Question paper, Set 5

Let $x$ be the pre-image of $11$. Then $f(x) = 11$

$\Rightarrow 3 x - 4 = 11$

$\Rightarrow 3 x = 15$

$\Rightarrow x = 5$

Answer: $5$

$f(-1) = (-1)^{2} + 1 = 2$, $f(0) = (0)^{2} + 1 = 1$, $f(1) = (1)^{2} + 1 = 2$, $f(2) = (2)^{2} + 1 = 5$

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

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

Range $= f(A) = \{1, 2, 5\}$

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

The highest power of $x$ is in the term $9 x^{4}$ (coefficient $9 \ne 0$)

$\therefore$ degree of $f(x)$ is $4$

Answer: $4$

put $x = 0$ $\Rightarrow$ $y = -9$: y-intercept $(0, -9)$

put $y = 0$ $\Rightarrow$ $- 3 x - 9 = 0 \Rightarrow x = -3$: x-intercept $(-3, 0)$

$a = -3 < 0$, so the inclination is greater than $90^\circ$

Answer: x-axis at $(-3, 0)$, y-axis at $(0, -9)$; inclination greater than $90^\circ$

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

$\Delta = b^2 - 4ac = 2^2 - 4(3)(5) = -56$

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

Answer: $\Delta = -56$; roots are not real (imaginary)

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

$\displaystyle \alpha + \beta = -\frac{b}{a} = - \frac{7}{3}$

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

$\Delta = b^2 - 4ac = 7^2 - 4(3)(-6) = 121$

$\displaystyle |\alpha - \beta| = \frac{\sqrt{\Delta}}{|a|} = \frac{\sqrt{121}}{3} = \frac{11}{3}$

Answer: $\displaystyle \alpha + \beta = - \frac{7}{3}, \ \alpha\beta = -2, \ |\alpha - \beta| = \frac{11}{3}$

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

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(2, \frac{-4 - 16}{-4}\right) = (2, 5)$

$y = a(x - h)^2 + k = -(x - 2)^2 + 5$

$a < 0$, so the parabola opens downward: maximum value $= 5$ at $x = 2$

Answer: $y = -(x - 2)^2 + 5$; vertex $(2, 5)$; maximum value $5$

$a = 1 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 4^2 - 4(1)(6) = -8$

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

$\Delta < 0$: the graph does not meet the x-axis

put $x = 0 \Rightarrow y = 6$: cuts the y-axis at $(0, 6)$

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

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

Answer: Upward parabola; does not meet the x-axis; cuts the y-axis at $(0, 6)$; vertex $(-2, 2)$

$y = (x - 2)^2 - 3 = x^{2} - 4 x + 1$

$a = 1 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = (-4)^2 - 4(1)(1) = 12$

$\displaystyle y = 0 \Rightarrow x = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$

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

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

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

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

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

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

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

$\log_2 64 = 6 \Leftrightarrow 2^6 = 64$

Answer: $2^6 = 64$

$\log_b a \cdot \log_c b = \log_c a$

$\log_5 81 \cdot \log_3 5 = \log_3 81$

$= \log_3 3^4 = 4$

Answer: $4$

$= (2.5)^{\log_{2.5} 8} \cdot (2.5)^1$

$a^{\log_a N} = N$, $a^{m + n} = a^m \cdot a^n$

$= 8 \times 2.5 = 20$

Answer: $20$

$\displaystyle \frac{|-15|}{|3|} = \frac{15}{3} = 5$

$\displaystyle \left|\frac{-15}{3}\right| = |-5| = 5$

(property $\displaystyle |\frac{x}{y}| = \frac{|x|}{|y|}$)

Answer: $5, \ 5$

$5 - x \le -3$ (or) $5 - x \ge 3$

$\Rightarrow x \le 2$ (or) $x \ge 8$ (dividing by a negative number turns the signs)

$\Rightarrow x \in (-\infty, 2] \cup [8, \infty)$

Answer: $x \in (-\infty, 2] \cup [8, \infty)$

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

$2 x - 1 = \pm (x^{2} + 2 x - 4)$

$2 x - 1 = x^{2} + 2 x - 4 \Rightarrow x^{2} - 3 = 0 \Rightarrow x = \pm \sqrt{3}$

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

$\therefore x \in \{-5, - \sqrt{3}, 1, \sqrt{3}\}$

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

Critical points: $x + 1 = 0 \Rightarrow x = -1$, $x - 3 = 0 \Rightarrow x = 3$

Case (i) Suppose $x < -1$ ($x + 1 < 0, x - 3 < 0$): $(- x - 1) - (3 - x) < x - 3 \Rightarrow - x - 1 < 0 \Rightarrow x \in (-1, \infty)$: no common value

Case (ii) Suppose $-1 \le x \le 3$ ($x + 1 \ge 0, x - 3 \le 0$): $(x + 1) - (3 - x) < x - 3 \Rightarrow x + 1 < 0 \Rightarrow x \in (-\infty, -1)$: no common value

Case (iii) Suppose $x > 3$ ($x + 1 > 0, x - 3 > 0$): $(x + 1) - (x - 3) < x - 3 \Rightarrow 7 - x < 0 \Rightarrow x \in (7, \infty)$: common values $(7, \infty)$

Ans: $x \in (7, \infty)$

Answer: $x \in (7, \infty)$

$-16 < x^{2} + 6 x < 16$

$\Rightarrow x^{2} + 6 x + 16 > 0 \text{ and } x^{2} + 6 x - 16 < 0$

$x^{2} + 6 x + 16 > 0$: $a > 0, \Delta = -28 < 0 \Rightarrow x \in \mathbb{R}$

$x^{2} + 6 x - 16 < 0$: $\displaystyle \Delta = 100, \ x = \frac{-6 \pm 10}{2} = -8, 2 \Rightarrow x \in (-8, 2)$

Common values: $x \in (-8, 2)$

Answer: $x \in (-8, 2)$

$\sqrt{x^{2} + x - 6}$ is defined when $x^{2} + x - 6 \ge 0$

$\Rightarrow (x - 2)(x + 3) \ge 0$

Domain $= (-\infty, -3] \cup [2, \infty)$

Answer: $(-\infty, -3] \cup [2, \infty)$

$[x]$ is an integer: $[x] = 3, 4$ or $5$

$[x] = 3$ for $3 \le x < 4$, ..., $[x] = 5$ for $5 \le x < 6$

$\therefore x \in [3, 6)$

Answer: $x \in [3, 6)$

$\displaystyle \frac{1}{x^2}$ is not defined at $x = 0$: Domain $= \mathbb{R} - \{0\}$

$x^2 > 0$, so $\displaystyle \frac{1}{x^2} > 0$: Range $= (0, \infty)$

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