JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 14)
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:
$$\int\limits_0^3 {\left( {g(x) - f(x)} \right)} dx$$ is equal to:
1
0
$${1 \over 2}$$
$${3 \over 2}$$
Explanation
$$f(x) = |x - 2|$$
$$ \therefore $$ $$f(f(x)) = \left| {|x - 2| - 2} \right| = g(x)$$
$$ \Rightarrow g(x) = \left| {|x - 2| - 2} \right| = \left\{ {\matrix{ {|x - 4|} & {if\,x \ge 2} \cr {| - x|} & {if\,x < 2} \cr } } \right.$$
$$ \therefore $$ $$\int\limits_0^3 {(g(x) - f(x))dx} $$
$$ = \int\limits_0^3 {g(x)dx} - \int\limits_0^3 {f(x)dx} $$
$$ = \int\limits_0^2 {| - x|dx} + \int\limits_2^3 { - (x - 4)dx} - \int\limits_0^2 { - (x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $$
$$ = \int\limits_0^2 {x\,dx} - \int\limits_2^3 {(x - 4)dx} + \int\limits_0^2 {(x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $$
$$ = \left[ {{{{x^2}} \over 2}} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 4x} \right]_2^3 + \left[ {{{{x^2}} \over 2} - 2x} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 2x} \right]_2^3$$
$$ = 2 - \left\{ {{9 \over 2} - 12 - 2 + 8} \right\} + \{ 2 - 4\} - \left\{ {{9 \over 2} - 6 - 2 + 4} \right\}$$
$$ = 2 - \left\{ {{9 \over 2} - 6} \right\} - 2 - \left\{ {{9 \over 2} - 4} \right\}$$
$$ = 10 - {9 \over 2} - {9 \over 2} = {{20 - 9 - 9} \over 2} = {2 \over 2} = 1$$
$$ \therefore $$ $$f(f(x)) = \left| {|x - 2| - 2} \right| = g(x)$$
$$ \Rightarrow g(x) = \left| {|x - 2| - 2} \right| = \left\{ {\matrix{ {|x - 4|} & {if\,x \ge 2} \cr {| - x|} & {if\,x < 2} \cr } } \right.$$
$$ \therefore $$ $$\int\limits_0^3 {(g(x) - f(x))dx} $$
$$ = \int\limits_0^3 {g(x)dx} - \int\limits_0^3 {f(x)dx} $$
$$ = \int\limits_0^2 {| - x|dx} + \int\limits_2^3 { - (x - 4)dx} - \int\limits_0^2 { - (x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $$
$$ = \int\limits_0^2 {x\,dx} - \int\limits_2^3 {(x - 4)dx} + \int\limits_0^2 {(x - 2)dx} - \int\limits_2^3 {(x - 2)dx} $$
$$ = \left[ {{{{x^2}} \over 2}} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 4x} \right]_2^3 + \left[ {{{{x^2}} \over 2} - 2x} \right]_0^2 - \left[ {{{{x^2}} \over 2} - 2x} \right]_2^3$$
$$ = 2 - \left\{ {{9 \over 2} - 12 - 2 + 8} \right\} + \{ 2 - 4\} - \left\{ {{9 \over 2} - 6 - 2 + 4} \right\}$$
$$ = 2 - \left\{ {{9 \over 2} - 6} \right\} - 2 - \left\{ {{9 \over 2} - 4} \right\}$$
$$ = 10 - {9 \over 2} - {9 \over 2} = {{20 - 9 - 9} \over 2} = {2 \over 2} = 1$$
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