Relations and Functions

JEE Main 2025 (Online) 29th January Evening ShiftMCQMedium📈 Topic in 34% papers+4−1

Ques. 1JEE Main 2025, 29 January (Evening Shift)

If the domain of the function 5(18x - x2 - 77) is ( , ) and the domain of the function (x-1) ( 2x2 + 3x - 2x2 - 3x - 4 ) is ( , ), then 2 + 2 + 2 is equal to:

Answer

(1) 186

Solution

needs argument >0, base >0, base 1

First: 18x-x2-77>0 ⇒ (x-7)(x-11)<0 ⇒ (7,11) ⇒ =7, =11

Second: base x-1>0, 1 ⇒ x>1, x 2; (2x-1)(x+2)(x-4)(x+1)>0 — for x>1 only (x-4) can be negative ⇒ x>4 ⇒ =4

[figure]

∴ 2+ 2+ 2=49+121+16=186

Answer: (1)

Keep practising — Relations and Functions

Ques. 2JEE Main 2025, 28 January (Morning Shift)

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

Ques. 3JEE Main 2025, 24 January (Evening Shift)

The number of real solution(s) of the equation x2+3x+2= \|x-3|,|x+2|\ is :

Ques. 4JEE Main 2025, 23 January (Morning Shift)

Let f(x)= e x and g(x)= x4-2x3+3x2-2x+22x2-2x+1. Then the domain of f g is

(0, )
[1, )
R
[0, )

Ques. 5JEE Main 2025, 22 January (Morning Shift)

The product of all solutions of the equation e5( ex)2+3=x8, x>0, is :

e2
e
e6/5
e8/5

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