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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)

satisfies the relation

f(x)=e^{x}+\displaystyle\int_{0}^{1}e^{x}f(t)\,dt

, then

f(x)=

\dfrac{e^{x}}{2-e}

correct

\dfrac{e^{x}}{2}

2e^{x}

(e-2)e^{x}

Solution

Step 1: The integral

\displaystyle\int_{0}^{1}f(t)\,dt

is a constant; call it

K

. Since

e^{x}

factors out of the integral over

t

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

K

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

K

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+\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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