JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 6)
Let $$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ be a differentiable function for all x$$\in$$R. Then f(x) equals :
$${e^{({e^{x - 1}})}}$$
$$2{e^{{e^x}}} - 1$$
$$2{e^{{e^x} - 1}} - 1$$
$${e^{{e^x}}} - 1$$
Explanation
$$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ .... (1)
Differentiating both sides w.r.t. x
$$f'(x) = {e^x}.f(x) + {e^x}$$ (Using Newton L:eibnitz Theorem)
$$ \Rightarrow {{f'(x)} \over {f(x) + 1}} = {e^x}$$
Integrating w.r.t. x
$$\int {{{f'(x)} \over {f(x) + 1}}dx = \int {{e^x}dx} } $$
$$ \Rightarrow \ln (f(x) + 1) = {e^x} + c$$
Put x = 0
ln 2 = 1 + c ($$ \because $$ f(0) = 1, from equation (1))
$$ \therefore $$ $$\ln (f(x) + 1) = {e^x} + \ln 2 - 1$$
$$ \Rightarrow f(x) + 1 = 2.\,{e^{{e^x} - 1}}$$
$$ \Rightarrow f(x) = 2{e^{{e^x} - 1}} - 1$$
Differentiating both sides w.r.t. x
$$f'(x) = {e^x}.f(x) + {e^x}$$ (Using Newton L:eibnitz Theorem)
$$ \Rightarrow {{f'(x)} \over {f(x) + 1}} = {e^x}$$
Integrating w.r.t. x
$$\int {{{f'(x)} \over {f(x) + 1}}dx = \int {{e^x}dx} } $$
$$ \Rightarrow \ln (f(x) + 1) = {e^x} + c$$
Put x = 0
ln 2 = 1 + c ($$ \because $$ f(0) = 1, from equation (1))
$$ \therefore $$ $$\ln (f(x) + 1) = {e^x} + \ln 2 - 1$$
$$ \Rightarrow f(x) + 1 = 2.\,{e^{{e^x} - 1}}$$
$$ \Rightarrow f(x) = 2{e^{{e^x} - 1}} - 1$$
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