JEE Advanced
Mathematics
Continuity and Differentiability
2020
JEE Advanced 2020 (Paper 2)
JEE Advanced Mathematics Question (2020) — Solution
Question
Let f : ℝ → ℝ and g : ℝ → ℝ be functions satisfying f x + y = f x + f y + f x f y  and  f x = x g x for all x , y ∈ ℝ . If lim x → 0 g x = 1 , then which of the following statements is/are TRUE?
Options
- A. f is differentiable at every x ∈ ℝ
- B. If g 0 = 1 , then g is differentiable at every x ∈ ℝ
- C. The derivative f ' 1 is equal to 1
- D. The derivative f ' 0 is equal to 1
Answer
D. The derivative f ' 0 is equal to 1
Step-by-step solution
Given f x + y = f x + f y + f x f y Put x = y = 0 in given relation. ⇒    f 0 = f 0 + f 0 + f 2 0 ⇒    f 0 = 0 or - 1 ∵    f x + y = f x + f y + f x · f y ⇒    f x + y - f x y = f y 1 + f x y ⇒    lim y → 0 f ( x + y ) - f ( x ) y = lim y → 0 1 + f x · f y y ∵    lim x → 0   g x = lim x → 0 f x x = 1 ⇒    f ' x = 1 + f x ⇒    f ' 0 = 1 + f 0 ⇒    f ' 0 = 1 + 0 ⇒    f ' 0 = 1 Again f ' x 1 + f x = 1 ⇒ ∫ f ' x d x 1 + f x d x = ∫ d x ⇒ l n 1 + f x = x + C ⇒ l n   1 + f x = x         ∵ C = 0 ⇒ 1 + f x = e x ⇒ f x = e x - 1   ⇒   f ' x = e x ⇒ f ' 1 = e Also, f x is differentiable for every x ∈ R . g x = f x x = e x - 1 x    y ' 0 + = lim h → 0 g 0 + h - g 0 h If g 0 = 1 then g ' 0 + = lim h → 0 e h - 1 h - 1 h = lim h → 0 e h - 1 - h h 2 = 1 2 g ' 0 - = lim h → 0 g 0 - h - g 0 - h = lim h → 0 e - h - 1 - h - h lim h → 0 e - h - 1 + h h 2 = 1 2 g x is differentiable for every x ∈ R .
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