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Solve f(x)=eˣ+∫₀¹eˣf(t)dt for f(x) | JEE

JEE Maths question with a full step-by-step solution.

Question
If a function f(x)f(x) satisfies the relation f(x)=ex+∫01exf(t) dtf(x)=e^{x}+\displaystyle\int_{0}^{1}e^{x}f(t)\,dt, then f(x)=f(x)=
Aex2−e\dfrac{e^{x}}{2-e}correct
Bex2\dfrac{e^{x}}{2}
C2ex2e^{x}
D(e−2)ex(e-2)e^{x}
Solution
Step 1: The integral ∫01f(t) dt\displaystyle\int_{0}^{1}f(t)\,dt is a constant; call it KK. Since exe^{x} factors out of the integral over tt,
f(x)=ex+ex∫01f(t) dt=ex(1+K).f(x)=e^{x}+e^{x}\int_{0}^{1}f(t)\,dt=e^{x}(1+K).
 Step 2: Substitute this form into the definition of KK:
K=∫01f(t) dt=∫01et(1+K) dt=(1+K)∫01et dt=(1+K)(e−1).K=\int_{0}^{1}f(t)\,dt=\int_{0}^{1}e^{t}(1+K)\,dt=(1+K)\int_{0}^{1}e^{t}\,dt=(1+K)(e-1).
 Step 3: Solve the linear equation for KK:
K=(1+K)(e−1) ⇒ K=(e−1)+K(e−1) ⇒ K[1−(e−1)]=e−1 ⇒ K=e−12−e.K=(1+K)(e-1)\ \Rightarrow\ K=(e-1)+K(e-1)\ \Rightarrow\ K\big[1-(e-1)\big]=e-1\ \Rightarrow\ K=\dfrac{e-1}{2-e}.
 Step 4: Therefore
1+K=1+e−12−e=(2−e)+(e−1)2−e=12−e ⇒ f(x)=ex2−e.1+K=1+\dfrac{e-1}{2-e}=\dfrac{(2-e)+(e-1)}{2-e}=\dfrac{1}{2-e}\ \Rightarrow\ f(x)=\dfrac{e^{x}}{2-e}.
 Correct answer: (1)
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