Limits, Continuity & Differentiability
Continuity and Differentiability
nta_pyq_2025_jan
Grade 12
Question:
Let f (x) be a real differentiable function such that f (0) = 1 and f (x + y) = f (x)f (y) + f (x)f (y) for all ′ ′ x, y \in R . Then \sum 100 n=1 log e f (n) is equal to :
Step-by-Step Solution
Key Concept: Apply the core result for continuity and differentiability at a point and simplify using the given constraints.
∵ f (x + y) = f (x) ⋅ f (y) + f (x) ⋅ f (y), \forallx, y \in R ....(i) (1) And f (0) = 1 ....(ii) Now replace x by zero and y by zero we get f (0) = f (0)f (0) + f (0)f (0) 1 = f (0) + f (0) 1 ′ \therefore f (0) = . . . (iii) 2 Now replace y by zero in equation (i), we get 1 ′ f (x) = f (x) + f (x) 2 or, 1 2 f (x) = f (x) ′ ′ f (x) then f (x) = 1 2 hence ln |f (x)| = x 2 + c Put x = 0, we get c = 0 x \therefore ln |f (x)| = 2 Then \sum 100 n=1 ln(f ($\eta$)) = ( 1 2 + 2 2 + 3 2 + \ldots + 100 2 ) 5050 = = 2525 2
Correct Answer: 1