JEE Main 2016 (26)

Calculus Level 4

y = ∣ x − 3 ∣ + ∣ x + 1 ∣ ∣ x + 3 ∣ + ∣ x − 1 ∣ \large y=\frac{|x-3|+|x+1|}{|x+3|+|x-1|} The area under the above curve, x x -axis and the ordinate at x = − 3 , x = 1 x=-3,x=1 is

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Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Ayush Verma
Mar 24, 2016

∣ x + 3 ∣ + ∣ x − 1 ∣ = ( x + 3 ) + ( 1 − x ) = 4 , ( a s a s − 3 ≤ x ≤ 1 ) ∴ y = ∣ x − 3 ∣ + ∣ x + 1 ∣ 4 i f − 3 ≤ x ≤ − 1 , y = ( 3 − x ) + ( − x − 1 ) 4 = 1 − x 2 i f − 1 ≤ x ≤ 1 , y = ( 3 − x ) + ( x + 1 ) 4 = 1 \left| x+3 \right| +\left| x-1 \right| =\left( x+3 \right) +\left( 1-x \right) =4,\left( as\quad as\quad -3\le x\le 1 \right) \\ \\ \therefore \quad y=\cfrac { \left| x-3 \right| +\left| x+1 \right| }{ 4 } \\ \\ if\quad -3\le x\le -1,\quad y=\cfrac { \left( 3-x \right) +\left( -x-1 \right) }{ 4 } =\cfrac { 1-x }{ 2 } \\ \\ if\quad -1\le x\le 1,\quad y=\cfrac { \left( 3-x \right) +\left( x+1 \right) }{ 4 } =1 A r e a = A B C D E A = ( 1 ) [ 1 − ( − 3 ) ] + 1 2 [ ( − 1 ) − ( − 3 ) ] ( 1 ) = 5 A n s = A r e a + 2 = 5 + 2 = 7 Area=ABCDEA=\left( 1 \right) \left[ 1-\left( -3 \right) \right] +\cfrac { 1 }{ 2 } \left[ \left( -1 \right) -\left( -3 \right) \right] \left( 1 \right) =5\\ \\ Ans=Area+2=5+2=7

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