Differentiating (f(x))2=25+∫0x((f(t))2+(f′(t))2)dt:
2ff′=f2+(f′)2⇒(f−f′)2=0⇒f′=f.
f(x)=Cex. At x=0: f(0)2=25⇒f(0)=5 (non-negative), so f(x)=5ex.
f(logek)=5k for k=1,2,…,625.
Mean =6255(1+2+⋯+625)=2×6255×625×626=25×626=1565.
JEE Main 2026 — Mathematics Calculus
Let f be a twice differentiable non-negative function such that (f(x))2=25+∫0x((f(t))2+(f′(t))2)dt. Then the mean of f(loge(1)),f(loge(2)),…..,f(loge(625)) is equal to ____.
Held on 23 Jan 2026 · Verified 6 Jul 2026.
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