QJEEM20Ma3f8b: JEE Main 2020 Mathematics - Class 12 JEEM20Ma3f8b Solution: If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed…

Question

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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:

Accepted Answer

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}

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