QJEEM20Ma3eb3: JEE Main 2020 Mathematics - Class 12 JEEM20Ma3eb3 Solution: Let and g(x) = f(f(x)), . Then is equal to:

Question

Verified Solution • 2025–26 Session moderate

Let

f(x) = \left| {x - 2} \right|

and g(x) = f(f(x)),

x \in \left[ {0,4} \right]

. Then

\int\limits_0^3 {\left( {g(x) - f(x)} \right)} dx

is equal to:

Accepted Answer

f(x) = |x - 2|

\therefore

f(f(x)) = \left| {|x - 2| - 2} \right| = g(x)

\Rightarrow g (x) = | | x - 2 | - 2 | = {\begin{matrix} | x - 4 | & i f x \geq 2 \\ | - x | & i f x < 2 \end{matrix}

\therefore

\int\limits_0^3 {(g(x) - f(x))dx}

= \int\limits_0^3 {g(x)dx} - \int\limits_0^3 {f(x)dx}

= \int\limits_0^2 {| - x|dx} + \int\limits_2^3 { - (x - 4)dx} - \int\limits_0^2 { - (x - 2)dx} - \int\limits_2^3 {(x - 2)dx}

= \int\limits_0^2 {x\,dx} - \int\limits_2^3 {(x - 4)dx} + \int\limits_0^2 {(x - 2)dx} - \int\limits_2^3 {(x - 2)dx}

= \left[ {{{{x^2}} \over 2}} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 4x} \right]_2^3 + \left[ {{{{x^2}} \over 2} - 2x} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 2x} \right]_2^3

= 2 - \left\{ {{9 \over 2} - 12 - 2 + 8} \right\} + \{ 2 - 4\} - \left\{ {{9 \over 2} - 6 - 2 + 4} \right\}

= 2 - \left\{ {{9 \over 2} - 6} \right\} - 2 - \left\{ {{9 \over 2} - 4} \right\}

= 10 - {9 \over 2} - {9 \over 2} = {{20 - 9 - 9} \over 2} = {2 \over 2} = 1

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