Sets, Relations & Functions
Functions
nta_pyq_2025_jan
Grade 11
Question:
Let f : R - {0} \to R be a function such that f (x) - 6f ( 1 ) = 35 - 5 . x 3x 2 If the lim x\to0 ( 1 \alphax + f (x)) = \beta; \alpha, \beta \in R , then \alpha + 2\beta is equal to
Step-by-Step Solution
Key Concept: Apply the core result for domains, ranges and functional equations and simplify using the given constraints.
f (x) - 6f ( 1 ) = 35 - 5 . . . (1) 2 (3) x 3x 1 35x 5 6 (f ( ) - 6f (x) = - ) x 3 2 1 210x 30 6f ( ) - 36f (x) = - . . . (2) x 3 2 (1) + (2) 35 1 5 - 35f (x) = [ + 6x] - (1 + 6) 3 x 2 1 1 1 - f (x) = ( + 6x) - 3 x 2 1 1 f (x) = - - 2x + 3x 2 1 1 1 lim [ - - 2x + ] = \beta x\to0 \alphax 3x 2 \Rightarrow \alpha = 3 1 \beta + 2\beta = 3 + 2 \times = 4 2 x
Correct Answer: 3