A range proof needs the degenerate cases

(x+5)/(x-1)>9/x

Total marks: 9

Solution

(a)i

Rational inequalities

Without a calculator, solve $\dfrac{x+5}{x-1}>\dfrac9x$ .

Explore this idea

Hint

Move $\frac9x$ across and form a single fraction.
Complete the square on the numerator: does it have real roots?
If not, only the denominator decides the sign.

Answer · (a)i

x<0

Explore the idea

The numerator never changes sign

The factor (x − 2)² + 5 is always positive, so the sign of the fraction comes only from x and x − 1.

\frac{(x-2)^2+5}{x(x-1)}>0

Choose an interval to test it. Shading marks the solution intervals.

Factor
x	negative	positive	positive
x − 1	negative	negative	positive
product	positive	negative	positive

x	x − 1	product
negative	negative	positive
positive	negative	negative
positive	positive	positive

Test value of xx = 0.50

Try an example

(x)+(x − 1)−⇒ product negative

Is it > 0?

x is not a solution

The shaded strips are where the expression satisfies the inequality; they match the green-tinted columns of the chart.

Fraction, and where it is positive
Your test value

Key idea

An always-positive factor can be ignored in the sign analysis. Only the denominator's sign mattered here.

Ideas to try

Test $x=0.5$ : why is it not a solution?
Is there any $x$ where the numerator is zero?

(a)ii

Exponential substitution

Hence solve $\dfrac{e^x+5}{e^x-1}>9e^{-x}$ .

Explore this idea

Hint

Write $9e^{-x}=\frac{9}{e^x}$ .
Compare with (i): $e^x$ plays the role of $x$ .
Keep only positive values of $e^x$ .

Answer · (a)ii

x>0

Explore the idea

e^x > 1 exactly when x > 0

The exponential is always above the x-axis, so e^x < 0 never happens, and it is above 1 only to the right of 0.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

Y = eˣ
Y = 1

Ideas to try

Is the curve ever below the x-axis?

(b)

Range by discriminant

For the function $y=\dfrac{1+kx}{x^2-1}$ , with real $x\ne\pm1$ and $-1\le k\le1$ , find the range of $y$ in terms of $k$ (for $0<|k|<1$ ).

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Hint

A value $y$ is in the range exactly when the equation $y=\frac{1+kx}{x^2-1}$ has a real solution $x$ .
Rearrange into a quadratic in $x$ and require discriminant $\ge0$ .
Solve the resulting inequality in $y$ .

Answer · (b)

y\le(-1-\sqrt{1-k^2})/2

Explore the idea

Which heights does the curve reach?

With k = 0.6 the boundary heights are y = (−1 ± 0.8)/2, i.e. −0.9 and −0.1. Between them the horizontal line misses the curve.

Height of the horizontal linek = -0.5

Try an example

Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct intersections

No intersection in the gap −0.9 < y < −0.1.

There are no isolated intersections at this value. A touching point counts once; hollow endpoints are excluded.

Y = (1 + 0.6x)/(x² − 1)
Y = h
Intersections
Intersection selected in the coordinate table

Key idea

The range is the set of heights the curve reaches; the discriminant of the quadratic in $x$ finds exactly where a horizontal line stops meeting it.

Ideas to try

Find where the intersection count drops to zero. Does it match $-0.9$ and $-0.1$ ?
Try $h=0$ : how many times does the line meet the curve?

A range proof needs the degenerate cases

Solution

Rational inequalities

One side zero

Single fraction

The numerator is always positive

So the denominator decides

The numerator never changes sign

Exponential substitution

Match the form

Use the earlier answer

Reject the impossible

e^x > 1 exactly when x > 0

Range by discriminant

Quadratic in $x$

Real roots need discriminant $\ge0$

Roots of the boundary

Outside the roots

Which heights does the curve reach?

One side zero

Single fraction

The numerator is always positive

So the denominator decides

Match the form

Use the earlier answer

Reject the impossible

Quadratic in xxx

Real roots need discriminant ≥0\ge0≥0

Roots of the boundary

Outside the roots

Quadratic in $x$

Real roots need discriminant $\ge0$