Sonia, we reject the boundary cases a = -5/4 and a = 0 as these lead to straight-line equations with no turning points at all.

dy/dx

= -a + (4a+5)/(2 - x)²

At the turning points, dy/dx = 0

-a + (4a+5)/(2 - x)² = 0

a - (4a+5)/(2 - x)² = 0

a(2 - x)² - (4a+5) = 0

4a - 4ax + ax² - 4a - 5 = 0

ax² - 4ax - 5 = 0


In order to have turning points, there must be real and unequal roots for dy/dx = 0.

Discriminant > 0 (b² - 4ac > 0)

(-4a)² - 4a(-5) > 0

16a² + 20a > 0

4a² + 5a > 0

a(4a + 5) >0

This is a downward sloping curve. We need the region above the x-axis


a > 0 or 4a + 5 < 0

4a < -5

a < -5/4

Sonia, it may be clearer if I type here.

y = -ax - (2a + 4) + (4a + 5) / (2 - x)

dy/dx = -a + (4a + 5) / (2 - x)²

At the stationary point(s), if any,

dy/dx = 0
-a + (4a + 5) / (2 - x)² = 0
a = (4a + 5) / (2 - x)²
a (2 - x)² = 4a + 5
a (4 - 4x + x²) = 4a + 5
4a - 4ax + ax² = 4a + 5
ax² - 4ax - 5 = 0

Since there exists at least one stationary point, the equation dy/dx = 0 must have real roots.

(dy/dx can be thought of as an ordinary graph like "y = ax² - 4ax - 5" except that it's called dy/dx here, so if we draw the graph of dy/dx against x, a cutting along the x-axis would result in real roots to the quadratic equation)

Discriminant >= 0
(-4a)² - 4 (a) (-5) >= 0
16a² + 20a >= 0
4a² + 5a >= 0
a (4a + 5) >= 0
a <= -5/4 or a >= 0

BUT...

At the boundary condition a = -5/4, the equation for y transposes to

y = 5x/4 - 3/2

which is a straight line equation with no turning point at all, contrary to what the equation for dy/dx would suggest.

At the boundary condition a = 0, the equation for y transposes to

y = 5 / (2 - x)

which is another curve with no stationary points.

As such, we reject these boundary cases and therefore a < -5/4 or a > 0.

We don't have to consider the case of real and equal roots (i.e discriminant = 0) because the question already says the curve has turning points.

That phrasing means there are at least 2 turning points so there will be real and unequal roots for dy/dx = 0

A more intuitive way as to why a cannot be 0 and -5/4 :


dy/dx = -a + (4a+5)/(2 - x)²

if a = 0,

dy/dx = 5/(2 - x)² , which > 0 for all real x

(The denominator is always positive since it's squared and we know it cannot be 0 since dividing by 0 is undefined. The numerator is 5 which is positive)

if a = -5/4, 4a + 5 = 0

dy/dx = -5/4 which ≠ 0

We can never get a turning point for these cases.

Depends on what the examiner actually intends the question to be.

Some questions I come across put “find the values of x for which...” only to end up having one value of x (and I therefore felt cheated for the question). Because of that, I do not assume that all setters are good in English and I do not assume that this question automatically means “more than one stationary point”.

For this example, it is quite obvious/intuitive from the equation.

Let's say that one starts this question with the interpretation that 'there is at least one turning point.'

When he/she gets to dy/dx = -a + (4a+5)/(2 - x)², the typical action would be to set it to 0 and solve for discriminant ≥ 0

But looking at the equation, this is just a reciprocal squared graph with horizontal asymptote -a.

For such graphs and knowledge of the topics of graphing techniques (both A level content and also from A Math), there will always be 2 different values of x for any value of y.

(The graph is symmetrical about the vertical asymptote)

This already suggests for any possible value of dy/dx (be it 0 or not), there can only be 2 distinct values of x. (Here dy/dx is the dependent variable)

This matches the phrasing 'has turning points' and naturally the option with only 1 turning point doesn't make sense. Neither can there be more than 2.



Furthermore, since the graph never reaches the asymptote -a , it is quite evident that a cannot be 0 since that would mean dy/dx would never reach 0. And also a has to be above 0 in order to have -a < 0, such that there's intersection with the x-axis (where y = 0), given that a > 0 means the coefficient 4a + 5 is positive and then curve is sloping upwards.