JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 5)
Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 $$-$$ x) for all x$$ \in $$ (0, 2), f(0) = 1 and f(2) = e2. Then the value of $$\int\limits_0^2 {f(x)} dx$$ is :
1 + e2
2(1 + e2)
1 $$-$$ e2
2(1 $$-$$ e2)
Explanation
f'(x) = f'(2 $$-$$ x)
On integrating both side f(x) = $$-$$f(2 $$-$$ x) + c
put x = 0
f(0) + f(2) = c $$ \Rightarrow $$ c = 1 + e2
$$ \Rightarrow $$ f(x) + f(2 $$-$$ x) = 1 + e2 ..... (i)
$$I = \int\limits_0^2 {f(x)dx} = \int\limits_0^1 {\{ f(x) + f(2 - x)\} dx = (1 + {e^2})} $$
On integrating both side f(x) = $$-$$f(2 $$-$$ x) + c
put x = 0
f(0) + f(2) = c $$ \Rightarrow $$ c = 1 + e2
$$ \Rightarrow $$ f(x) + f(2 $$-$$ x) = 1 + e2 ..... (i)
$$I = \int\limits_0^2 {f(x)dx} = \int\limits_0^1 {\{ f(x) + f(2 - x)\} dx = (1 + {e^2})} $$
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