Mathematics · Differential Equations

JEE Main 2026 — 23 January, Morning Shift — Question 23

Let f be a twice differentiable non-negative function such that (f(x))2=25+∫0x((f(t))2+(f′(t))2)dt(\mathrm{f}(\mathrm{x}))^{2}=25+\int_{0}^{\mathrm{x}}\left((\mathrm{f}(\mathrm{t}))^{2}+\left(\mathrm{f}^{\prime}(\mathrm{t})\right)^{2}\right) \mathrm{dt}. Then the mean of f(log⁡e(1)),f(log⁡e(2)),……,f(log⁡e(625))f\left(\log _{e}(1)\right), f\left(\log _{e}(2)\right), \ldots \ldots, f\left(\log _{e}(625)\right) is equal to ____\_\_\_\_。

Answer: 1565

Numerical answer — enter this value.

Step-by-step solution

2f(x)f′(x)=f2(x)+(f′(x))22 f(x) f^{\prime}(x)=f^{2}(x)+\left(f^{\prime}(x)\right)^{2}

⇒(f(x)−f′(x))2=0\Rightarrow\left(\mathrm{f}(\mathrm{x})-\mathrm{f}^{\prime}(\mathrm{x})\right)^{2}=0

⇒f(x)=f′(x)\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x}) ⇒ln⁡(f(x))=x+c⇒f(x)=c′ex\Rightarrow \ln (\mathrm{f}(\mathrm{x}))=\mathrm{x}+\mathrm{c} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{c}^{\prime} \mathrm{e}^{\mathrm{x}}

f(0)=5⇒f(x)=5ex\mathrm{f}(0)=5 \Rightarrow \mathrm{f}(\mathrm{x})=5 \mathrm{e}^{\mathrm{x}}

Mean =f(ℓn1)+f(ℓn2)+……….+f(ℓn625)625=\frac{\mathrm{f}(\ell \mathrm{n} 1)+\mathrm{f}(\ell \mathrm{n} 2)+\ldots \ldots \ldots .+\mathrm{f}(\ell \mathrm{n} 625)}{625} =5[1+2+…………+625]625=1565=\frac{5[1+2+\ldots \ldots \ldots \ldots+625]}{625}=1565

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Mathematics
Chapter
Differential Equations
Topic
Introduction to Differential Equations
Let f be a twice differentiable non-negative function such that ( f (… | JEE Main 2026 PYQ with Solution · DhiX AI