JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 16)
Let f : R $$ \to $$ R be a continuous function such that f(x) + f(x + 1) = 2, for all x$$\in$$R.
If $${I_1} = \int\limits_0^8 {f(x)dx} $$ and $${I_2} = \int\limits_{ - 1}^3 {f(x)dx} $$, then the value of I1 + 2I2 is equal to ____________.
If $${I_1} = \int\limits_0^8 {f(x)dx} $$ and $${I_2} = \int\limits_{ - 1}^3 {f(x)dx} $$, then the value of I1 + 2I2 is equal to ____________.
Answer
16
Explanation
$$f(x) + f(x + 1) = 2$$ .... (i)
$$x \to (x + 1)$$
$$f(x + 1) + f(x + 2) = 2$$ .... (ii)
by (i) & (ii)
$$f(x) - f(x + 2) = 0$$
$$f(x + 2) = f(x)$$
$$ \therefore $$ f(x) is periodic with T = 2
$${I_1} = \int_0^{2 \times 4} {f(x)dx} = 4\int_0^2 {f(x)dx} $$
$${I_2} = \int_{ - 1}^3 {f(x)dx} = \int_0^4 {f(x + 1)dx} = \int_0^4 {(2 - f(x))dx} $$
$$ \Rightarrow $$ $${I_2} = 8 - 2\int_0^2 {f(x)dx} $$
$$ \Rightarrow $$ $${I_2} = 8 - $$$${{{I_1}} \over 2}$$
$$ \Rightarrow $$ $${I_1} + 2{I_2} = 16$$
$$x \to (x + 1)$$
$$f(x + 1) + f(x + 2) = 2$$ .... (ii)
by (i) & (ii)
$$f(x) - f(x + 2) = 0$$
$$f(x + 2) = f(x)$$
$$ \therefore $$ f(x) is periodic with T = 2
$${I_1} = \int_0^{2 \times 4} {f(x)dx} = 4\int_0^2 {f(x)dx} $$
$${I_2} = \int_{ - 1}^3 {f(x)dx} = \int_0^4 {f(x + 1)dx} = \int_0^4 {(2 - f(x))dx} $$
$$ \Rightarrow $$ $${I_2} = 8 - 2\int_0^2 {f(x)dx} $$
$$ \Rightarrow $$ $${I_2} = 8 - $$$${{{I_1}} \over 2}$$
$$ \Rightarrow $$ $${I_1} + 2{I_2} = 16$$
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