Verified Solution • 2025–26 Session moderate

QJEEM21Maa4bf: JEE Main 2021 Mathematics - Class 12 JEEM21Maa4bf Solution: Let be a differentiable function for all x R. Then f(x) equals :

Let f(x)=∫0xetf(t)dt+exf(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} be a differentiable function for all x ∈\in R. Then f(x) equals :

Multiple Choice Options

  • A e(ex−1){e^{({e^{x - 1}})}}
  • B 2eex−12{e^{{e^x}}} - 1
  • C 2eex−1−12{e^{{e^x} - 1}} - 1
  • D eex−1{e^{{e^x}}} - 1

Step-by-Step Verified Solution

f(x)=∫0xetf(t)dt+exf(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} .... (1)

Differentiating both sides w.r.t. x

f′(x)=ex.f(x)+exf'(x) = {e^x}.f(x) + {e^x} (Using Newton L:eibnitz Theorem)

⇒f′(x)f(x)+1=ex\Rightarrow {{f'(x)} \over {f(x) + 1}} = {e^x}

Integrating w.r.t. x

∫f′(x)f(x)+1dx=∫exdx\int {{{f'(x)} \over {f(x) + 1}}dx = \int {{e^x}dx} }

⇒ln⁡(f(x)+1)=ex+c\Rightarrow \ln (f(x) + 1) = {e^x} + c

Put x = 0

ln 2 = 1 + c ( ∵\because f(0) = 1, from equation (1))

∴\therefore ln⁡(f(x)+1)=ex+ln⁡2−1\ln (f(x) + 1) = {e^x} + \ln 2 - 1

⇒f(x)+1=2. eex−1\Rightarrow f(x) + 1 = 2.\,{e^{{e^x} - 1}}

⇒f(x)=2eex−1−1\Rightarrow f(x) = 2{e^{{e^x} - 1}} - 1

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