JEE Advanced Model Paper 18 with solutions for online practice

x+√5

∫

x+√2

f (x) dx = ax + b ∀ x ∊ R
⇒ f (x + √5) - f (x + √2) = a ∀ x ∊ R
⇒ f (x + √5) = f (x + √2) = a ∀ x ∊ R
⇒ f' f (x) (x) is a periodic function but is polynomial ⇒ f' (x) is a polynomial
⇒ f' (x) = k (constant)
⇒ f' (x) = kx + c
Now f (0) = 3 and f (1) = 5
⇒ f (x) = 2x + 3