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Relations and Functions

JEE Main 2024 (Online) 1st February Evening ShiftMCQEasy📈 Topic in 31% papers+4−1
Ques. 1JEE Main 2024, 1 February (Evening Shift)

If the domain of the function f(x)= x2-25 (4-x2 )+ 10 (x2+2x-15 ) is (- , ) [ , ), then 2+ 3 is equal to :

  1. 140
  2. 175
  3. 125
  4. 150

Answer

(4) 150

Solution

x2-25 0 x (- ,-5] [5, )

4-x2 0 x 2

x2+2x-15>0 (x+5)(x-3)>0 x (- ,-5) (3, )

Domain (common part) =(- ,-5) [5, )

at x=-5 the log argument is 0, so -5 is excluded

=-5, =5

2+ 3=25+125=150

Answer: (4)

Keep practising — Relations and Functions

Ques. 2JEE Main 2024, 30 January (Morning Shift)

If the domain of the function f(x)= -1 ( 2-|x|4 )+ \ e(3-x) \-1 is [- , )-\ \, then + + is equal to :

  1. 11
  2. 12
  3. 9
  4. 8
Ques. 3JEE Main 2024, 30 January (Evening Shift)

If the domain of the function f(x)= e ( 2 x+34 x2+x-3 )+ -1 ( 2 x-1x+2 ) is ( , ], then the value of 5 -4 is equal to

  1. 9
  2. 12
  3. 11
  4. 10

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