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2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(3)

This problem appears at the Qualifying Examinations for Applicants for Japanese Government (MEXT) Scholarships 2020. There are two mathematics exams: one for biology-related natural sciences (Mathematics A), and another for physics- and engineering-related natural sciences (Mathematics B). This problem is from the 2020 Mathematics (B) questionnaire. The official answer key is here.

Problem 1(3)

Let $a$ and $b$ be real numbers with $b \geq 0$. When the equation $$ x^4 + ax^2 + b = 0 $$ has exactly two real solutions, the minimum value of $a + 2b$ is $\fbox{ A }$, and the maximum value of $\lceil a - b \rceil$ is $\fbox{ B }$. Here, $\lceil r \rceil$ denotes the smallest integer that is larger than or equal to the real number $r$.

Solution

The most important hint is the phrase "has exactly two real solutions". This means that there are no complex solutions, and at least one or both of the real solutions have a multiplicity greater than $1$.

From the quadratic formula, we know that $$ x^2 = \frac{-a \pm \sqrt{a^2 - 4b}}{2}, $$ and therefore, $$ x = \pm\sqrt{\frac{-a \pm \sqrt{a^2 - 4b}}{2}}. $$

We notice that to avoid complex roots, the radicand $a^2 - 4b$ must either be positive or $0$. However, if it were positive, then we will still have $4$ roots. Therefore, the only option is for this quantity to be zero, i.e. $$ （★）a^2 - 4b = 0 \Rightarrow a = \pm 2\sqrt{b}, $$ so that the solutions are $$ （★★）x = \pm\sqrt{-\frac{a}{2}}, $$ which also implies that $$ （★★★）a \leq 0 \Rightarrow a = -2\sqrt{b}(\because（★）), $$ so that the term inside the radical is positive.

First, we are interested in the minimum value of $a +2b$. From the previous equations, we have $$\begin{align} a^2 &= 4b\\ a &= -2 \sqrt{b} \hbox{ } \because a\leq 0\\ a +2b&= -2 \sqrt{b} + 2b \\ &=2\left(b - \sqrt{b}\right)\\ &=2\left(\sqrt{b }^2- \sqrt{b}\right)\\ &=2\left(\sqrt{b }^2- \sqrt{b} + \frac{1}{4} - \frac{1}{4}\right)\\ &=2\left(\sqrt{b }^2- \sqrt{b} + \frac{1}{4}\right) - 2\cdot\frac{1}{4}\\ &=2\left(\sqrt{b }^2- \sqrt{b} + \frac{1}{4}\right) - \frac{1}{2}\\ a +2b &=2\left(\sqrt{b}- \frac{1}{2}\right)^2 - \frac{1}{2}. \end{align}$$ Looking at the right-hand side of the last line, we notice that this is a parabola in $b$ opening upwards and with the minimum at the vertex $\left(\frac{1}{2},-\frac{1}{2}\right)$. Thus the minimum value of $a + 2b$ is $-\frac{1}{2}$, which occurs at $b=\frac{1}{2}$. An illustration of this is in the figure below. The purple region is excluded from the possible values because in that region, $\sqrt{b} \lt 0$, which does not make sense especially because the problem stipulates that $b \geq 0$.

Now, we turn to the expression $\lceil a-b\rceil$. From the previous paragraph, we have $$\begin{align} a &= -2 \sqrt{b} \\ a - b &= -2 \sqrt{b} - b \\ &= - \left(b + 2\sqrt{b}\right) \\ &= - \left(\sqrt{b}^2 + 2\sqrt{b}+1-1\right) \\ &= - \left(\sqrt{b}^2 + 2\sqrt{b} + 1\right) + 1 \\ a -b &= - \left(\sqrt{b} + 1\right)^2 + 1. \end{align}$$ Looking at the right-hand side of the last line, we notice that this is a parabola in $\sqrt{b}$ opening downwards and with the maximum at the vertex $\left(-1,1\right)$. Thus the maximum value of $a - b$ is $1$, which occurs at $\sqrt{b} = -1$. From the second half of （★★★）, $\sqrt{a} = (-2)(-1) = 2>0$, which contradicts the first half of （★★★）. Furthermore, if we substitute this value into （★★）, we find that $x$ becomes complex! Therefore, $\sqrt{b} \neq -1$.

To see what is going on, we sketch the graph of $a-b = - \left(\sqrt{b} + 1\right)^2 + 1$, letting the vertical axis be $a-b$ and the horizontal axis be $\sqrt{b}$. This is shown below, where the purple region is the region not considered because in that region, $\sqrt{b} \lt 0$ and $a > 0$.

Working from the given ($b \geq 0$), we have $$\begin{align} \sqrt{b} &\geq 0\\ \sqrt{b} + \frac{1}{2} &\geq \frac{1}{2}\\ \left(\sqrt{b} + \frac{1}{2}\right)^2 &\geq \frac{1}{4}\\ -2\left(\sqrt{b} + \frac{1}{2}\right)^2 &\leq -\frac{1}{2}\\ -2\left(\sqrt{b} + \frac{1}{2}\right)^2 + \frac{1}{2}&\leq -\frac{1}{2}+ \frac{1}{2}\\ a -b=-2\left(\sqrt{b} + \frac{1}{2}\right)^2 + \frac{1}{2}&\leq 0\\ a -b&\leq 0.\\ \lceil a -b \rceil&\leq \lceil 0 \rceil = 0.\\ \end{align}$$

Therefore, the minimum value of $a+2b$ is $-\frac{1}{2}$ and the maximum value of $\lceil a-b\rceil$ is $0$. ■

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