Verified Solution • 2025–26 Session moderate

QJEEM20Ma2048: JEE Main 2020 Mathematics - Class 12 JEEM20Ma2048 Solution: Let be a differentiable function such that f(1) = e and . If…

Let f:(0,∞)→(0,∞)f:\left( {0,\infty } \right) \to \left( {0,\infty } \right) be a differentiable function such that f(1) = e and
lim⁡t→xt2f2(x)−x2f2(t)t−x=0\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0 . If f(x) = 1, then x is equal to :

Multiple Choice Options

  • A 1e{1 \over e}
  • B e
  • C 12e{1 \over 2e}
  • D 2e

Step-by-Step Verified Solution

lim⁡t→xt2f2(x)−x2f2(t)t−x=0\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0

(Using L'Hospital's Rule)

⇒lim⁡t→x2tf2(x)−2x2f(t).f′(t)1=0\Rightarrow \mathop {\lim }\limits_{t \to x} {{2t{f^2}(x) - 2{x^2}f(t).f'(t)} \over 1} = 0

⇒\Rightarrow 2xf2(x) - 2x2.f(x).f'(x) = 0

⇒\Rightarrow 2xf(x){f(x) - xf'(x)} = 0

[x ≠\ne 0, f(x) ≠\ne 0 as given
function f:(0,∞)→(0,∞)f:\left( {0,\infty } \right) \to \left( {0,\infty } \right) only takes positive value as input and output]

⇒f(x)=xf′(x)\Rightarrow f(x) = xf'(x)

⇒f′(x)f(x)=1x\Rightarrow {{f'(x)} \over {f(x)}} = {1 \over x}

Integrating w.r.t x, we get

⇒ln f(x)=ln x+ln C\Rightarrow ln\,f(x) = ln\,x + ln\,C

⇒f(x)=Cx\Rightarrow f(x) = Cx

∵\because f(1) = e

⇒C=e; so f(x)=ex\Rightarrow C = e;\,so\,f(x) = ex

When f(x) = 1 = ex

⇒x=1e\Rightarrow x = {1 \over e}

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