JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 16)
If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then
$${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$$ is equal to:
$${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$$ is equal to:
$$\int_{a - 1}^{b - 1} {f(x+1)dx} $$
$$\int_{a + 1}^{b + 1} {f(x + 1)dx} $$
$$\int_{a - 1}^{b - 1} {f(x)dx} $$
$$\int_{a + 1}^{b + 1} {f(x)dx} $$
Explanation
I = $${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$$ ...(1)
x $$ \to $$ a + b - x
I = $${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {a + b - x} \right) + f\left( {a + b - x + 1} \right)} \right)dx} $$
I = $${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $$ .....(2)
[As ƒ(x) = ƒ(a + b + 1 - x)
$$ \Rightarrow $$ ƒ(x + 1) = ƒ(a + b - x)]
Adding (1) and (2) we get
2I = $${{a + b} \over {a + b}}\int\limits_a^b {\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $$
$$ \Rightarrow $$ I = $${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {x + 1} \right)dx} $$
$$ \Rightarrow $$ I = $${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {a + b - x + 1} \right)dx} $$
$$ \Rightarrow $$ I = $${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( x \right)dx} $$
$$ \Rightarrow $$ I = $$\int\limits_a^b {f\left( x \right)dx} $$
Let x = t + 1
$$ \therefore $$ I = $$\int\limits_{a - 1}^{b - 1} {f\left( {t + 1} \right)dt} $$
= $$\int\limits_{a - 1}^{b - 1} {f\left( {x + 1} \right)dx} $$
x $$ \to $$ a + b - x
I = $${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {a + b - x} \right) + f\left( {a + b - x + 1} \right)} \right)dx} $$
I = $${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $$ .....(2)
[As ƒ(x) = ƒ(a + b + 1 - x)
$$ \Rightarrow $$ ƒ(x + 1) = ƒ(a + b - x)]
Adding (1) and (2) we get
2I = $${{a + b} \over {a + b}}\int\limits_a^b {\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $$
$$ \Rightarrow $$ I = $${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {x + 1} \right)dx} $$
$$ \Rightarrow $$ I = $${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {a + b - x + 1} \right)dx} $$
$$ \Rightarrow $$ I = $${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( x \right)dx} $$
$$ \Rightarrow $$ I = $$\int\limits_a^b {f\left( x \right)dx} $$
Let x = t + 1
$$ \therefore $$ I = $$\int\limits_{a - 1}^{b - 1} {f\left( {t + 1} \right)dt} $$
= $$\int\limits_{a - 1}^{b - 1} {f\left( {x + 1} \right)dx} $$
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