Question paper, Set 4

Image of $-2$ is $f(-2)$

$f(-2) = 5 (-2) + 1 = -9$

Answer: $-9$

$f(1) = 3 (1) - 1 = 2$, $f(2) = 3 (2) - 1 = 5$, $f(3) = 3 (3) - 1 = 8$

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

Codomain $= B = \{1, 2, 5, 8, 11\}$

Range $= f(A) = \{2, 5, 8\}$

Answer: Domain $= \{1, 2, 3\}$, codomain $= \{1, 2, 5, 8, 11\}$, range $= \{2, 5, 8\}$

$I_A(x) = x$ for every $x \in A$

$I_A = \{(2, 2), (5, 5), (7, 7)\}$

Answer: $I_A = \{(2, 2), (5, 5), (7, 7)\}$

put $x = 0$ $\Rightarrow$ $y = -7$: y-intercept $(0, -7)$

put $y = 0$ $\Rightarrow$ $x - 7 = 0 \Rightarrow x = 7$: x-intercept $(7, 0)$

$a = 1 > 0$, so the inclination is less than $90^\circ$

Answer: x-axis at $(7, 0)$, y-axis at $(0, -7)$; inclination less than $90^\circ$

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

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

$\displaystyle x = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$

Answer: $x = 1 - \sqrt{2}, 1 + \sqrt{2}$

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

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

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

$\displaystyle 2 x^{2} - x - 6 = 2(x + \frac{3}{2})(x - 2)$

Answer: $\displaystyle 2(x + \frac{3}{2})(x - 2)$ $= (x - 2)(2 x + 3)$

$a = 2, b = 4, c = -1$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(-1, \frac{-8 - 16}{8}\right) = (-1, -3)$

$a > 0$: minimum value is $-3$

$\therefore$ Range $= [-3, \infty)$

Answer: Range $= [-3, \infty)$

$a = 1 > 0$ (upward parabola)

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

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

$\Delta = 0$: touches the x-axis at $(2, 0)$

put $x = 0 \Rightarrow y = 4$: cuts the y-axis at $(0, 4)$

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

Plot these points and draw the upward parabola through them (minimum value $0$ at $x = 2$)

Answer: Upward parabola; touches the x-axis at $(2, 0)$; cuts the y-axis at $(0, 4)$; vertex $(2, 0)$

$a = 1 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 3^2 - 4(1)(5) = -11$

$a > 0$, $\Delta < 0$: $y > 0$ for all $x \in \mathbb{R}$

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

(ii) $y \le 0 \Rightarrow x \in \phi$

(iii) $y > 0 \Rightarrow x \in \mathbb{R}$

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

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

$\displaystyle \left(\frac{1}{2}\right)^{-3} = (2^{-1})^{-3} = 2^3 = 8$

$\therefore \log_{1/2} 8 = -3$

Answer: $-3$

Base $\displaystyle 0 < \frac{1}{3} < 1$: $\log_a T > 0 \Leftrightarrow 0 < T < 1$

$0 < x - 1 < 1 \Rightarrow 1 < x < 2$

$x \in (1, 2)$

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

$\displaystyle \log_5 \frac{x^2 - 2x}{x - 2} = 1$

$\displaystyle \log_5 \frac{x(x - 2)}{x - 2} = 1 \Rightarrow \log_5 x = 1$

$x = 5^1 = 5$

Answer: $5$

$|x + y| = |-3| = 3$

$|x| + |y| = 4 + 7 = 11$

$xy < 0$, so $|x + y| < |x| + |y|$

Answer: $|x + y| = 3, \ |x| + |y| = 11$; $|x| + |y|$ is bigger

$x - 6 < -2$ (or) $x - 6 > 2$

$\Rightarrow x < 4$ (or) $x > 8$

$\Rightarrow x \in (-\infty, 4) \cup (8, \infty)$

Answer: $x \in (-\infty, 4) \cup (8, \infty)$

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

$x^{2} - 2 x - 5 = \pm 3$

$\displaystyle x^{2} - 2 x - 5 = 3 \Rightarrow x^{2} - 2 x - 8 = 0 \Rightarrow x = \frac{2 \pm 6}{2} = -2, 4$

$\displaystyle x^{2} - 2 x - 5 = -3 \Rightarrow x^{2} - 2 x - 2 = 0 \Rightarrow x = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$

$\therefore x \in \{-2, 1 - \sqrt{3}, 1 + \sqrt{3}, 4\}$

Answer: $x \in \{-2, 1 - \sqrt{3}, 1 + \sqrt{3}, 4\}$

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

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

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

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

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

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

Case (i) $x < 2$: $|x - 2| = 2 - x$, $\displaystyle \frac{2 - x}{x + 3} - 1 > 0 \Rightarrow \frac{- 2 x - 1}{x + 3} > 0 \Rightarrow x \in (-3, - \frac{1}{2})$

Case (ii) $x \ge 2$: $|x - 2| = x - 2$, $\displaystyle \frac{x - 2}{x + 3} - 1 > 0 \Rightarrow \frac{-5}{x + 3} > 0$: no common value

So $\displaystyle x \in (-3, - \frac{1}{2})$; integers: $-2, -1$

Number of integers $= 2$

Answer: $\displaystyle x \in (-3, - \frac{1}{2})$; number of integers $= 2$

$\sqrt{|x| - x}$ is defined when $|x| - x \ge 0$, i.e. $|x| \ge x$

For $x \ge 0$: $|x| = x$; for $x < 0$: $|x| = -x > x$

So $|x| \ge x$ for every real $x$: Domain $= \mathbb{R}$

Answer: $\mathbb{R}$

$[5.99] = 5$

$-6 \le -5.99 < -5 \Rightarrow [-5.99] = -6$

$5 - (-6) = 11$

Answer: $11$

$\{5.25\} = 5.25 - 5 = 0.25$

$\{-5.25\} = -5.25 - (-6) = 0.75$

$0.25 + 0.75 = 1$

Answer: $1$