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Relations and Functions · Question 18

JEE Main 2025 (Online) 2nd April Evening ShiftMCQEasy📈 Topic in 35% papers+4−1
1

If the domain of the function f(x) = 1 10 + 3x - x2 + 1 x + |x| is (a, b), then (1 + a)2 + b2 is equal to :

  1. 29
  2. 30
  3. 25
  4. 26

Answer

(4) 26

Solution

10 + 3x - x2 > 0 x2 - 3x - 10 < 0

(x - 5)(x + 2) < 0 -2 < x < 5

x + |x| > 0: for x 0, x + |x| = 0; for x > 0, x + |x| = 2x > 0 x > 0

domain = (0, 5) a = 0, b = 5

(1 + a)2 + b2 = 1 + 25 = 26

Answer: (4)

JEE Main 2025 (Online) 2nd April Evening Shift

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2

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
JEE Main 2025 (Online) 28th January Morning Shift

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