JEE ADVANCED MATHS QUESTION SEPTEMBER 2020
Answers
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 ?
solution : let's put x = y = 0 in the given relation.
f(0 + 0) = f(0) + f(0) + f(0) f(0)
⇒f(0) = 2f(0) + f(0)²
⇒f(0) = 0 or - 1 .......(1)
now f(x + y) = f(x) + f(y) + f(x) f(y)
⇒f(x + y) - f(x) = f(y) + f(x) f(y)
⇒{f(x + y) - f(x)} /y = {f(y) + f(x) f(y)}/y = f(y){1 + f(x)}/y
now let's take lim(y→0) both sides
lim(y→0) {f(x + y) - f(x)}/y = lim(y→0) f(y){1 + f(x)}/y
we know, f'(x) = lim(y→0) [f(x + y) - f(x)]/y
so, f'(x) = (1 + f(x) lim(y→0) f(y)/y
from equation (1), lim(y→0) f(y)/y = 1
so, f'(x) = 1 + f(x) ......(2)
at x = 0,
f'(0) = 1 + f(0) = 1 + 0
f'(0) = 1 , option (D) is correct choice.
given, f(x) = x g(x)
g(x) = f(x)/x
so, lim(x→0) g(x) = lim(x→0) f(x)/x = 1 [from eq (1) ]
from equation (2),
f'(x) = 1 + f(x)
⇒f'(x)/[1 + f(x)] = 1
⇒∫f'(x)/[1 + f(x)] dx = ∫ dx
⇒ln(1 + f(x)) = x + c
at x = 0,
ln(1 + f(0)) = 0 + c
⇒c = 0
so, 1 + f(x) = e^x ⇒f(x) = e^x - 1 , so f is differentiable at every x ∈ R
now f'(x) = e^x ⇒f'(1) = e¹ = e , option (C) is incorrect.
g(x) = f(x)/x = (e^x - 1)/x
if g(x) is differentiable at x = 0, then it will be differentiable for all x ∈ R , option (A) is correct choice.
so, let's check it.
g'(0⁺) = lim(h→0) [g(0 + h) - g(0)]/h
if we assume g(0) = 1
g'(0⁺) = lim(h→0) [(e^h - 1)/h - 1]/h = (e^h - 1 - h)/h² = 1/2
g'(0¯) = lim(h→0) = [(e^-h-1)/-h - 1]/-h = (e^-h -1 + h)/h² = 1/2
therefore g(x) is differentiable for all x ∈ R, so option (B) is correct choice.
Therefore options (A), (B) and (D) are correct choices.