Question paper, Set 1

Image of $4$ is $f(4)$

$f(4) = 3 (4) - 2 = 10$

Answer: $10$

$f(-2) = (-2)^{2} = 4$, $f(-1) = (-1)^{2} = 1$, $f(0) = (0)^{2} = 0$, $f(1) = (1)^{2} = 1$, $f(2) = (2)^{2} = 4$

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

Codomain $= B = \{0, 1, 2, 3, 4, 5\}$

Range $= f(A) = \{0, 1, 4\}$

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

The highest power of $x$ is in the term $- 3 x^{5}$ (coefficient $-3 \ne 0$)

$\therefore$ degree of $f(x)$ is $5$

Answer: $5$

put $x = 0$ $\Rightarrow$ $y = -6$: y-intercept $(0, -6)$

put $y = 0$ $\Rightarrow$ $3 x - 6 = 0 \Rightarrow x = 2$: x-intercept $(2, 0)$

$a = 3 > 0$, so the inclination is less than $90^\circ$

Answer: x-axis at $(2, 0)$, y-axis at $(0, -6)$; inclination less than $90^\circ$

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

$\Delta = b^2 - 4ac = (-5)^2 - 4(2)(3) = 1$

$\Delta > 0$ $\Rightarrow$ roots are real and distinct

Answer: $\Delta = 1$; roots are real and distinct

$a = 2, b = -7, c = 3$

$\displaystyle \alpha + \beta = -\frac{b}{a} = \frac{7}{2}$

$\displaystyle \alpha\beta = \frac{c}{a} = \frac{3}{2}$

$\Delta = b^2 - 4ac = (-7)^2 - 4(2)(3) = 25$

$\displaystyle |\alpha - \beta| = \frac{\sqrt{\Delta}}{|a|} = \frac{\sqrt{25}}{2} = \frac{5}{2}$

Answer: $\displaystyle \alpha + \beta = \frac{7}{2}, \ \alpha\beta = \frac{3}{2}, \ |\alpha - \beta| = \frac{5}{2}$

$a = 2, b = -8, c = 5$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(2, \frac{40 - 64}{8}\right) = (2, -3)$

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

$a > 0$, so the parabola opens upward: minimum value $= -3$ at $x = 2$

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

$y = (x + 3)(x - 1) = x^{2} + 2 x - 3$

$a = 1 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 2^2 - 4(1)(-3) = 16$

$\displaystyle y = 0 \Rightarrow x = \frac{-2 \pm 4}{2} = -3, 1$: cuts the x-axis at $(-3, 0)$, $(1, 0)$

put $x = 0 \Rightarrow y = -3$: cuts the y-axis at $(0, -3)$

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

Plot these points and draw the upward parabola through them (minimum value $-4$ at $x = -1$)

Answer: Upward parabola; cuts the x-axis at $(-3, 0)$ and $(1, 0)$; cuts the y-axis at $(0, -3)$; vertex $(-1, -4)$

$a = 2 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 1^2 - 4(2)(-6) = 49$

$\displaystyle y = 0 \Rightarrow x = \frac{-1 \pm 7}{4} = -2, \frac{3}{2}$

Sign: $+$ outside $-2$ and $\displaystyle \frac{3}{2}$, $-$ between them

(i) $\displaystyle y < 0 \Rightarrow x \in (-2, \frac{3}{2})$

(ii) $\displaystyle y \le 0 \Rightarrow x \in [-2, \frac{3}{2}]$

(iii) $\displaystyle y > 0 \Rightarrow x \in (-\infty, -2) \cup (\frac{3}{2}, \infty)$

(iv) $\displaystyle y \ge 0 \Rightarrow x \in (-\infty, -2] \cup [\frac{3}{2}, \infty)$

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

$a^x = N \Leftrightarrow x = \log_a N$

$3^4 = 81 \Leftrightarrow 4 = \log_3 81$

Answer: $4 = \log_3 81$

$\displaystyle \log_a m - \log_a n = \log_a \frac{m}{n}$

$\displaystyle \log_2 48 - \log_2 3 = \log_2 \frac{48}{3} = \log_2 16$

$= \log_2 2^4 = 4 \log_2 2 = 4$

Answer: $4$

$\log_3 x(x + 6) = 3 \Rightarrow x(x + 6) = 3^3$

$x^2 + 6x - 27 = 0 \Rightarrow (x + 9)(x - 3) = 0$

$x = 3 \text{ (or) } x = -9$

For $x = -9$, $\log_3 (-9)$ is not defined

$\therefore x = 3$

Answer: $3$

$|-8| = -(-8) = 8, \ |3| = 3, \ |-5| = 5$

$8 + 3 - 5 = 6$

Answer: $6$

$-7 < 2 x - 5 < 7$

$\Rightarrow -7 + 5 < 2 x < 7 + 5$

$\Rightarrow -1 < x < 6$

$\Rightarrow x \in (-1, 6)$

Answer: $x \in (-1, 6)$

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

$x - 3 = \pm (5 - 2 x)$

$\displaystyle x - 3 = 5 - 2 x \Rightarrow 3 x - 8 = 0 \Rightarrow x = \frac{8}{3}$

$x - 3 = 2 x - 5 \Rightarrow x - 2 = 0 \Rightarrow x = 2$

$\displaystyle \therefore x \in \{2, \frac{8}{3}\}$

Answer: $\displaystyle x \in \{2, \frac{8}{3}\}$

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

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

Case (ii) Suppose $\displaystyle - \frac{5}{2} \le x \le 2$ ($x - 2 \le 0, 2 x + 5 \ge 0$): $(2 - x) + (2 x + 5) = 9 \Rightarrow x + 7 = 9$ $\Rightarrow$ $x = 2$, which is in this case

Case (iii) Suppose $x > 2$ ($x - 2 > 0, 2 x + 5 > 0$): $(x - 2) + (2 x + 5) = 9 \Rightarrow 3 x + 3 = 9$ $\Rightarrow$ $x = 2$, not in this case: no solution

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

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

$x = 0$ is the critical point

Case (i) Suppose $x < 0$: $|x| = -x$, $x^{2} + 7 x + 10 < 0 \Rightarrow (x + 2)(x + 5) < 0 \Rightarrow x \in (-5, -2)$; common values $(-5, -2)$

Case (ii) Suppose $x \ge 0$: $|x| = x$, $x^{2} - 7 x + 10 < 0 \Rightarrow (x - 5)(x - 2) < 0 \Rightarrow x \in (2, 5)$; common values $(2, 5)$

From (i), (ii): $x \in (-5, -2) \cup (2, 5)$

Answer: $x \in (-5, -2) \cup (2, 5)$

$\sqrt{x^{2} - 7 x + 12}$ is defined when $x^{2} - 7 x + 12 \ge 0$

$\Rightarrow (x - 4)(x - 3) \ge 0$

Domain $= (-\infty, 3] \cup [4, \infty)$

Answer: $(-\infty, 3] \cup [4, \infty)$

$[x]$ is an integer: $[x] = -1, 0, 1$ or $2$

$[x] = -1$ for $-1 \le x < 0$, ..., $[x] = 2$ for $2 \le x < 3$

$\therefore x \in [-1, 3)$

Answer: $x \in [-1, 3)$

$3^x$ is defined for every real $x$: Domain $= \mathbb{R}$

$3^x > 0$ always and takes every positive value: Range $= (0, \infty)$

Answer: Domain $= \mathbb{R}$, Range $= (0, \infty)$