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Relations and Functions · Question 19

JEE Main 2025 (Online) 28th January Morning ShiftMCQMedium📈 Topic in 3% papers+4−1
1

Let f : R R be a function defined by f(x) = (2 + 3a)x2 + ( a+2a-1 )x + b, a 1. If f(x + y) = f(x) + f(y) + 1 - 27xy, then the value of 28 i=15|f(i)| is

  1. 735
  2. 675
  3. 715
  4. 545

Answer

(2) 675

Solution

Let f(x) = px2 + qx + b, where p = 2 + 3a, q = a+2a-1.

f(x + y) = p(x + y)2 + q(x + y) + b

= f(x) + f(y) + 2pxy - b

Comparing with f(x) + f(y) + 1 - 27xy:

2p = - 27 p = - 17 and -b = 1 b = -1

2 + 3a = - 17 a = - 57

q = - 57 + 2- 57 - 1 = 9/7-12/7 = - 34

f(x) = - ( x27 + 3x4 + 1 ), which is negative for x > 0

|f(i)| = i27 + 3i4 + 1

i=15|f(i)| = 557 + 34(15) + 5 = 557 + 454 + 5

28 i=15|f(i)| = 220 + 315 + 140 = 675

Answer: (2)

JEE Main 2025 (Online) 28th January Morning Shift

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