<p>Let \(f\) be continuous on \([0,1]\) and \(\int_0^1 f(x)e^x\,dx=\int_0^1 f(x)e^{-x}\,dx=0\). Which must hold? [JEE Advanced 2016]</p>
Step-by-Step Solution
Key Concept: Adding: \intf \cdot cosh(x)dx=0. Subtracting: \intf \cdot sinh(x)dx=0. By Rolle's argument applied twice, f has \geq2 roots in (0,1).
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<p>Let $F(t)=\int_0^1 f(x)e^{tx}dx$. $F(1)=0$, $F(-1)=0$.</p>
<p>By Rolle's theorem, $\exists\,c\in(-1,1): F'(c)=\int_0^1 xf(x)e^{cx}dx=0$.</p>
<p>More precisely: $\int_0^1 f(x)\cosh x\,dx=0$ and $\int_0^1 f(x)\sinh x\,dx=0$.</p>
<p>If $f$ has fewer than 2 zeros in $(0,1)$, then $f$ is of one sign on most of $(0,1)$, contradicting both integrals being zero. Therefore $f$ has at least 2 zeros. ✓</p>
Correct Answer: A