In a modulus inequality, why does a rejected critical value appear at all, when it genuinely solves the equation that produced it?

Because that equation is only the correct one inside its own case , and solving it treats the formula as though it held everywhere. On a sketch the rejected value is where the curve's continuation beyond its range would have met the other graph, so it is a real intersection of two extended curves rather than of the actual ones.

A LevelFurther MathsEdexcelFurther Pure 1FlashcardsInequalitiesInequalities

Inequalities (Edexcel A Level Further Maths: Further Pure 1): Flashcards

Exam code: 9FM0

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FrontSolving Inequalities involving Algebraic Fractions
True or False?
An inequality involving algebraic fractions can be solved by multiplying both sides by the lowest common denominator.

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Cards in this collection (15)

True or False?
An inequality involving algebraic fractions can be solved by multiplying both sides by the lowest common denominator.
False.
An algebraic denominator is positive for some values of $x$ and negative for others, and multiplying an inequality by a negative quantity reverses the sign.
Multiplying $\frac{1}{x - 3} < 1$ by $x - 3$ gives $1 < x - 3$ , whose solution $x > 4$ has silently lost the whole of $x < 3$ .
When solving an inequality with algebraic fractions, what should you multiply both sides by instead of the lowest common denominator?
Multiply by the squares of the denominators, such as ${(x - 2)}^{2} {(x - 1)}^{2}$ .
A square is never negative whatever $x$ is, so the sign of the inequality is safe, and each denominator still cancels once against its own numerator.
After multiplying by the squared denominators and bringing everything to one side, why should you not expand the brackets?
Because those brackets are already the factors needed to find the critical values, and expanding would only mean factorising the same expression again.
Take out every common bracket instead, so that $4 (x - 2) {(x - 1)}^{2} - x {(x - 2)}^{2} (x - 1) \geq 0$ becomes $(x - 2) (x - 1) [4 (x - 1) - x (x - 2)] \geq 0$ .
The working reaches $(x - 2) (x - 1) (- x^{2} + 6 x - 4) \geq 0$ . Why multiply through by $- 1$ , and what must you remember?
Multiplying by $- 1$ turns $- x^{2}$ into $+ x^{2}$ , so the quartic is a positive one and its sketch has the familiar shape with both ends rising.
Because $- 1$ is definitely negative you know to reverse the inequality, giving $(x - 2) (x - 1) (x^{2} - 6 x + 4) \leq 0$ .
The polynomial inequality gives $3 - \sqrt{5} \leq x \leq 1$ or $2 \leq x \leq 3 + \sqrt{5}$ , and the original inequality had denominators $x - 2$ and $x - 1$ . Which endpoints must change?
The values $x = 1$ and $x = 2$ must be excluded, because each makes an original denominator zero and the original expression is undefined there.
The answer becomes $3 - \sqrt{5} \leq x < 1$ or $2 < x \leq 3 + \sqrt{5}$ , and the question never arises if the original inequality was strict.
The reciprocal graph $y = \frac{a x + b}{c x + d}$ has one vertical and one horizontal asymptote. Complete both equations:
The vertical one is $x =______$ and the horizontal one is $y =______$ for every such graph.
The completed asymptotes are:
The vertical one is $x = - \frac{d}{c}$ and the horizontal one is $y = \frac{a}{c}$ for every such graph.
Set the denominator equal to zero for the first, and divide top and bottom by $x$ and let $x \to \infty$ for the second.
Why is it useful to rewrite $y = \frac{2 x - 1}{x - 1}$ as $y = 2 + \frac{1}{x - 1}$ ?
The rewritten form shows the curve is a translation of $y = \frac{1}{x}$ , here by $[\begin{matrix} 1 \\ 2 \end{matrix}]$ , so it can be sketched from a shape you already know.
Getting there needs either polynomial division or the trick of writing the numerator as $2 (x - 1) + 1$ .
How can $\frac{4}{x - 2} \geq \frac{x}{x - 1}$ be solved by sketching rather than by algebra?
Sketch $y = \frac{4}{x - 2}$ and $y = \frac{x}{x - 1}$ on the same axes and read off the ranges of $x$ in which the first curve is vertically above the second.
Solving the two equations simultaneously supplies the $x$ -coordinates of the intersections, and those are the ends of the ranges.
How do you sketch $y = | f (x) |$ ?
Sketch $y = f (x)$ first, then reflect in the $x$ -axis every part of it that lies below the axis.
For $y = | x^{2} - 25 |$ that turns the dip between $x = - 5$ and $x = 5$ into a hump, so the curve has corners at $- 5$ and $5$ and never goes below the axis.
The modulus $| x^{2} - 25 |$ splits into two cases. Complete the one that applies between the roots:
$| x^{2} - 25 | =______for - 5 < x < 5$
The completed case is:
$| x^{2} - 25 | = - (x^{2} - 25) for - 5 < x < 5$
Between the roots $x^{2} - 25$ is negative, so the modulus is its negative, which may equally be written $25 - x^{2}$ .
Outside that interval the modulus is simply $x^{2} - 25$ itself.
Solving in the case $x \leq - 5$ or $x \geq 5$ gives $x = 1 \pm 3 \sqrt{3}$ . What must you then do with those two roots?
Check each one against the range of its own case, and reject any that falls outside it.
Here $1 + 3 \sqrt{3} = 6.196 \dots$ satisfies $x \geq 5$ and is kept, while $1 - 3 \sqrt{3} = - 4.196 \dots$ satisfies neither condition and is thrown away.
In a modulus inequality, why does a rejected critical value appear at all, when it genuinely solves the equation that produced it?
Because that equation is only the correct one inside its own case, and solving it treats the formula as though it held everywhere.
On a sketch the rejected value is where the curve's continuation beyond its range would have met the other graph, so it is a real intersection of two extended curves rather than of the actual ones.
True or False?
Putting a modulus on the other side of the inequality as well can create extra critical values.
True.
Each modulus breaks its own graph into two pieces, so two moduli give more pairs of pieces that can meet.
Changing $| x^{2} - 25 | < 2 x + 1$ to $| x^{2} - 25 | < | 2 x + 1 |$ brings in the line $y = - (2 x + 1)$ , and with it two further critical values.
Why does $y = \frac{x - 1}{2 - | x - 1 |}$ have two vertical asymptotes?
Setting the denominator equal to zero gives $| x - 1 | = 2$ , and a modulus equation of that kind has two solutions, $x - 1 = 2$ and $x - 1 = - 2$ .
Those give $x = 3$ and $x = - 1$ , so one modulus in the denominator produces a pair of asymptotes rather than a single one.
For the curve $y = \frac{x - 1}{2 - | x - 1 |}$ , what does the case $x \geq 1$ turn the right-hand side into?
With $x \geq 1$ the modulus is $x - 1$ itself, so the denominator becomes $2 - (x - 1) = 3 - x$ and the curve is $y = \frac{x - 1}{3 - x}$ over that range.
Each case therefore gives a different rational function, which is why the graph is built out of separate pieces.

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