JEE Main · 2010 · 11 Apr 2010 · Offline · Continuity and Differentiability

Write $f(x)=\frac{1}{e^x+2e^{-x}}$. By the arithmetic-geometric mean inequality $e^x+2e^{-x}\ge 2\sqrt{2}$, with equality when $e^x=\sqrt{2}$, so the denominator is at least $2\sqrt{2}$ and $f$ reaches a greatest value $\frac{1}{2\sqrt{2}}$; it is always positive, giving $0<f(x)\le \frac{1}{2\sqrt{2}}$, so Statement-2 is true. Since $\frac{1}{3}$ lies between $0$ and $\frac{1}{2\sqrt{2}}\approx 0.354$ and $f$ is continuous, the intermediate value theorem guarantees some $c$ with $f(c)=\frac{1}{3}$, so Statement-1 is true. It is exactly the range bound of Statement-2 that places $\frac{1}{3}$ within reach of $f$, so Statement-2 is the correct explanation of Statement-1.