TMUA Official 2022 Paper 2 · Question 17

A student answered the following question:

$a$ and $b$ are non-zero real numbers.

Prove that the equation $x^3+ax^2+b=0$ has three distinct real roots if

27b\left(b+\frac{4a^3}{27}\right)<0.

Here is the student's solution:

I. We differentiate $y=x^3+ax^2+b$ to get $\dfrac{\dd y}{\dd x}=3x^2+2ax=x(3x+2a)$ .

Solving $\dfrac{\dd y}{\dd x}=0$ shows that the stationary points are at $(0,b)$ and $\left(-\dfrac{2a}3,b+\dfrac{4a^3}{27}\right)$ .
II. If $27b\left(b+\dfrac{4a^3}{27}\right)<0$ , then $b$ and $b+\dfrac{4a^3}{27}$ must have opposite signs, and so one of the stationary points is above the $x$ -axis and one is below.
III. If the cubic has three distinct real roots, then one of the stationary points is above the $x$ -axis and one is below.
IV. Hence if $27b\left(b+\dfrac{4a^3}{27}\right)<0$ , then the equation has three distinct real roots.

Which one of the following options best describes the student's solution?

AIt is a completely correct solution.

BThe student has instead proved the converse of the statement in the question.

CThe solution is wrong, because the student should have stated step II after step III.

DThe solution is wrong, because the student should have shown the converse of the result in step II.

EThe solution is wrong, because the student should have shown the converse of the result in step III.

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