If ₀^ 2 ( ^4 x+ ^4 x ) d x=K ₀^ ^2 x d x+L ₀^ 2 ^2 x d x and K, L N , then the number of possible ordered pairs ( K , L ) is

Step-by-step solution

L.H.S. = ₀^ 2 ( ^4 x+ ^4 x ) d x aligned & = ₀^ 2 ^4 x d x+ ₀^ 2 ^4 x d x=2 ₀^ ^4 x d x+2 ₀^ ^4 x d x & =2 ₀^ / 2 ^4 x d x+4 ₀^ / 2 ^4 x d x & =4 ( 3 1 4 2 ) 2 +4 ( 3 1 4 2 ) 2 = 3 2 & R.H.S. =K ₀^ ^2 x d x+L ₀^ / 2 ^2 x d x & =2 K ₀^ / 2 ^2 x d x+L ₀^ / 2 ^2 x d x & =2 K ( 1 2 2 )+L ( 1 2 2 )=(K) 2 + ( L 2 ) 2 & For R.H.S. = L.H.S. & 3 2 =(K) 2 + ( L 2 ) 2 3=K+ L 2 aligned Since, K, L N possible values are (K, L)=(2,2),(1,4) So, two ordered pairs of (K, L) are possible.