Sec 3 A Math: Quadratic Functions & Inequalities, practice questions & worked solutions

$$\begin{aligned} 3x^2+15x+20 &= 3(x^2+5x)+20\\ &= 3\!\left[(x+2.5)^2-(2.5)^2\right]+20\\ &= 3(x+2.5)^2+1.25 \end{aligned}$$ Since $3(x+2.5)^2+1.25\geq 1.25$, $3x^2+15x+20$ cannot be smaller than 1.

$$\begin{aligned} 4x^2 - 6xy - 6y &= 4 \text{----(1)}\\ y - x &= 4 \text{------(2)} \end{aligned}$$ Sub (2) into (1): $$\begin{aligned} 4x^2 - 6x(4 + x) - 6(4 + x) &= 4\\ 4x^2 - 24x - 6x^2 - 24 - 6x &= 4\\ 2x^2 + 30x + 28 &= 0\\ x^2 + 15x + 14 &= 0\\ (x + 1)(x + 14) &= 0\\ x = -14 \ &\text{or} \ x = -1\\ y = -10 \ &\text{or} \ y = 3 \end{aligned}$$

(a)

$a<0$ for maximum curve $$\begin{aligned} (2b)^2-4a(-1) &< 0 \text{for no roots}\\ 4b^2+4a &< 0\\ b^2+a &< 0 (\text{or } a<-b^2) \end{aligned}$$

(b)

$$a=-5,\qquad b=1$$

(a)

$$\begin{aligned} -4x^2+12x+2 &= -4(x^2-3x)+2\\ &= -4\!\left[x^2-3x+\left(\tfrac{3}{2}\right)^2-\left(\tfrac{3}{2}\right)^2\right]+2\\ &= -4\!\left(x-\tfrac{3}{2}\right)^2+9+2\\ &= -4\!\left(x-\tfrac{3}{2}\right)^2+11 \end{aligned}$$ The turning point is $\left(1\tfrac{1}{2},\,11\right)$.

(b)

For all real values of $x$, $\left(x-\tfrac{3}{2}\right)^2\geq 0$, $-4\!\left(x-\tfrac{3}{2}\right)^2\leq 0$, $-4\!\left(x-\tfrac{3}{2}\right)^2+11\leq 11$. Since coefficient of $x^2$ is negative, $\left(1\tfrac{1}{2},\,11\right)$ is a maximum point.

(a)

$$\begin{aligned} P &= 120x - 40 - 30x^2\\ &= -30(x^2 - 4x) - 40\\ &= -30\left[x^2 - 4x + (-2)^2 - (-2)^2\right] - 40\\ &= -30[x - 2]^2 + 80 \end{aligned}$$

(b)

No. As he will be making a loss of $40 if he accept the order.

(a)

$$\begin{aligned} x^2 - x - 2 &= px + 2p\\ x^2 + (-1 - p)x - 2 - 2p &= 0 \end{aligned}$$ $$\begin{aligned} &\Rightarrow \text{discriminant} < 0\\ &\Rightarrow (-1 - p)^2 - 4(1)(-2 - 2p) < 0\\ &\Rightarrow 1 + 2p + p^2 + 8 + 8p < 0\\ &\Rightarrow p^2 + 10p + 9 < 0\\ &\Rightarrow (p + 9)(p + 1) < 0\\ &\Rightarrow -9 < p < -1 \end{aligned}$$

(b)

Line $y = p(x + 2)$ does not intersect curve $y = (x + 1)(x - 2)$ when $p$ is in the range $-9 < p < -1$. For $p = 3$, line intersects curve at 2 points.

$$\begin{aligned} y &= \frac{7}{2}x - 24\\ \sqrt{\left(\tfrac{7}{2}x - 24 + 11\right)^2 + x^2} &= 17\\ \frac{53}{4}x^2 - 91x - 120 &= 0 \end{aligned}$$ $$\begin{aligned} x &= \frac{364 \pm \sqrt{(-364)^2 - 4(53)(-480)}}{2 \times 53}\\ x &= \frac{364 \pm \sqrt{254256}}{106}\\ &= 8 \text{or} -1\frac{7}{53}\ (\text{or } -1.13)\\ y &= 4 \text{or} -27\frac{51}{53}\ (\text{or } -28.0) \end{aligned}$$

(a)

Since Side $> 0$, $$\begin{aligned} 3x - 5 > 0 \ &\text{and}\ x - 10 > 0\\ x > \frac{5}{3} \ &\text{and}\ x > 10 \text{------(1)} \end{aligned}$$ $$\begin{aligned} \text{Area of rectangle} &\leq 200\\ (3x - 5)(x - 10) &\leq 200\\ 3x^2 - 35x + 50 &\leq 200\\ 3x^2 - 35x - 150 &\leq 0\\ (3x + 10)(x - 15) &\leq 0\\ -\frac{10}{3} \leq x &\leq 15 \text{------(2)} \end{aligned}$$ From (1) & (2), $$10 < x \leq 15$$

(b)

To enclose the field, $$\begin{aligned} \text{Perimeter} &\leq 98\\ 2(3x - 5) + 2(x - 10) &\leq 98\\ 2(4x - 15) &\leq 98\\ 4x - 15 &\leq 49\\ x &\leq 16 \text{------(3)} \end{aligned}$$ From part (a) $10 < x \leq 15$ and (3) $x \leq 16$, it is enough to enclose the field.

(a)

$$\begin{aligned} (2x + 7)(x - 2) &> 0\\ 2x^2 + 3x - 14 &> 0\\ x^2 + ax - b &> 0\\ 2x^2 + 2ax - 2b &> 0\\ 2a &= 3\\ a &= 1.5\\ 2b &= 14\\ b &= 7 \end{aligned}$$ $$x^2 + 1.5x > 7$$

(b)

$$\begin{aligned} b^2 - 4ac &< 0\\ (-p - 3)^2 - 4(1)(-p) &< 0\\ p^2 + 6p + 9 + 4p &< 0\\ p^2 + 10p + 9 &< 0\\ (p + 1)(p + 9) &< 0\\ -9 < p &< -1 \end{aligned}$$

(c)

$$\begin{aligned} -5x - 2 &= kx^2 + 3\\ kx^2 + 5x + 5 &= 0\\ b^2 - 4ac &= 25 - 4k(5)\\ &= 25 - 20k\\ k &< 1\\ -20k &> -20\\ 25 - 20k &> 5\\ &> 0 \end{aligned}$$ Since $b^2 - 4ac > 0$, for $k < 1$, the line intersects the curve.

(a)

$$\begin{aligned} 10x^2 - nx + 2m &= m - 6x\\ 10x^2 + (6 - n)x + m &= 0 \end{aligned}$$ Since line is tangent to curve, $$\begin{aligned} (6 - n)^2 - 4(10)m &= 0\\ (6 - n)^2 &= 40m \end{aligned}$$ Since $(6 - n)^2 \geq 0$, $\ 40m \geq 0$, $\ m \geq 0$.

Hence $m$ cannot be negative.

(b)

(i) $$\begin{aligned} 2x^2 - 6x + 1 + k &> 0\\ (-6)^2 - 4(2)(1 + k) &< 0\\ 28 - 8k &< 0\\ k &> 3.5 \end{aligned}$$

(ii) $$\begin{aligned} kx^2 + (k + 1)x + k &< 0\\ \text{Discriminant} &< 0\\ (k + 1)^2 - 4k^2 &< 0\\ -3k^2 + 2k + 1 &< 0\\ 3k^2 - 2k - 1 &> 0\\ (3k + 1)(k - 1) &> 0\\ k < -\tfrac{1}{3} &\text{or} k > 1 \end{aligned}$$ Since $k < 0$, $\ k < -\dfrac{1}{3}$.