Question paper, Set 7

Let $x$ be the pre-image of $-13$. Then $f(x) = -13$

$\Rightarrow 3 - 4 x = -13$

$\Rightarrow - 4 x = -16$

$\Rightarrow x = 4$

Answer: $4$

$f(-1) = (-1)^{3} = -1$, $f(0) = (0)^{3} = 0$, $f(1) = (1)^{3} = 1$, $f(2) = (2)^{3} = 8$

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

Codomain $= B = \{-1, 0, 1, 2, 4, 8\}$

Range $= f(A) = \{-1, 0, 1, 8\}$

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

$f(3) = 3 (3)^{2} - 5 (3) + 2$

$= 14$

Answer: $14$

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

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

$a = -2 < 0$, so the inclination is greater than $90^\circ$

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

$a = 1, b = -3, c = -10$

$\Delta = b^2 - 4ac = (-3)^2 - 4(1)(-10) = 49$

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

Answer: $\Delta = 49$; roots are real and distinct

$a = 4, b = -8, c = 1$

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

$\displaystyle \alpha\beta = \frac{c}{a} = \frac{1}{4}$

$\Delta = b^2 - 4ac = (-8)^2 - 4(4)(1) = 48$

$\displaystyle |\alpha - \beta| = \frac{\sqrt{\Delta}}{|a|} = \frac{\sqrt{48}}{4} = \sqrt{3}$

Answer: $\displaystyle \alpha + \beta = 2, \ \alpha\beta = \frac{1}{4}, \ |\alpha - \beta| = \sqrt{3}$

$a = 3, b = -6, c = 7$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(1, \frac{84 - 36}{12}\right) = (1, 4)$

$y = a(x - h)^2 + k = 3(x - 1)^2 + 4$

$a > 0$, so the parabola opens upward: minimum value $= 4$ at $x = 1$

Answer: $y = 3(x - 1)^2 + 4$; vertex $(1, 4)$; minimum value $4$

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

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

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

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

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 downward parabola through them (maximum value $4$ at $x = 1$)

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

$a = -3 < 0$ (downward parabola)

$\Delta = b^2 - 4ac = 2^2 - 4(-3)(-4) = -44$

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

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

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

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

(iv) $y \ge 0 \Rightarrow x \in \phi$

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

$\log_4 (x - 3) = -1 \Leftrightarrow x - 3 = 4^{-1}$

$\displaystyle x - 3 = \frac{1}{4}$

$\displaystyle x = \frac{13}{4}$

Answer: $\displaystyle \frac{13}{4}$

$a^{\log_a N} = N$

$6^{\log_6 7} = 7, \ 2^{\log_2 3} = 3$

$7 \times 3 = 21$

Answer: $21$

$\log_{1/3} x = \log_{3^{-1}} x = -\log_3 x = \log_3 x^{-1}$

$\displaystyle \log_3 (2x + 1) = \log_3 \frac{1}{x} \Rightarrow 2x + 1 = \frac{1}{x}$

$\displaystyle 2x^2 + x - 1 = 0 \Rightarrow (2x - 1)(x + 1) = 0 \Rightarrow x = \frac{1}{2} \text{ (or) } x = -1$

For $x = -1$, $\log_{1/3} x$ is not defined

$\displaystyle \therefore x = \frac{1}{2}$

Answer: $\displaystyle \frac{1}{2}$

$|(-2)^3| = |-8| = 8$

$|-2|^3 = 2^3 = 8$

(property $|x^n| = |x|^n$)

Answer: $8, \ 8$

$2 x + 1 < -9$ (or) $2 x + 1 > 9$

$\Rightarrow x < -5$ (or) $x > 4$

$\Rightarrow x \in (-\infty, -5) \cup (4, \infty)$

Answer: $x \in (-\infty, -5) \cup (4, \infty)$

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

$4 x - 3 = \pm (2 x + 5)$

$4 x - 3 = 2 x + 5 \Rightarrow 2 x - 8 = 0 \Rightarrow x = 4$

$\displaystyle 4 x - 3 = - 2 x - 5 \Rightarrow 6 x + 2 = 0 \Rightarrow x = - \frac{1}{3}$

$\displaystyle \therefore x \in \{- \frac{1}{3}, 4\}$

Answer: $\displaystyle x \in \{- \frac{1}{3}, 4\}$

Critical points: $\displaystyle 4 x - 3 = 0 \Rightarrow x = \frac{3}{4}$

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

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

Ans: $x \in \{-3, 2\}$

Answer: $x \in \{-3, 2\}$

Case (i) $x < 1$: $|x - 1| = 1 - x$, $\displaystyle \frac{1 - x}{x + 3} - 2 > 0 \Rightarrow \frac{- 3 x - 5}{x + 3} > 0 \Rightarrow x \in (-3, - \frac{5}{3})$

Case (ii) $x \ge 1$: $|x - 1| = x - 1$, $\displaystyle \frac{x - 1}{x + 3} - 2 > 0 \Rightarrow \frac{- x - 7}{x + 3} > 0$: no common value

So $\displaystyle x \in (-3, - \frac{5}{3})$; integers: $-2$

Number of integers $= 1$

Answer: $\displaystyle x \in (-3, - \frac{5}{3})$; number of integers $= 1$

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

$\Rightarrow - 2 x \ge -5$

$\displaystyle \Rightarrow x \le \frac{5}{2}$ (dividing by a negative number)

Domain $\displaystyle = (-\infty, \frac{5}{2}]$

Answer: $\displaystyle (-\infty, \frac{5}{2}]$

$[x + 3] = [0.4] = 0$

$[x] + 3 = [-2.6] + 3 = -3 + 3 = 0$

(property $[x + k] = [x] + k$ for $k \in \mathbb{Z}$)

Answer: $0, \ 0$

$\displaystyle \left(\frac{1}{2}\right)^x = 2^{-x}$ is defined for all real $x$: Domain $= \mathbb{R}$

It is always positive (a decreasing curve through $(0, 1)$): Range $= (0, \infty)$

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