Sets, Relations & Functions
Functions
nta_pyq_2025_jan
Grade 11
Question:
Let f : R \to R be a function defined by a+2 2 f (x) = (2 + 3a)x + ( ) x + b, a \ne 1. If a-1 , then the value of 28 \sum is 2 5 f (x + y) = f (x) + f (y) + 1 - xy |f (i)| 7 i=1
Step-by-Step Solution
Key Concept: Apply the core result for domains, ranges and functional equations and simplify using the given constraints.
Put y = 0 f (x) = f (0) + f (x) + 1 - 0 (4) f (0) = -1 f (0) = 0 + 0 + b \Rightarrow b = -1 2 f (-1 + 1) = f (-1) + f (1) + 1 + 7 9 f (0) = f (-1) + f (1) + 7 a + 2 -1 = (2 + 3a) + ( ) (-1) + b + (2 + 3a) a - 1 a + 2 9 + + b + a - 1 7 9 - 1 = 4 + 6a - 2 + 7 9 - 1 = 2 + + 6a 7 9 6a = -1 - 2 - 7 -5 a = 7 9 2 -x 7 f (x) = + x - 1 -12 7 7 2 -x 3 f (x) = - x - 1 7 4 5 1 5 \times 6 \times 11 3 5 \times 6 \sum f (i) = - ( ) - ( ) - 5 7 6 4 2 i=1 -55 45 = - - 5 7 4 675 = 28 ∣ 5 ∣ ∣ ∣ \Rightarrow 28 \sum f (i) = 675 ∣ ∣ ∣ i-1 ∣
Correct Answer: 4