f(x)=ex+2e−x1=e2x+2ex f′(x)=(e2x+2)2(e2x+2)ex−2e2x⋅ex f′(x)=0⇒e2x+2=2e2x e2x=2⇒ex=2 maximum f(x)=42=221 0<f(x)≤221∀x∈R Since 0<31<221⇒ for some c∈R f(c)=31
JEE Main 2010 — Mathematics Calculus
Let f:R→R be a continuous function defined by f(x)=ex+2e−x1. Statement-1: f(c)=31, for some c∈R. Statement-2: 0<f(x)≤221, for all x∈R
Held on 30 Apr 2010 · Verified 6 Jul 2026.
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
Statement-1 is true, Statement-2 is false
Statement-1 is false, Statement-2 is true
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1
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