Question paper, Set 3

Each element of $A$ has exactly one image: $1 \to 4$, $2 \to 5$, $3 \to 4$

Two elements having the same image is allowed

$\therefore f$ is a function

Answer: Yes

$f(3) = (3) - 3 = 0$, $f(4) = (4) - 3 = 1$, $f(5) = (5) - 3 = 2$, $f(6) = (6) - 3 = 3$

Domain $= A = \{3, 4, 5, 6\}$

Codomain $= B = \{0, 1, 2, 3, 4\}$

Range $= f(A) = \{0, 1, 2, 3\}$

Answer: Domain $= \{3, 4, 5, 6\}$, codomain $= \{0, 1, 2, 3, 4\}$, range $= \{0, 1, 2, 3\}$

$f(-2) = 2 (-2)^{3} - (-2)^{2} + 4$

$= -16$

Answer: $-16$

put $x = 0$ $\Rightarrow$ $y = 8$: y-intercept $(0, 8)$

put $y = 0$ $\Rightarrow$ $8 - 4 x = 0 \Rightarrow x = 2$: x-intercept $(2, 0)$

$a = -4 < 0$, so the inclination is greater than $90^\circ$

Answer: x-axis at $(2, 0)$, y-axis at $(0, 8)$; inclination greater than $90^\circ$

$a = 1, b = 8, c = 16$

$\Delta = b^2 - 4ac = 8^2 - 4(1)(16) = 0$

$\Delta = 0$ $\Rightarrow$ roots are real and equal

Answer: $\Delta = 0$; roots are real and equal

$a = 1, b = -6, c = 4$

$\displaystyle \alpha + \beta = -\frac{b}{a} = 6$

$\displaystyle \alpha\beta = \frac{c}{a} = 4$

$\Delta = b^2 - 4ac = (-6)^2 - 4(1)(4) = 20$

$\displaystyle |\alpha - \beta| = \frac{\sqrt{\Delta}}{|a|} = \sqrt{20} = 2 \sqrt{5}$

Answer: $\alpha + \beta = 6, \ \alpha\beta = 4, \ |\alpha - \beta| = 2 \sqrt{5}$

$a = 1, b = 6, c = 4$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(-3, \frac{16 - 36}{4}\right) = (-3, -5)$

$y = a(x - h)^2 + k = (x + 3)^2 - 5$

$a > 0$, so the parabola opens upward: minimum value $= -5$ at $x = -3$

Answer: $y = (x + 3)^2 - 5$; vertex $(-3, -5)$; minimum value $-5$

$y = 2(x + 2)^2 + 1 = 2 x^{2} + 8 x + 9$

$a = 2 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 8^2 - 4(2)(9) = -8$

$\Delta < 0$: the graph does not meet the x-axis

put $x = 0 \Rightarrow y = 9$: cuts the y-axis at $(0, 9)$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = (-2, 1)$

Plot these points and draw the upward parabola through them (minimum value $1$ at $x = -2$)

Answer: Upward parabola; does not meet the x-axis; cuts the y-axis at $(0, 9)$; vertex $(-2, 1)$

$y = (3x + 4)^2 = 9 x^{2} + 24 x + 16$

$a = 9 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 24^2 - 4(9)(16) = 0$

$\displaystyle \Delta = 0 \Rightarrow y = 9(x + \frac{4}{3})^2$

$y > 0$ for all $\displaystyle x \ne - \frac{4}{3}$, and $y = 0$ at $\displaystyle x = - \frac{4}{3}$

(i) $y < 0 \Rightarrow x \in \phi$

(ii) $\displaystyle y \le 0 \Rightarrow x \in \{- \frac{4}{3}\}$

(iii) $\displaystyle y > 0 \Rightarrow x \in \mathbb{R} - \{- \frac{4}{3}\}$

(iv) $y \ge 0 \Rightarrow x \in \mathbb{R}$

Answer: (i) $x \in \phi$; (ii) $\displaystyle x \in \{- \frac{4}{3}\}$; (iii) $\displaystyle x \in \mathbb{R} - \{- \frac{4}{3}\}$; (iv) $x \in \mathbb{R}$

$\log_3 x = 4 \Leftrightarrow x = 3^4$

$x = 81$

Answer: $81$

$\log_8 32 = \log_{2^3} 2^5$

$\displaystyle \log_{b^n} a^m = \frac{m}{n} \log_b a$

$\displaystyle = \frac{5}{3} \log_2 2 = \frac{5}{3}$

Answer: $\displaystyle \frac{5}{3}$

$\log_b a \cdot \log_c b \cdot \log_d c = \log_d a$

$\displaystyle = \log_{16} 2 = \log_{2^4} 2^1 = \frac{1}{4}$

Answer: $\displaystyle \frac{1}{4}$

$|xy| = |-12| = 12$

$|x||y| = 6 \times 2 = 12$

(so $|xy| = |x||y|$)

Answer: $|xy| = 12, \ |x||y| = 12$

$-5 \le 3 x - 1 \le 5$

$\Rightarrow -5 + 1 \le 3 x \le 5 + 1$

$\displaystyle \Rightarrow - \frac{4}{3} \le x \le 2$

$\displaystyle \Rightarrow x \in [- \frac{4}{3}, 2]$

Answer: $\displaystyle x \in [- \frac{4}{3}, 2]$

$|x| = |y| \Rightarrow x = \pm y$

$3 x - 2 = \pm (x + 6)$

$3 x - 2 = x + 6 \Rightarrow 2 x - 8 = 0 \Rightarrow x = 4$

$3 x - 2 = - x - 6 \Rightarrow 4 x + 4 = 0 \Rightarrow x = -1$

$\therefore x \in \{-1, 4\}$

Answer: $x \in \{-1, 4\}$

Critical points: $\displaystyle 2 x - 5 = 0 \Rightarrow x = \frac{5}{2}$

Case (i) Suppose $\displaystyle x < \frac{5}{2}$ ($2 x - 5 < 0$): $(5 - 2 x) + x = 7 \Rightarrow 5 - x = 7$ $\Rightarrow$ $x = -2$, which is in this case

Case (ii) Suppose $\displaystyle x \ge \frac{5}{2}$ ($2 x - 5 \ge 0$): $(2 x - 5) + x = 7 \Rightarrow 3 x - 5 = 7$ $\Rightarrow$ $x = 4$, which is in this case

Ans: $x \in \{-2, 4\}$

Answer: $x \in \{-2, 4\}$

$x = 0$ is the critical point

Case (i) Suppose $x < 0$: $|x| = -x$, $x^{2} + 6 x + 5 \le 0 \Rightarrow (x + 1)(x + 5) \le 0 \Rightarrow x \in [-5, -1]$; common values $[-5, -1]$

Case (ii) Suppose $x \ge 0$: $|x| = x$, $x^{2} - 6 x + 5 \le 0 \Rightarrow (x - 5)(x - 1) \le 0 \Rightarrow x \in [1, 5]$; common values $[1, 5]$

From (i), (ii): $x \in [-5, -1] \cup [1, 5]$

Answer: $x \in [-5, -1] \cup [1, 5]$

$\sqrt{4 - x^{2}}$ is defined when $4 - x^{2} \ge 0$

$\Rightarrow x^{2} - 4 \le 0$

$\Rightarrow (x - 2)(x + 2) \le 0$

Domain $= [-2, 2]$

Answer: $[-2, 2]$

$x = 6.4 \notin \mathbb{Z}$: $[6.4] + [-6.4] = 6 + (-7) = -1$

$x = 6 \in \mathbb{Z}$: $[6] + [-6] = 6 - 6 = 0$

Answer: $-1, \ 0$

For $-4 \le x < 0$: $\displaystyle \frac{1}{x} \le -\frac{1}{4}$

For $0 < x \le 2$: $\displaystyle \frac{1}{x} \ge \frac{1}{2}$

$\displaystyle \frac{1}{x} \in \left(-\infty, -\frac{1}{4}\right] \cup \left[\frac{1}{2}, \infty\right)$

Answer: $\displaystyle (-\infty, - \frac{1}{4}] \cup [\frac{1}{2}, \infty)$