If $f(x)=2px^2 + qx - p + 1\geq 0$ when $|x|\leq 1$ then what is the maximum value of $p+q$.

Question

If I understand the problem correctly, then $f(x)$ is 'exactly' positive whenever $|x|\leq 1$. Since this is a parabola (or quadratic equation) then there seems to be only one option. The x-intercepts are -1 and 1, and the parabola is downward facing. The vertex is obtained for $f(0) = -p+1$, and $f(1)= 0$, $f(-1) = 0$.

Thus $$f(1)= 2p+q -p +1 =0,$$ therefore $p+q=-1$. $$f(-1) = 2p - q - p + 1= 0,$$ therefore, $p-q=-1$.

We have two simultaneous equations.

$$p+q = -1$$

$$p-q = -1.$$

Thus, $2p = -2$, hence $p = -1$, and $q = 0$.

But then the sum $p+q = -1$.

I believe the answer should be $2$. Is there something that I misunderstood?

Answer 1

Of course, the step where you said "there seems to be only one option" is a massive leap that is not justified. The condition that $2px^2+qx-p+1\ge 0$ when $|x|\le 1$ allows for many values of $p$ and $q$. After all, $p=q=0$ works just as well!

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Let's write $f(x)$ as $f(x)=p(2x^2-1)+qx+1\ge 0$. Now, the idea is to find a suitable $x$ where the terms multiplying $p$ and $q$ will be equal, so that, with $2x^2-1=x=c$ for some value $c$ (hopefully in $[-1,1]$), you would have $c(p+q)+1\ge 0$, and this will produce a condition on $p+q$.

So, let's solve the quadratic equation $2x^2-1=x$, i.e. $2x^2-x-1=0$. The solutions are: $x=1$ or $x=-\frac{1}{2}$ and both are in the range $[-1,1]$.

The solution $x=1$ is fruitless: it gives $c=1$ i.e. $p+q+1\ge 0$, i.e. $p+q\ge -1$, which is nice but is a lower bound for $p+q$.

With $x=-\frac{1}{2}$, however, the situation is diferent. It gives $c=-\frac{1}{2}$ and so $-\frac{1}{2}(p+q)+1\ge 0$, i.e. $-\frac{1}{2}(p+q)\ge -1$ i.e. $p+q\le 2$.

Note I could've cheated above and just said "set $x=-\frac{1}{2}$ and this directly gives you $p+q\le 2$", but this won't really tell you where that value $-\frac{1}{2}$ is coming from.

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The proof would be concluded if you can find an example of $p,q$ such that $p+q=2$. This is necessary because, so far, we have found an upper bound for $p+q$, but we have not found the maximum yet.

So let's set $q=2-p$ and see if we can find $p$ so that $f(x)$ satisfies the given conditions. The function now looks like this: $f(x)=p(2x^2-1)+(2-p)x+1=p(2x^2-x-1)+2x+1=\left(x+\frac{1}{2}\right)(2p(x-1)+2)$. This is a quadratic which has a zero at $x=-\frac{1}{2}$, and it will stay non-negative if the other factor $2p(x-1)+2$ has a zero at $x=-\frac{1}{2}$. So, $2p\left(-\frac{1}{2}-1\right)+2=0$, which is true for $p=\frac{2}{3}$. This gives $q=2-p=\frac{4}{3}$.

Note Again I could've cheated and just said "try out $p=\frac{2}{3}, q=\frac{4}{3}$ but again this won't tell you where those values would be coming from.

Answer 2

I think you have misunderstood the question. The answer is 2 if you take it to be asking for the maximum value of $p+q$, with $f(x)$ definitely non-negative when $|x|\le1$, but with unconstrained sign outside that region. I find $p+q$ is then maximised when $p>0$ and the minimum value of the quadratic is zero, giving $p=\frac{2}{3}$ and $q=\frac{4}{3}$. Do you want more detail or is this enough of a hint?

Answer 3

It is weird to ask what is the maximum value of $p + q$ if $(p,q)$ can only have one value. The title you wrote only says that $f(x) \geqslant 0$ when $|x| \leqslant 1$ but not that $f(x) < 0$ when $|x| > 1$.

First case : $p < 0$. In that case, we indeed have a downward facing parabola and $f(x) \geqslant 0$ on $[-1,1]$ if and only if $f(-1) \geqslant 0$ and $f(1) \geqslant 0$ i.e. $p + q \geqslant -1$ and $p - q \geqslant -1$ so $q \leqslant p - 1$ thus $p + q \leqslant 2p + 1 < 1$ since $p < 0$.

Second case : $p = 0$. In that case, $f(x) = qx + 1$ is affine, thus $f(x) \geqslant 0$ on $[-1,1]$ if and only if $q + 1 \geqslant 0$ and $-q + 1 \geqslant 0$, which means that $|q| \leqslant 1$. $p + q = q \leqslant 1$ and this value is attained when $q = 1$.

Third case : $p > 0$. In this last case, $f$ is facing up so you can check (draw it you will see it is quite obvious) that $f(x) \geqslant 0$ on $[-1,1]$ if and only if $f(-1) \geqslant 0$, $f(1) \geqslant 0$ (i.e. $p + q \geqslant -1$ and $p - q \geqslant -1$, which can be sumurised by $|q| \leqslant p + 1$) and one of the two following conditions is fullfilled,

$*$ The minimum of $f$ is non-negtaive i.e. the sign of $f$ doesn't change i.e. $\Delta = q^2 + 8p(p - 1) \leqslant 0$ i.e. $p \leqslant 1$ and $|q| \leqslant 2\sqrt{2p(1 - p)}$,

$*$ The minimum of $f$ is not in the interval $[-1,1]$. Since this minimum is attained at $x_0 = \frac{-q}{4p}$, it means that $|q| \geqslant 4p$.

Notice that all the conditions on $q$ are actually conditions on $|q|$ (which is normal because $[-1,1]$ is symmetric so you can always replace $x$ by $-x$, which changes $q$ into $-q$). Therefore, since we want to maximize $p + q$, we can choose $q \geqslant 0$ whithout loss of generality.

If we want the first condition to be checked, we must have $p \leqslant 1$, $q \leqslant p + 1$ and $q \leqslant 2\sqrt{2p(1 - p)}$. I let you verify that $q \leqslant 2\sqrt{2p(1 - p)} \Rightarrow q \leqslant q + 1$ and $p + q = p + 2\sqrt{2p(1 - p)}$ (in the best case) has for maximum $2$ (study the function $x \mapsto x + 2\sqrt{2x(1 - x)}$) and I let you find for which $(p,q)$ this maximum is reached.

If we want the second condition to be checked, we must have $4p \leqslant q \leqslant p + 1$, so $p \leqslant \frac{1}{3}$ and $q \leqslant p + 1$, thus $p + q$ reaches its maximum when $p = \frac{1}{3}$ and $q = \frac{4}{3}$ which implies $p + q = \frac{5}{3} < 2$.

You deduce that $\max\{p + q\} = 2$.