Question paper, Set 10

Let $x$ be the pre-image of $-11$. Then $f(x) = -11$

$\Rightarrow 6 x + 1 = -11$

$\Rightarrow 6 x = -12$

$\Rightarrow x = -2$

Answer: $-2$

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

Domain $= A = \{-1, 0, 1, 2\}$

Codomain $= B = \{0, 2, 4, 6, 8\}$

Range $= f(A) = \{0, 2, 4, 6\}$

Answer: Domain $= \{-1, 0, 1, 2\}$, codomain $= \{0, 2, 4, 6, 8\}$, range $= \{0, 2, 4, 6\}$

$f(1) = f(3) = f(5) = 8$

$f = \{(1, 8), (3, 8), (5, 8)\}$

Range $= \{8\}$ (only one element, so $f$ is a constant function)

Answer: $f = \{(1, 8), (3, 8), (5, 8)\}$, range $= \{8\}$

put $x = 0$ $\Rightarrow$ $y = -10$: y-intercept $(0, -10)$

put $y = 0$ $\Rightarrow$ $- 5 x - 10 = 0 \Rightarrow x = -2$: x-intercept $(-2, 0)$

$a = -5 < 0$, so the inclination is greater than $90^\circ$

Answer: x-axis at $(-2, 0)$, y-axis at $(0, -10)$; inclination greater than $90^\circ$

$a = 2, b = -9, c = 4$

$\Delta = b^2 - 4ac = (-9)^2 - 4(2)(4) = 49$

$\displaystyle x = \frac{9 \pm 7}{4} = \frac{1}{2}, 4$

Answer: $\displaystyle x = \frac{1}{2}, 4$

$a = 5, b = -2, c = -3$

$\displaystyle \alpha + \beta = -\frac{b}{a} = \frac{2}{5}$

$\displaystyle \alpha\beta = \frac{c}{a} = - \frac{3}{5}$

$\Delta = b^2 - 4ac = (-2)^2 - 4(5)(-3) = 64$

$\displaystyle |\alpha - \beta| = \frac{\sqrt{\Delta}}{|a|} = \frac{\sqrt{64}}{5} = \frac{8}{5}$

Answer: $\displaystyle \alpha + \beta = \frac{2}{5}, \ \alpha\beta = - \frac{3}{5}, \ |\alpha - \beta| = \frac{8}{5}$

$a = 1, b = -5, c = 4$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(\frac{5}{2}, \frac{16 - 25}{4}\right) = (\frac{5}{2}, - \frac{9}{4})$

$\displaystyle y = a(x - h)^2 + k = \left(x - \frac{5}{2}\right)^2 - \frac{9}{4}$

$a > 0$, so the parabola opens upward: minimum value $\displaystyle = - \frac{9}{4}$ at $\displaystyle x = \frac{5}{2}$

Answer: $\displaystyle y = \left(x - \frac{5}{2}\right)^2 - \frac{9}{4}$; vertex $\displaystyle (\frac{5}{2}, - \frac{9}{4})$; minimum value $\displaystyle - \frac{9}{4}$

$y = -2(x + 1)^2 + 8 = - 2 x^{2} - 4 x + 6$

$a = -2 < 0$ (downward parabola)

$\Delta = b^2 - 4ac = (-4)^2 - 4(-2)(6) = 64$

$\displaystyle y = 0 \Rightarrow x = \frac{4 \pm 8}{-4} = -3, 1$: cuts the x-axis at $(-3, 0)$, $(1, 0)$

put $x = 0 \Rightarrow y = 6$: cuts the y-axis at $(0, 6)$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = (-1, 8)$

Plot these points and draw the downward parabola through them (maximum value $8$ at $x = -1$)

Answer: Downward parabola; cuts the x-axis at $(-3, 0)$ and $(1, 0)$; cuts the y-axis at $(0, 6)$; vertex $(-1, 8)$

$y = (x + 1)^2 - 5 = x^{2} + 2 x - 4$

$a = 1 > 0$ (upward parabola)

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

$\displaystyle y = 0 \Rightarrow x = \frac{-2 \pm \sqrt{20}}{2} = -1 \pm \sqrt{5}$

Sign: $+$ outside $-1 - \sqrt{5}$ and $-1 + \sqrt{5}$, $-$ between them

(i) $y < 0 \Rightarrow x \in (-1 - \sqrt{5}, -1 + \sqrt{5})$

(ii) $y \le 0 \Rightarrow x \in [-1 - \sqrt{5}, -1 + \sqrt{5}]$

(iii) $y > 0 \Rightarrow x \in (-\infty, -1 - \sqrt{5}) \cup (-1 + \sqrt{5}, \infty)$

(iv) $y \ge 0 \Rightarrow x \in (-\infty, -1 - \sqrt{5}] \cup [-1 + \sqrt{5}, \infty)$

Answer: (i) $x \in (-1 - \sqrt{5}, -1 + \sqrt{5})$; (ii) $x \in [-1 - \sqrt{5}, -1 + \sqrt{5}]$; (iii) $x \in (-\infty, -1 - \sqrt{5}) \cup (-1 + \sqrt{5}, \infty)$; (iv) $x \in (-\infty, -1 - \sqrt{5}] \cup [-1 + \sqrt{5}, \infty)$

$2x + 1 = 6^2 = 36$

$\displaystyle 2x = 35 \Rightarrow x = \frac{35}{2}$

Answer: $\displaystyle \frac{35}{2}$

$\displaystyle \log_3 \frac{1}{9} = \log_3 3^{-2} = -2, \ \log_3 \frac{1}{3} = -1$

$\log_3 1 = 0, \ \log_3 3 = 1, \ \log_3 9 = 2$

Points: $\displaystyle (\frac{1}{9}, -2), (\frac{1}{3}, -1), (1, 0), (3, 1), (9, 2)$ (an increasing curve, base $3 > 1$)

Answer: $\displaystyle \left(\frac{1}{9}, -2\right), \left(\frac{1}{3}, -1\right), (1, 0), (3, 1), (9, 2)$

$\displaystyle \log_{3^2} 2^3 \cdot \log_{5^2} 3^3 \cdot \log_{2^2} 5^3 = \frac{3}{2}\log_3 2 \cdot \frac{3}{2}\log_5 3 \cdot \frac{3}{2}\log_2 5$

$\displaystyle = \frac{27}{8} \log_3 2 \cdot \log_5 3 \cdot \log_2 5 = \frac{27}{8} \log_5 2 \cdot \log_2 5 = \frac{27}{8}$

Answer: $\displaystyle \frac{27}{8}$

$|3 - 10| = |-7| = 7, \ |10 - 3| = 7$

(property $|-x| = |x|$)

$7 + 7 = 14$

Answer: $14$

$3 x + 4 \le -2$ (or) $3 x + 4 \ge 2$

$\Rightarrow x \le -2$ (or) $\displaystyle x \ge - \frac{2}{3}$

$\displaystyle \Rightarrow x \in (-\infty, -2] \cup [- \frac{2}{3}, \infty)$

Answer: $\displaystyle x \in (-\infty, -2] \cup [- \frac{2}{3}, \infty)$

$|x| = k \Rightarrow x = \pm k$

$x^{2} - 3 x = \pm 4$

$\displaystyle x^{2} - 3 x = 4 \Rightarrow x^{2} - 3 x - 4 = 0 \Rightarrow x = \frac{3 \pm 5}{2} = -1, 4$

$x^{2} - 3 x = -4 \Rightarrow x^{2} - 3 x + 4 = 0$, $\Delta = -7 < 0$: no real $x$

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

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

Critical points: $x + 4 = 0 \Rightarrow x = -4$, $\displaystyle 2 x + 5 = 0 \Rightarrow x = - \frac{5}{2}$

Case (i) Suppose $x < -4$ ($x + 4 < 0, 2 x + 5 < 0$): $(- x - 4) + (- 2 x - 5) = 6 \Rightarrow - 3 x - 9 = 6$ $\Rightarrow$ $x = -5$, which is in this case

Case (ii) Suppose $\displaystyle -4 \le x \le - \frac{5}{2}$ ($x + 4 \ge 0, 2 x + 5 \le 0$): $(x + 4) + (- 2 x - 5) = 6 \Rightarrow - x - 1 = 6$ $\Rightarrow$ $x = -7$, not in this case: no solution

Case (iii) Suppose $\displaystyle x > - \frac{5}{2}$ ($x + 4 > 0, 2 x + 5 > 0$): $(x + 4) + (2 x + 5) = 6 \Rightarrow 3 x + 9 = 6$ $\Rightarrow$ $x = -1$, which is in this case

Ans: $x \in \{-5, -1\}$

Answer: $x \in \{-5, -1\}$

$x = 0$ is the critical point

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

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

From (i), (ii): $x \in (-1, 1)$

Answer: $x \in (-1, 1)$

$\sqrt{2 x^{2} - 5 x + 2}$ is defined when $2 x^{2} - 5 x + 2 \ge 0$

$\Rightarrow (x - 2)(2 x - 1) \ge 0$

Domain $\displaystyle = (-\infty, \frac{1}{2}] \cup [2, \infty)$

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

$[-2.5] = -3$, $[2.5] = 2$

$-3 - 2 = -5$

Answer: $-5$

$[-0.45] = -1$

$x - [x] = -0.45 - (-1) = 0.55$

Answer: $0.55$