Given:
3f(x)+2f(x1)=x1−10....(1)
Replace x→x1
3f(x1)+2f(x)=x−10....(2)
Eliminating f(x1) from (1)&(2), we get
⇒5f(x)=x3−2x−10
Let us replace x with 3
⇒5f(3)=33−2(3)−10
⇒f(3)=−3
⇒5f′(x)=x2−3−2
⇒5f′(41)=−48−2
⇒f′(41)=−10
⇒∣f(3)+f′(41)∣=13
Hence this is the correct option.