Q46 / 46 · NTA Abhyas (JEE Main) 2020 · Mathematics · Differentiation

Let fx is a differentiable function on x∈R, such that fx+y=fxfy for all x,y∈R where f0≠0. If f5=10, f'0=6, then the value of  f ′ 5 is equal to

Correct answer: 60

Solution

f ′ 5 = lim h→0 f 5+h −f 5 h
=limh→0f5+h-f5+0h
=limh→0f5.fh-f5.f0h
[∵ fx+y=fx.fy for all x,y]
=limh→0fh-f0h.f5
= f ′ 0 .f 5 =6×10
=60

Hence, the correct answer is 60.

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