Question paper, Set 6

Image of $3$ is $f(3)$

$f(3) = 7 - 2 (3) = 1$

Answer: $1$

$f(2) = 10 - (2) = 8$, $f(4) = 10 - (4) = 6$, $f(6) = 10 - (6) = 4$, $f(8) = 10 - (8) = 2$

Domain $= A = \{2, 4, 6, 8\}$

Codomain $= B = \{2, 4, 6, 8, 10\}$

Range $= f(A) = \{2, 4, 6, 8\}$

Answer: Domain $= \{2, 4, 6, 8\}$, codomain $= \{2, 4, 6, 8, 10\}$, range $= \{2, 4, 6, 8\}$

$y = -4$ is a constant function: $y$ is $-4$ for every $x$

Its graph is a line parallel to the x-axis, $4$ units below it

Points: $(0, -4)$, $(1, -4)$

Answer: Parallel to the x-axis; for example $(0, -4)$ and $(1, -4)$

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

put $y = 0$ $\Rightarrow$ $\displaystyle 4 x + 6 = 0 \Rightarrow x = - \frac{3}{2}$: x-intercept $\displaystyle (- \frac{3}{2}, 0)$

$a = 4 > 0$, so the inclination is less than $90^\circ$

Answer: x-axis at $\displaystyle (- \frac{3}{2}, 0)$, y-axis at $(0, 6)$; inclination less than $90^\circ$

$a = 6, b = -1, c = -2$

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

$\displaystyle x = \frac{1 \pm 7}{12} = - \frac{1}{2}, \frac{2}{3}$

Answer: $\displaystyle x = - \frac{1}{2}, \frac{2}{3}$

$\Delta = b^2 - 4ac = (-10)^2 - 4(3)(3) = 64$

$\displaystyle x = \frac{10 \pm 8}{6} = \frac{1}{3}, 3$

$\displaystyle \alpha = \frac{1}{3}, \beta = 3, a = 3$

$\displaystyle 3 x^{2} - 10 x + 3 = 3(x - \frac{1}{3})(x - 3)$

Answer: $\displaystyle 3(x - \frac{1}{3})(x - 3)$ $= (x - 3)(3 x - 1)$

$a = -3, b = 12, c = -7$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(2, \frac{84 - 144}{-12}\right) = (2, 5)$

$a < 0$: maximum value is $5$

$\therefore$ Range $= (-\infty, 5]$

Answer: Range $= (-\infty, 5]$

$a = 2 > 0$ (upward parabola)

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

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

$\displaystyle y = 0 \Rightarrow x = \frac{3 \pm 5}{4} = - \frac{1}{2}, 2$: cuts the x-axis at $\displaystyle (- \frac{1}{2}, 0)$, $(2, 0)$

put $x = 0 \Rightarrow y = -2$: cuts the y-axis at $(0, -2)$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = (\frac{3}{4}, - \frac{25}{8})$

Plot these points and draw the upward parabola through them (minimum value $\displaystyle - \frac{25}{8}$ at $\displaystyle x = \frac{3}{4}$)

Answer: Upward parabola; cuts the x-axis at $\displaystyle (- \frac{1}{2}, 0)$ and $(2, 0)$; cuts the y-axis at $(0, -2)$; vertex $\displaystyle (\frac{3}{4}, - \frac{25}{8})$

$a = -1 < 0$ (downward parabola)

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

$\Delta = 0 \Rightarrow y = -(x - 2)^2$

$y < 0$ for all $x \ne 2$, and $y = 0$ at $x = 2$

(i) $y < 0 \Rightarrow x \in \mathbb{R} - \{2\}$

(ii) $y \le 0 \Rightarrow x \in \mathbb{R}$

(iii) $y > 0 \Rightarrow x \in \phi$

(iv) $y \ge 0 \Rightarrow x \in \{2\}$

Answer: (i) $x \in \mathbb{R} - \{2\}$; (ii) $x \in \mathbb{R}$; (iii) $x \in \phi$; (iv) $x \in \{2\}$

$\displaystyle 7^{-2} = \frac{1}{7^2} = \frac{1}{49}$

$\displaystyle \therefore \log_7 \frac{1}{49} = -2$

Answer: $-2$

Here $0 < a = 0.4 < 1$ and $0 < x = 0.7 < 1$

$\log_a x > 0$ when $0 < a < 1$, $0 < x < 1$

$\therefore \log_{0.4} 0.7 > 0$

Answer: Positive

$\displaystyle 3^{\log_{3^2} 7^2} = 3^{\frac{2}{2}\log_3 7} = 3^{\log_3 7} = 7$

Answer: $7$

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

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

$|x| \ge 0$ and it takes every value $\ge 0$: Range $= [0, \infty)$

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

$-6 < 4 x + 2 < 6$

$\Rightarrow -6 - 2 < 4 x < 6 - 2$

$\Rightarrow -2 < x < 1$

$\Rightarrow x \in (-2, 1)$

Answer: $x \in (-2, 1)$

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

$x^{2} + 3 x + 1 = \pm 3$

$\displaystyle x^{2} + 3 x + 1 = 3 \Rightarrow x^{2} + 3 x - 2 = 0 \Rightarrow x = \frac{-3 \pm \sqrt{17}}{2} = \frac{-3 \pm \sqrt{17}}{2}$

$x^{2} + 3 x + 1 = -3 \Rightarrow x^{2} + 3 x + 4 = 0$, $\Delta = -7 < 0$: no real $x$

$\displaystyle \therefore x \in \{\frac{-3 - \sqrt{17}}{2}, \frac{-3 + \sqrt{17}}{2}\}$

Answer: $\displaystyle x \in \{\frac{-3 - \sqrt{17}}{2}, \frac{-3 + \sqrt{17}}{2}\}$

Critical points: $x - 3 = 0 \Rightarrow x = 3$, $4 - 2 x = 0 \Rightarrow x = 2$

Case (i) Suppose $x < 2$ ($x - 3 < 0, 4 - 2 x > 0$): $(3 - x) + (4 - 2 x) = 8 \Rightarrow 7 - 3 x = 8$ $\Rightarrow$ $\displaystyle x = - \frac{1}{3}$, which is in this case

Case (ii) Suppose $2 \le x \le 3$ ($x - 3 \le 0, 4 - 2 x \le 0$): $(3 - x) + (2 x - 4) = 8 \Rightarrow x - 1 = 8$ $\Rightarrow$ $x = 9$, not in this case: no solution

Case (iii) Suppose $x > 3$ ($x - 3 > 0, 4 - 2 x < 0$): $(x - 3) + (2 x - 4) = 8 \Rightarrow 3 x - 7 = 8$ $\Rightarrow$ $x = 5$, which is in this case

Ans: $\displaystyle x \in \{- \frac{1}{3}, 5\}$

Answer: $\displaystyle x \in \{- \frac{1}{3}, 5\}$

$x = 0$ is the critical point

Case (i) Suppose $x < 0$: $|x| = -x$, $x^{2} + 3 x - 4 > 0 \Rightarrow (x - 1)(x + 4) > 0 \Rightarrow x \in (-\infty, -4) \cup (1, \infty)$; common values $(-\infty, -4)$

Case (ii) Suppose $x \ge 0$: $|x| = x$, $x^{2} - 3 x - 4 > 0 \Rightarrow (x - 4)(x + 1) > 0 \Rightarrow x \in (-\infty, -1) \cup (4, \infty)$; common values $(4, \infty)$

From (i), (ii): $x \in (-\infty, -4) \cup (4, \infty)$

Answer: $x \in (-\infty, -4) \cup (4, \infty)$

$\displaystyle \frac{1}{\sqrt{x^{2} - 9}}$ is defined when $x^{2} - 9 > 0$

$\Rightarrow (x - 3)(x + 3) > 0$

Domain $= (-\infty, -3) \cup (3, \infty)$

Answer: $(-\infty, -3) \cup (3, \infty)$

$[-0.3] = -1$, $[0.3] = 0$, $[-1] = -1$

$-1 + 0 - 1 = -2$

Answer: $-2$

$\{7\} = 7 - 7 = 0$

$\{2.8\} = 2.8 - 2 = 0.8$

$0 + 0.8 = 0.8$

Answer: $0.8$