Question paper, Set 9

Image of $5$ is $f(5)$

$f(5) = 4 (5) - 9 = 11$

Answer: $11$

$f(0) = (0)^{2} - (0) = 0$, $f(1) = (1)^{2} - (1) = 0$, $f(2) = (2)^{2} - (2) = 2$, $f(3) = (3)^{2} - (3) = 6$

Domain $= A = \{0, 1, 2, 3\}$

Codomain $= B = \{0, 1, 2, 4, 6\}$

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

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

The highest power of $x$ is in the term $2 x^{6}$ (coefficient $2 \ne 0$)

$\therefore$ degree of $f(x)$ is $6$

Answer: $6$

put $x = 0$ $\Rightarrow$ $y = -21$: y-intercept $(0, -21)$

put $y = 0$ $\Rightarrow$ $\displaystyle 6 x - 21 = 0 \Rightarrow x = \frac{7}{2}$: x-intercept $\displaystyle (\frac{7}{2}, 0)$

$a = 6 > 0$, so the inclination is less than $90^\circ$

Answer: x-axis at $\displaystyle (\frac{7}{2}, 0)$, y-axis at $(0, -21)$; inclination less than $90^\circ$

$a = 2, b = 3, c = 4$

$\Delta = b^2 - 4ac = 3^2 - 4(2)(4) = -23$

$\Delta < 0$ $\Rightarrow$ roots are not real (imaginary)

Answer: $\Delta = -23$; roots are not real (imaginary)

$a = 1, b = 3, c = -5$

$\displaystyle \alpha + \beta = -\frac{b}{a} = -3$

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

$\Delta = b^2 - 4ac = 3^2 - 4(1)(-5) = 29$

$\displaystyle |\alpha - \beta| = \frac{\sqrt{\Delta}}{|a|} = \sqrt{29} = \sqrt{29}$

Answer: $\alpha + \beta = -3, \ \alpha\beta = -5, \ |\alpha - \beta| = \sqrt{29}$

$a = -2, b = -8, c = -11$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(-2, \frac{88 - 64}{-8}\right) = (-2, -3)$

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

$a < 0$, so the parabola opens downward: maximum value $= -3$ at $x = -2$

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

$y = (2x + 5)^2 = 4 x^{2} + 20 x + 25$

$a = 4 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 20^2 - 4(4)(25) = 0$

$\Delta = 0$: touches the x-axis at $\displaystyle (- \frac{5}{2}, 0)$

put $x = 0 \Rightarrow y = 25$: cuts the y-axis at $(0, 25)$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = (- \frac{5}{2}, 0)$

Plot these points and draw the upward parabola through them (minimum value $0$ at $\displaystyle x = - \frac{5}{2}$)

Answer: Upward parabola; touches the x-axis at $\displaystyle (- \frac{5}{2}, 0)$; cuts the y-axis at $(0, 25)$; vertex $\displaystyle (- \frac{5}{2}, 0)$

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

$\Delta = b^2 - 4ac = 5^2 - 4(-2)(3) = 49$

$\displaystyle y = 0 \Rightarrow x = \frac{-5 \pm 7}{-4} = - \frac{1}{2}, 3$

Sign: $-$ outside $\displaystyle - \frac{1}{2}$ and $3$, $+$ between them

(i) $\displaystyle y < 0 \Rightarrow x \in (-\infty, - \frac{1}{2}) \cup (3, \infty)$

(ii) $\displaystyle y \le 0 \Rightarrow x \in (-\infty, - \frac{1}{2}] \cup [3, \infty)$

(iii) $\displaystyle y > 0 \Rightarrow x \in (- \frac{1}{2}, 3)$

(iv) $\displaystyle y \ge 0 \Rightarrow x \in [- \frac{1}{2}, 3]$

Answer: (i) $\displaystyle x \in (-\infty, - \frac{1}{2}) \cup (3, \infty)$; (ii) $\displaystyle x \in (-\infty, - \frac{1}{2}] \cup [3, \infty)$; (iii) $\displaystyle x \in (- \frac{1}{2}, 3)$; (iv) $\displaystyle x \in [- \frac{1}{2}, 3]$

$\log_x 64 = 3 \Leftrightarrow x^3 = 64$

$x^3 = 4^3 \Rightarrow x = 4$

Answer: $4$

$\displaystyle \log_6 9 + \log_6 8 - \log_6 2 = \log_6 \frac{9 \times 8}{2}$

$= \log_6 36 = \log_6 6^2 = 2$

Answer: $2$

$\log_6 x(x - 5) = 2 \Rightarrow x(x - 5) = 36$

$x^2 - 5x - 36 = 0 \Rightarrow (x - 9)(x + 4) = 0$

$x = 9 \text{ (or) } x = -4$

For $x = -4$, $\log_6 (-4)$ is not defined

$\therefore x = 9$

Answer: $9$

$|x| = x$ for $x \ge 0$ and $|x| = -x \ne x$ for $x < 0$

$\therefore x \ge 0, \text{ i.e. } x \in [0, \infty)$

Answer: $x \ge 0$, i.e. $x \in [0, \infty)$

$-3 < 7 - 2 x < 3$

$\Rightarrow -3 - 7 < - 2 x < 3 - 7$

$\Rightarrow 2 < x < 5$ (dividing by a negative number turns the signs)

$\Rightarrow x \in (2, 5)$

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

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

$x - 1 = \pm (3 x + 5)$

$x - 1 = 3 x + 5 \Rightarrow 2 x + 6 = 0 \Rightarrow x = -3$

$x - 1 = - 3 x - 5 \Rightarrow 4 x + 4 = 0 \Rightarrow x = -1$

$\therefore x \in \{-3, -1\}$

Answer: $x \in \{-3, -1\}$

Critical points: $x + 1 = 0 \Rightarrow x = -1$, $x - 2 = 0 \Rightarrow x = 2$

Case (i) Suppose $x < -1$ ($x + 1 < 0, x - 2 < 0$): $\displaystyle (- x - 1) - (2 - x) < x - \frac{1}{2} \Rightarrow - x - \frac{5}{2} < 0 \Rightarrow x \in (- \frac{5}{2}, \infty)$: common values $\displaystyle (- \frac{5}{2}, -1)$

Case (ii) Suppose $-1 \le x \le 2$ ($x + 1 \ge 0, x - 2 \le 0$): $\displaystyle (x + 1) - (2 - x) < x - \frac{1}{2} \Rightarrow x - \frac{1}{2} < 0 \Rightarrow x \in (-\infty, \frac{1}{2})$: common values $\displaystyle [-1, \frac{1}{2})$

Case (iii) Suppose $x > 2$ ($x + 1 > 0, x - 2 > 0$): $\displaystyle (x + 1) - (x - 2) < x - \frac{1}{2} \Rightarrow \frac{7}{2} - x < 0 \Rightarrow x \in (\frac{7}{2}, \infty)$: common values $\displaystyle (\frac{7}{2}, \infty)$

Ans: $\displaystyle x \in (- \frac{5}{2}, \frac{1}{2}) \cup (\frac{7}{2}, \infty)$

Answer: $\displaystyle x \in (- \frac{5}{2}, \frac{1}{2}) \cup (\frac{7}{2}, \infty)$

$\displaystyle \frac{20|x|}{4 x^{2} + 25} \le 1$ (the denominator is positive)

(i) $x < 0$: $-20 x \le 4 x^{2} + 25 \Rightarrow 4 x^{2} + 20 x + 25 \ge 0 \Rightarrow (2 x + 5)^2 \ge 0$: true for all $x < 0$

(ii) $x \ge 0$: $20 x \le 4 x^{2} + 25 \Rightarrow 4 x^{2} - 20 x + 25 \ge 0 \Rightarrow (2 x - 5)^2 \ge 0$: true for all $x \ge 0$

From (i), (ii): $x \in (-\infty, 0) \cup [0, \infty) = \mathbb{R}$

Answer: $x \in \mathbb{R}$

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

$\Rightarrow (x - 2)^{2} \ge 0$: a perfect square

Domain $= \mathbb{R}$

Answer: $\mathbb{R}$

$[-3.7] = -4$

$[[-3.7]] = [-4] = -4$ (property $[[x]] = [x]$)

Answer: $-4$

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

It takes every real value (negative for $x < 0$): Range $= \mathbb{R}$

Answer: Domain $= \mathbb{R}$, Range $= \mathbb{R}$