Functions
Functions
Allen Star Batch
Grade 12
Question:
MATCH THE FOLLOWING
Column 1:
(A) $f(x) = \frac{(x+3)^2}{x^2+1}$
(B) $A = \{(x,y): x, y \in \mathbb{R}, x^2 + y^2 \leq 25\}$, $B = \{(x,y): x, y \in \mathbb{R}, y \geq \frac{4x^2}{9}\}$ and let $w, f(x)) = A \cap B$
(C) $f(x) = \frac{9}{2 - \cos 3x}$
(D) $f(x) = 3\sqrt{2} \cdot \sin\left(\sqrt{\frac{\pi^2}{16} - x^2}\right)$
Column 2 (Range of f(x)):
(p) $[0, 3]$
(q) $[3, 9]$
(r) $[0, 10]$
(s) $[0, 5]$
Step-by-Step Solution
Key Concept: For rational and trigonometric functions, use the discriminant or inverse relationships to find range constraints.
(A) From $y = \frac{x^2 + 9 + 6x}{x^2 + 1}$, rearranging gives $(y-1)x^2 - 6x + (y-9) = 0$. For real $x$, the discriminant $D \geq 0$ yields $0 \leq y \leq 10$. (B) The range of $A$ is $[-5, 5]$ and range of $B$ is $\mathbb{R}^+ \cup \{0\}$, so the range of $f(x)$ is $[0, 5]$. (C) From $y = \frac{9}{2 - \cos 3x}$, solving gives $\cos 3x = \frac{2y - 9}{y} \in [-1, 1]$, which restricts $3 \leq y \leq 9$. (D) On the domain $[-\frac{\pi}{4}, \frac{\pi}{4}]$, we have $x^2 \in [0, \frac{\pi^2}{16}]$, giving $0 \leq t(x) \leq 3$.
Correct Answer: [A-r] [B-s] [C-p] [D-p]