithm) and therefore are not algebraic.

📖 Explanation: The square‑root in $f(x)=\sqrt{x^{2}-4}$ requires the radicand to be non‑negative, giving $x^{2}-4\ge0$ or $|x|\ge2$, so its domain is $(-\infty,-2]\cup[2,\infty)$. The polynomial expression for $g(x)$ is defined for every real $x$. Therefore the correct description is option C.

📖 Explanation: Zeros of a sine function occur when its argument equals an integer multiple of $\pi$: $Bx-C=n\pi$. Setting $x=0$ gives $-C=n_1\pi$ and $x=\pi/4$ gives $B(\pi/4)-C=n_2\pi$. Subtracting yields $B(\pi/4)= (n_2-n_1)\pi$, so $B=4(n_2-n_1)$. The smallest positive integer difference gives $B=4$, which matches option A.

📖 Explanation: At $x=3$ the radicand $3^{2}-4=5$ is positive, so both $p(3)=\sqrt5$ and $q(3)=1/\sqrt5$ exist and are finite. Each component function is continuous at $x=3$; the composition $p^{\,q}$ of continuous functions is also continuous there. Hence $r$ is continuous at $x=3$.

📖 Explanation: For a sinusoid $y=A\sin(Bx)$, the amplitude equals $|A|$ and the period equals $2\pi/|B|$. Here $A=2$ gives an amplitude of 2, and $B=4$ gives a period of $2\pi/4=\pi/2$. Therefore the correct description is option B.

📖 Explanation: Expanding the expression yields $f(x)=\frac{1}{3}x^{4}+ \frac{4}{3}x^{3}+ \frac{4}{3}x^{2}$. The leading term $\frac{1}{3}x^{4}$ dominates for large $|x|$ and is positive, so as $x\to -\infty$ the function grows without bound in the positive direction, i.e., $f(x)\to +\infty$.

📖 Explanation: When $x$ approaches 2 from the right, the radicand $x^{2}-4$ approaches $0^{+}$. The square‑root of a quantity tending to zero from above approaches zero. Hence $\lim_{x\to 2^{+}}\sqrt{x^{2}-4}=0$, which corresponds to option A.

📖 Explanation: Amplitude measures the maximum vertical displacement from the midline and is given by the absolute value of the coefficient multiplying the sine (or cosine) term. Therefore for $y=A\sin(Bx)$ the amplitude equals $|A|$, matching option A.

📖 Explanation: The period of a sinusoid depends on the absolute value of the angular frequency: $\text{period}=2\pi/|B|$. Changing the sign of $B$ reflects the graph horizontally but does not alter the distance required for one full cycle. Hence the period remains $2\pi/|B|$, which is option B.

📖 Explanation: Amplitude gives $|A|=5$. Period $\pi$ implies $2\pi/|B|=\pi$ so $|B|=2$. A rightward shift of $\pi/6$ means $C/B=\pi/6$; with $B=2$ this yields $C=2\cdot\pi/6=\pi/3$. Thus the triple $(5,2,\pi/3)$ meets all requirements, which is option A.

📖 Explanation: Inside the outer square‑root we have $\bigl(\frac{x^{2}}{3}(x+2)^{2}\bigr)^{2}-4$. For the expression to be non‑negative we need $|\frac{x^{2}}{3}(x+2)^{2}|\ge2$. Since $\frac{x^{2}}{3}(x+2)^{2}\ge0$ for all real $x$, the condition reduces to $\frac{x^{2}}{3}(x+2)^{2}\ge2$. Hence the domain is precisely the set of $x$ satisfying that inequality, which is described by option B.

📖 Explanation: Within the common domain $|x|\ge2$, the product simplifies algebraically: $f(x)g(x)=\sqrt{x^{2}-4}\cdot\frac{1}{\sqrt{x^{2}-4}}=1$. The result is constant 1 wherever the original functions are defined, so option A accurately describes the product.