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Ex 2C
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Solutions
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(a)
$$ -4 < x < 7 $$
(b)
$$ x < -0.4 \text{ or } x > 3 $$
(c)
\begin{align*}
2x^2 + 7x & \ge 4 \\
2x^2 + 7x - 4 & \ge 0 \\
(x + 4)(2x - 1) & \ge 0
\end{align*}
$$ x \le -4 \text{ or } x \ge 0.5 $$
(d)
\begin{align*}
3x^2 & \le x^2 - x + 3 \\
2x^2 + x - 3 & \le 0 \\
(2x + 3)(x - 1) & \le 0
\end{align*}
$$ -1.5 \le x \le 1 $$
Question 2 - Expression is always positive/negative
(i)
\begin{align*}
2px^2 & + 7x + 18p \\
\\
a & > 0 \phantom{00000} [\text{Minimum curve } \cup] \\
2p & > 0 \\
p & > {0 \over 2} \\
p & > 0 \phantom{00000} [\text{Condition #1}] \\
\\
b^2 - 4ac & = (7)^2 - 4(2p)(18p) \\
& = 49 - 8p(18p) \\
& = 49 - 144p^2 \\
\\
b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\
49 - 144p^2 & < 0 \\
144p^2 - 49 & > 0 \\
(12p)^2 - (7)^2 & > 0 \\
(12p + 7)(12p - 7) & > 0
\end{align*}
\begin{align*}
p < - {7 \over 12} & \text{ or } p > {7 \over 12} \phantom{00000} [\text{Condition #2}] \\
\\
\therefore p& > {7 \over 12} \phantom{00000} [\text{Satisfy conditions #1 & #2}]
\end{align*}
(ii)
\begin{align*} 2px^2 & + 7x + 18p \\ \\ a & < 0 \phantom{00000} [\text{Maximum curve } \cap] \\ 2p & < 0 \\ p & < {0 \over 2} \\ p & < 0 \phantom{00000} [\text{Condition #1}] \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ \\ [\text{Same as} & \text{ part (i)}] \\ \\ p < - {7 \over 12} & \text{ or } p > {7 \over 12} \phantom{00000} [\text{Condition #2}] \\ \\ \therefore p & < -{7 \over 12} \end{align*}
Question 3 - Expression is always negative
\begin{align*}
-x^2 & +qx + (3q + 8) \\
\\
a & = -1 <0 \phantom{00000} [\text{Maximum curve } \cap] \\
\\
b^2 - 4ac & = (q)^2 - 4(-1)(3q + 8) \\
& = q^2 + 4(3q + 8) \\
& = q^2 + 12q + 32 \\
\\
b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\
q^2 + 12q + 32 & < 0 \\
(q + 8)(q + 4) & < 0
\end{align*}
$$ -8 < q < -4 $$
$$ \therefore \text{Smallest integer value of } q = -7 $$
(i)
\begin{align*}
y & = x + k \phantom{000} \text{ --- (1)} \\
y & = 7x - kx^2 \phantom{000} \text{ --- (2)} \\
\\
\text{Substitute } & \text{(1) into (2),} \\
x + k & = 7x - kx^2 \\
kx^2 - 6x + k & = 0 \\
\\
b^2 - 4ac & = (-6)^2 - 4(k)(k) \\
& = 36 - 4k^2 \\
\\
b^2 - 4ac & \ge 0 \phantom{00000} [\text{Real roots since line intersects the curve - can be once or twice}] \\
36 - 4k^2 & \ge 0 \\
4k^2 - 36 & \le 0 \\
k^2 - 9 & \le 0 \\
(k + 3)(k - 3) & \le 0
\end{align*}
$$ -3 \le k \le 3 $$
(ii)
\begin{align*} b^2 - 4ac & = 0 \phantom{00000} [\text{1 real root since line is tangent to curve}] \\ 36 - 4k^2 & = 0 \\ -4k^2 & = -36 \\ k^2 & = {-36 \over -4} \\ k^2 & = 9 \\ k & = \pm \sqrt{9} \\ k & = \pm 3 \end{align*}
\begin{align*}
x(5x - 3) & < 2x(x - 4) + 2 \\
5x^2 - 3x & < 2x^2 - 8x + 2 \\
3x^2 + 5x - 2 & < 0 \\
(x + 2)(3x - 1) & < 0
\end{align*}
$$ -2 < x < {1 \over 3} $$
\begin{align*}
\text{Consider } (3x - 5)^2 - \left({3 \over 2}\right)^2 & = 0 \\
(3x - 5)^2 & = \left(3 \over 2\right)^2 \\
3x - 5 & = \pm \sqrt{\left(3 \over 2\right)^2} \\
3x - 5 & = \pm 1.5 \\
3x & = \pm 1.5 + 5 \\
3x & = 6.5 \text{ or } 3.5 \\
x & = {13 \over 6} \text{ or } {7 \over 6} \\
\\
\therefore (3x - 5)^2 - \left({3 \over 2}\right)^2 & \ge 0 \\
\left(x - {7 \over 6}\right) \left(x - {13 \over 6}\right) & \ge 0
\end{align*}
$$ x \le {7 \over 6} \text{ or } x \ge {13 \over 6} $$
Question 7 - Solve simultaneous inequality
$$ 1 - x < (x - 1)(5 - x) < 3 $$
\begin{array} {r l c r l }
1 - x & < (x - 1)(5 - x) & \phantom{0} \text{ or } \phantom{0} & (x - 1)(5 - x) & < 3 \\
1 - x & < 5x - x^2 - 5 + x & & 5x - x^2 - 5 + x & < 3 \\
0 & < -x^2 + 7x - 6 & & -x^2 + 6x - 8 & < 0 \\
0 & > x^2 - 7x + 6 & & x^2 - 6x + 8 & > 0 \\
0 & > (x - 1)(x - 6) & & (x - 2)(x - 4) & > 0
\end{array}
$$ 1 < x < 6 \text{ or } x < 2 \text{ or } x > 4 $$
Represent the two inequalities on a common number line and identify the overlapping region(s):
$$ 1 < x < 2 \text{ or } 4 < x < 6 $$
Question 8 - Expression is always positive
\begin{align*}
x^2 + 2k + 10 & = x - 3kx \\
x^2 + 3kx - x + 2k + 10 & = 0 \\
x^2 + (3k - 1)x + (2k + 10) & = 0 \\
\\
a & = 1 > 0 \phantom{00000} [\text{Minimum curve } \cup] \\
\\
b^2 - 4ac & = (3k - 1)^2 - 4(1)(2k + 10) \\
& = (3k)^2 - 2(3k)(1) + (1)^2 - 4(2k + 10) \\
& = 9k^2 - 6k + 1 - 8k - 40 \\
& = 9k^2 - 14k - 39 \\
\\
b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\
9k^2 - 14k - 39 & < 0 \\
(9k + 13)(k - 3) & < 0
\end{align*}
$$ -{13 \over 9} < k < 3 $$
(i)
\begin{align*}
(p + 1)x^2 + 4px & = 8x - 2p \\
(p + 1)x^2 + 4px - 8x + 2p & = 0 \\
(p + 1)x^2 + (4p - 8)x + 2p & = 0 \\
\\
b^2 - 4ac & = (4p - 8)^2 - 4(p + 1)(2p) \\
& = (4p)^2 - 2(4p)(8) + (8)^2 - (4p + 4)(2p) \\
& = 16p^2 - 64p + 64 - (8p^2 + 8p) \\
& = 16p^2 - 64p + 64 - 8p^2 - 8p \\
& = 8p^2 - 72p + 64 \\
\\
b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\
8p^2 - 72p + 64 & < 0 \\
p^2 - 9p + 8 & < 0 \\
(p - 1)(p - 8) & < 0
\end{align*}
$$ 1 < p < 8 $$
(ii)
\begin{align*} b^2 - 4ac & = 0 \\ (p - 1)(p - 8) & = 0 \\ \\ p & = 1 \text{ or } 8 \end{align*}
Question 10 - Line touches a curve
\begin{align*} y & = {1 \over 2}x + p \phantom{000} \text{ --- (1)} \\ x^2 + y^2 & = 8p \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + \left({1 \over 2}x + p \right)^2 & = 8p \\ x^2 + \left({1 \over 2}x\right)^2 + 2\left({1 \over 2}x\right)(p) + (p)^2 & = 8p \\ x^2 + {1 \over 4}x^2 + px + p^2 & = 8p \\ {5 \over 4}x^2 + px + p^2 - 8p & = 0 \\ 5x^2 + 4px + 4p^2 - 32p & = 0 \\ \\ b^2 - 4ac & = (4p)^2 - 4(5)(4p^2 - 32p) \\ & = 16p^2 - 20(4p^2 - 32p) \\ & = 16p^2 - 80p^2 + 640p \\ & = -64p^2 + 640p \\ \\ b^2 - 4ac & = 0 \phantom{00000} [\text{Real and equal roots since line touches the curve}] \\ -64p^2 + 640p & = 0 \\ -64p(p - 10) & = 0 \\ \\ p = 0 \phantom{0} & \text{ or } \phantom{0} p - 10 = 0 \\ & \phantom{00000000} p = 10 \end{align*}
(i)
\begin{align*} y & = mx + c \phantom{000} \text{ --- (1)} \\ x^2 + y^2 & = 4 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + (mx + c)^2 & = 4 \\ x^2 + (mx)^2 + 2(mx)(c) + (c)^2 & = 4 \\ x^2 + m^2 x^2 + 2mcx + c^2 & = 4 \\ (1 + m^2)x^2 + 2mcx + (c^2 - 4) & = 0 \\ \\ b^2 - 4ac & = (2mc)^2 - 4(1 + m^2)(c^2 - 4) \\ & = 4m^2 c^2 - 4(c^2 - 4 + m^2 c^2 - 4m^2) \\ & = 4m^2 c^2 - 4c^2 + 16 - 4m^2 c^2 + 16m^2 \\ & = 16m^2 - 4c^2 + 16 \\ \\ b^2 - 4ac & = 0 \phantom{00000} [\text{Real and equal roots since line is tangent to curve}] \\ 16m^2 - 4c^2 + 16 & = 0 \\ 16m^2 & = 4c^2 - 16 \\ 4m^2 & = c^2 - 4 \phantom{00000} \text{ (Proven)} \end{align*}
(ii)
\begin{align*} 4m^2 & = c^2 - 4 \\ \\ \text{Let } & c = 2, \\ 4m^2 & = (2)^2 - 4 \\ 4m^2 & = 0 \\ m^2 & = 0 \\ m & = \pm \sqrt{0} \\ m & = 0 \\ \\ y & = mx + c \\ y & = (0)x + (2) \\ y & = 2 \\ \\ \therefore \text{Equation of line: } & \phantom{.} y = 2 \end{align*}
Question 12 - Real-life problem
\begin{align*}
\text{Profit} & = \text{Revenue} - \text{Cost} \\
& = R(x) - C(x) \\
& = 5 - 3x - (17 - 8x - 2x^2) \\
& = 5 - 3x - 17 + 8x + 2x^2 \\
& = 2x^2 + 5x - 12 \\
\\
\text{Profit} & \le 0 \phantom{00000} [\text{No profit}] \\
2x^2 + 5x - 12 & \le 0 \\
(x + 4)(2x - 3) & \le 0
\end{align*}
$$ -4 \le x \le 1.5 $$
\begin{align*}
\therefore \text{No. of years} & = 1.5 \\
\\
\text{No. of months} & = 1.5 \text{ years} \times 12 \\
& = 18
\end{align*}
(i)
\begin{align*} y & = x^2 - 2(a + b)x + 2b \phantom{000} \text{ --- (1)} \\ y & = 2c \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 2c & = x^2 - 2(a + b)x + 2b \\ 0 & = x^2 + (-2a - 2b)x + 2b - 2c \\ \\ b^2 - 4ac & = (-2a - 2b)^2 - 4(1)(2b - 2c) \\ & = (-2a)^2 - 2(-2a)(2b) + (2b)^2 - 4(2b - 2c) \\ & = 4a^2 + 8ab + 4b^2 - 8b + 8c \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots since line does not intersect curve}] \\ 4a^2 + 8ab + 4b^2 - 8b + 8c & < 0 \\ a^2 + 2ab + b^2 - 2b + 2c & < 0 \\ (a + b)^2 - 2b + 2c & < 0 \\ (a + b)^2 & < 2b - 2c \end{align*}
(ii)
\begin{align*}
(a + b)^2 & < 2b - 2c \\
\\
\text{Let } & a = 1 \text{ and } b = 2, \\
(1 + 2)^2 & < 2(2) - 2c \\
(3)^2 & < 4 - 2c \\
9 & < 4 - 2c \\
2c & < 4 - 9 \\
2c & < - 5 \\
c & < {-5 \over 2} \\
c & < -2.5 \\
\\
c & = -3 \\
\\
\text{Line: } \phantom{.} y & = -6 \\
\text{Curve: } \phantom{.} y & = x^2 - 6x + 4
\end{align*}