The question I faced was:

Let $f(x)$ be a non-negative continuous and bounded function for all $x \ge 0$. If
$$(\cos x)f'(x) \le (\sin x - \cos x)f(x), \; \forall \; x \ge 0$$ then which of the following is/are correct?

(A) $f(6) + f(5) > 0$

(B) $x^2 - 3x + 2 + f(9) = 0$ has two distinct solutions

(C) $f(5)f(7) - f(6)f(5) = 0$

(D) $\lim\limits_{x \to 4} \dfrac{f(x) - \sin(\pi x)}{x-4} = 1$

The answer is:

(B), (C)

By observation, $f(x)=0$ satisfies the given conditions. But is it the only solution? If so, how to prove it is?

After rearranging the terms and combining them, I converted the inequality to this form: $$ \left( f(x)\,\cos x \right)' + f(x)\,\cos x \le 0 $$ Despite its allure, this inequality isn't getting me anywhere! It doesn't seem to have any information about $f(x)$, since it is stuck with a "$\cos x$". Even then, I don't see where I can go with it.

So how to solve this problem? Thank you.

  • 2
    $\begingroup$ Your condition is equivalent to $$\frac{\mathrm{d}}{\mathrm{d}x} \ f(x) \cos x \ e^x \le 0$$ $\endgroup$ – Crostul Apr 28 '17 at 10:56
  • $\begingroup$ @BST Actually the discriminant is $(-3)^2-4(2+f(9)) = 1-4f(9)$ which may be negative $\endgroup$ – Crostul Apr 28 '17 at 11:00
  • $\begingroup$ Sorry, you're right. My mistake. I'll remove the comment. I miscalculated the discriminant. $\endgroup$ – be5tan Apr 28 '17 at 11:02

We assume that $f(x)$ is bounded and non-negative, for $x \geqslant 0$. This means that there is a non-negative function $g(x)$ with the property that $f(x) = g(x)e^{-x}$ for $x \geqslant 0$.

Plugging this into the inequality you found gives, \begin{equation} 0 \geqslant (f(x)\cos (x))' + f(x) \cos (x) = e^{-x} (g(x) \cos(x))'. \end{equation} And since $e^{-x} > 0$, we have \begin{equation} 0 \geqslant (g(x) \cos(x))'. \end{equation} This means that the function $g(x) \cos(x)$ is weakly decreasing. Because any point $x \in \mathbb{R}$ is between two zeroes of $\cos(x)$, we have that $g(x) \cos(x) = 0$ for all $x$.

  • $\begingroup$ Instead of integrating the inequality, it gives more information simply to note that the derivative of $g(x)\cos x$ can never be positive. Since every $x$ is between two zeroes of $g(x)\cos x$, the mean value theorem forces it to be $0$ everywhere, not just where the cosine is positive. $\endgroup$ – hmakholm left over Monica Apr 28 '17 at 11:07
  • $\begingroup$ Ah, I missed that! Thank you, will edit my answer. $\endgroup$ – Peter Apr 28 '17 at 11:08
  • $\begingroup$ Also, I think it would be slicker simply to define $g(x)=f(x)e^x$ at the beginning. Then we don't need to appeal to "bounded and nonnegative", and the first inequality is still true. $\endgroup$ – hmakholm left over Monica Apr 28 '17 at 11:22
  • $\begingroup$ This is brilliant! How did you figure out the substitution $f=ge^{-x}$? $\endgroup$ – FreezingFire Apr 28 '17 at 11:28
  • 1
    $\begingroup$ I asked myself what function $h(x)$ would satisfy the strict version of your bound, i.e.~$h'(x) + h(x) = 0$. Evidently the function $h(x) = e^{-x}$ solves this equation. The rest was the result of a bit of experimentation with this function. $\endgroup$ – Peter Apr 28 '17 at 11:56

If you set $g(x)=f(x)\cos x$, your rearrangement tells you that $$ \tag{1} g'(x) \le -g(x) $$

Consider an interval $[2\pi k-\frac12\pi , 2\pi k +\frac12\pi]$ where $\cos x$ is $\ge 0$. Then you know that $g(x)\ge 0$ on this interval, and $g(x)=0$ at the start of the interval. Then, because of (1), $g(x)$ must be identically zero on that interval, and so must $f(x)$.

This settles at least (A) and (C), because $5$, $6$ and $7$ are all within $\pi/2$ of $2\pi$.

For (B) and (D) this argument doesn't tell you enough; I recommend Peter's slicker answer instead.


Notice that whenever some $f$ satisfies the hypothesis, then so does any positive multiple of $f$. In particular:

Note that (B) is equivalent to $$0<(-3)^2-4(2+f(9)) = 1-4f(9)$$ i.e. $f(9)<1/4$. Hence (B) can hold if and only if $f(9) =0$ for any $f$ satisfying the hypothesis.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.