A calculus of the absurd
4.8.4 Some examples

Example 4.8.1 Find the complete set of values satisfying the equation \(\abs {x2} \leqq \abs {2x  6}\). ^{26}^{26} This question came from the OCR A June 2019 Single Maths Pure and Mechanics Paper
Solution: We can use the definition \(\abs {x} = \sqrt {x^2}\) to rewrite the inequality above as
\(\seteqnumber{0}{4.}{44}\)\begin{equation} \sqrt {(x2)^2} \leqq \sqrt {(2x6)^2} \end{equation}
Squaring both sides gives the following result
\(\seteqnumber{0}{4.}{45}\)\begin{equation} (x2)^2 = (2x6)^2 \end{equation}
which can then be expanded to give
\(\seteqnumber{0}{4.}{46}\)\begin{equation} x^2  4x + 4 \leqq 4x^2  24x + 36 \end{equation}
and then, being careful not to subtract anything incorrectly (which I always do) the equation can be reduced to this quadratic
\(\seteqnumber{0}{4.}{47}\)\begin{equation} \label {inequality of modulus function} 0 \leqq 3x^2  20x + 32 \end{equation}
We can then sketch this to work out when it would be greater than zero (or zero would be less than the curve).
From the graph, the inequality is true whenever \(x\) is less than the first time it is zero, and whenever the inequality is greater than the second time it is zero. To find when \(3x^2  20x + 32=0\), we can use the quadratic formula, and thus obtain an equation for the values of \(x\).
\(\seteqnumber{0}{4.}{48}\)\begin{equation*} x = \frac {20 \pm \sqrt {20^2  4(3)(32)}}{2(3)} \end{equation*}
Subtracting a positive number from a positive number (note that \(\sqrt {20^2  4(3)(32)}\) is a positive number) gives a smaller value than adding something to a positive number. Therefore, we can deduce (with aid from a calculator) that the smaller value of \(x\) is \(\frac {3}{8}\) and the larger value of \(x\) is \(4\), leaving us with two regions in which the inequality is satisfied.
\(\seteqnumber{0}{4.}{48}\)\begin{equation*} x \leqq \frac {3}{8} \text { and } x \geqq 4 \end{equation*}
A slightly harder example: Find the complete set of values satisfying the inequality
\[\abs {\abs {x1}  5} < 3\]
^{27}^{27} This question came from https://madasmaths.com
Solution: This question is a bit fiddly. The first step is to square both sides (which is fine, because the values we’re taking square roots of are always positive).
\(\seteqnumber{0}{4.}{48}\)\begin{align} & (\abs {x1}  5)^2 < 3^2 \\ & (x1)^2  10\sqrt {(x1)^2} + 25 < 9 \\ & (x1)^2  10\sqrt {(x1)^2} + 16 < 0 \end{align}
This is just a quadratic ^{28}^{28} Review the section on hidden quadratics if you’re not sure what this is. TODO: write this section, and we can substitute \(z=\sqrt {(x1)^2}\) to get that
\(\seteqnumber{0}{4.}{51}\)\begin{align} & z^2  10z + 16 < 0 \\ & (z  8)(z  2) < 0 \end{align}
We can sketch this graph to work out where the inequality is true ^{29}^{29} The other way to do this is to think about where one of the brackets is positive, and the other negative, but in my experience it’s a more errorprone method.
From the graph, the inequality is true whenever \(2 < z < 8\) (i.e. \(z\) is between the roots).
Thus we have \(2 < \sqrt {(x1)^2} < 8\). This means that both of these inequalities are true:
\(\seteqnumber{0}{4.}{53}\)\begin{align} & \sqrt {(x1)^2} < 8 \\ & \sqrt {(x1)^2} > 2 \end{align}
If we square both sides, we get that
\(\seteqnumber{0}{4.}{55}\)\begin{align} & (x1)^2 < 64 \\ & (x1)^2 > 4 \end{align}
Here’s a time to be careful; \((x1)^2\) is always positive. When we take the square root, however, we can have either the positive or negative square root. Note that the negative square root is just \(1\) times \(\sqrt {\text {whatever}}\), and thus we need to flip the inequality.
Overall then we have these four inequalities.
\(\seteqnumber{0}{4.}{57}\)\begin{align} & x1 < 8 \\ & x1 > 8 \\ & x1 > 2 \\ & x1 < 2 \end{align}
Which can all be arranged to obtain that the following four inequalities must all be true for the inequality to hold
\(\seteqnumber{0}{4.}{61}\)\begin{align} & x < 9 \\ & x > 7 \\ & x > 3 \\ & x < 1 \end{align}
Examining these, it is clear that there are two regions in which this is true: \(7 < x < 1\) and \(3 < x < 9\). This is the complete set of regions which satisfy the inequality. We can write it using set theory notation as
\(\seteqnumber{0}{4.}{65}\)\begin{equation} \left \{7 < x <  1\right \} \cup \left \{3 < x < 9\right \} \end{equation}
where \(\cup \) denotes the "union" operator (which means that overall we have the set of objects in either \(\left \{7 < x <  1\right \}\), \(\left \{3 < x < 9\right \}\), or in both of them  which in this case is nothing, as the two ranges don’t overlap).

Example 4.8.2 Solve the equation
\(\seteqnumber{0}{4.}{66}\)\begin{equation} \abs {\frac {1}{9} (8t  9)} = \abs {\frac {1}{3}(2t11)} \end{equation}
Solution: this can be solved by using the fact that \(\abs {x} = \sqrt {x^2}\), but the numbers are not nice, and I made so many arithmetic errors (it was painful). The other way is to think about the definition of the modulus function.
Instead, we can think about the definition of the modulus function. For each side of the equation we have
\(\seteqnumber{0}{4.}{67}\)\begin{equation} \abs {\frac {1}{9}\left (8t9\right )} = \begin{cases} \frac {1}{9}\left (8t9\right ) & \frac {1}{9}\left (8t9\right ) \geqq 0 \\ \frac {1}{9}\left (8t9\right ) & \frac {1}{9}\left (8t9\right ) < 0 \end {cases} \end{equation}
\(\seteqnumber{0}{4.}{68}\)\begin{equation} \abs {\frac {1}{3}(2t11)} = \begin{cases} \frac {1}{3}(2t11) & \frac {1}{3}(2t11) \geqq 0 \\ \frac {1}{3}(2t11) & \frac {1}{3}(2t11) < 0 \end {cases} \end{equation}
Thinking about the cases, the graphs can intersect either

• Where the argument to the modulus function is less than zero, and thus has been multiplied by negative one, for both functions.

• Where the argument of one of the functions is less than zero, and the other is greater than zero.

• Where the argument of both functions is greater than zero.
In the first case, the values of \(t\) where the two curves intersect would be when
\(\seteqnumber{0}{4.}{69}\)\begin{equation} (1) \cdot \frac {1}{9}\left (8t9\right ) = (1) \cdot \frac {1}{3}(2t11) \end{equation}
this is just the same as
\(\seteqnumber{0}{4.}{70}\)\begin{equation} \label {case 1 and case 3} \frac {1}{9}\left (8t9\right ) = \frac {1}{3}(2t11) \end{equation}
Which is the same equation as for case three.
The only other case is when
\(\seteqnumber{0}{4.}{71}\)\begin{equation} \label {case 2 only}  \frac {1}{9}\left (8t9\right ) = \frac {1}{3}(2t11) \end{equation}
Note that it doesn’t matter which one is less than zero (and has thus been multiplied by \(1\)), as we can just multiply both sides by \(1\) to get from one to the other.
Thus solving Equation 4.71, we have
\(\seteqnumber{0}{4.}{72}\)\begin{align} & \frac {1}{9}\left (8t9\right ) = \frac {1}{3}(2t11) \\ & 8t9 = 3(2t11) \\ & 8t  9 = 6t  33 \\ & 2t = 24 \\ & t =  12 \end{align}
And solving Equation 4.72 we have
\(\seteqnumber{0}{4.}{77}\)\begin{align} &  \frac {1}{9}\left (8t9\right ) = \frac {1}{3}(2t11) \\ & (8t9) = 3(2t11) \\ & 8t + 9 = 6t  33 \\ & 14t = 42 \\ & t = 3 \end{align}