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QJEEM21Mac0f9: JEE Main 2021 Mathematics - Class 12 JEEM21Mac0f9 Solution: If , (a > 2) and [x] denotes the greatest integer x, then is…

If ∫−aa(∣x∣+∣x−2∣)dx=22\int\limits_{ - a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = 22 , (a > 2) and [x] denotes the greatest integer ≤\le x, then ∫−aa(x+[x])dx\int\limits_{ - a}^a {\left( {x + \left[ x \right]} \right)} dx is equal to _________.

Step-by-Step Verified Solution

∫−aa(∣x∣+∣x−2∣)dx=22\int\limits_{ - a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = 22

⇒\Rightarrow ∫−a0(−2x+2)dx+∫02(x+2−x)dx+∫2a(2x−2)dx=22\int\limits_{ - a}^0 {( - 2x + 2)dx} + \int\limits_0^2 {(x + 2 - x)dx} + \int\limits_2^a {(2x - 2)dx} = 22

⇒\Rightarrow x2−2x∣0−a+2x∣02+x2−2x∣2a=22{x^2} - 2x|_0^{ - a} + 2x|_0^2 + {x^2} - 2x|_2^a = 22

⇒\Rightarrow a2+2a+4+a2−2a−(4−4)=22{a^2} + 2a + 4 + {a^2} - 2a - (4 - 4) = 22

⇒\Rightarrow 2a2=18⇒a=32{a^2} = 18 \Rightarrow a = 3

∫3−3(x+[x])dx=−(∫−33(x+[x])dx)=−(∫−33[x]dx)\int\limits_3^{ - 3} {(x + [x])dx} = - \left( {\int\limits_{ - 3}^3 {(x + [x])dx} } \right) = - \left( {\int\limits_{ - 3}^3 {[x]dx} } \right)

= -( ∫−3−2[x]dx+∫−2−1[x]dx+∫−10[x]dx\int\limits_{ - 3}^{ - 2} {\left[ x \right]dx} + \int\limits_{ - 2}^{ - 1} {\left[ x \right]dx} + \int\limits_{ - 1}^0 {\left[ x \right]dx}

+ ∫01[x]dx+∫12[x]dx+∫23[x]dx\int\limits_0^1 {\left[ x \right]dx} + \int\limits_1^2 {\left[ x \right]dx} + \int\limits_2^3 {\left[ x \right]dx} )

=−(−3−2−1+0+1+2)=3= - ( - 3 - 2 - 1 + 0 + 1 + 2) = 3

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