Let f be a positive continuous function on the closed interval from 0 to π/2, and suppose that the integral of e to the f(x) times sin x times ln f(x), and the integral of e to the f(x) times cos x times ln f(x), are both zero. Prove there is a point x₁ in the open interval with f(x₁) = 1.

Step-by-step solution

1. 1

   Rephrase the target in terms of $\ln f$. Since $f>0$, the composition $g(x)=\ln f(x)$ is defined and continuous on the whole interval, and

   $f(x_1)=1\iff g(x_1)=0.$

   So the claim is that $g$ has a zero in $\left(0,\tfrac{\pi}{2}\right)$. Converting a "hits the value $1$" statement into a "has a root" statement is what lets the intermediate value theorem be used.

2. 2

   Identify the sign of the weight. On the open interval $\left(0,\tfrac{\pi}{2}\right)$ we have $\sin x>0$, and $e^{f(x)}>0$ everywhere because the exponential is always positive. Hence the weight

   $w(x)=e^{f(x)}\sin x$

   is strictly positive on $\left(0,\tfrac{\pi}{2}\right)$, and the sign of the integrand $w(x)g(x)$ is decided entirely by $g$.

3. 3

   Argue by contradiction using the intermediate value theorem. Suppose $g$ has no zero on $\left(0,\tfrac{\pi}{2}\right)$. A continuous function with no zero on an interval cannot change sign — if it took both a positive and a negative value the IVT would force a zero in between. So either $g>0$ throughout or $g<0$ throughout.

4. 4

   Derive the contradiction from $I_1=0$. In the first case $w g>0$ on the open interval, so the continuous integrand is strictly positive there and

   $I_1=\int_0^{\pi/2}w(x)g(x)\,dx>0.$

   In the second case the same reasoning gives $I_1<0$. Either way $I_1\ne0$, contradicting the hypothesis. Hence $g$ must vanish somewhere in $\left(0,\tfrac{\pi}{2}\right)$, i.e. $f(x_1)=1$ for some $x_1$. $\blacksquare$

5. 5

   Note what the second hypothesis is for. The proof used only $I_1=0$; the condition $I_2=0$ was never needed, because $\cos x>0$ on $\left(0,\tfrac{\pi}{2}\right)$ makes $I_2$ carry exactly the same information. Either integral alone forces the conclusion. What genuinely matters is that the weight keeps one sign on the open interval — replace $\sin x$ by $\cos 2x$, which changes sign at $\pi/4$, and the argument collapses: the constant $f\equiv e$ then gives $\int_0^{\pi/2}e^{e}\cos 2x\,dx=0$ without $f$ ever reaching $1$ at all.