Inverse Trigonometric Functions

JEE Main 2025 (Online) 4th April Evening ShiftMCQMedium📈 Topic in 16% papers+4−1

Ques. 1JEE Main 2025, 4 April (Evening Shift)

The sum of the infinite series -1 ( 74 )+ -1 ( 194 )+ -1 ( 394 )+ -1 ( 674 )+ is :

2+ -1 ( 12 )
2- -1 ( 12 )
2- -1 ( 12 )
2+ -1 ( 12 )

Answer

(3) 2- -1 ( 12 )

Solution

Numerators 7, 19, 39, 67, have second differences constant (12, 20, 28): they are 4n2+3

Tn= -1 ( 4n2+34 )= -1 (n2+ 34 )= -1 11+ (n2- 14 )

Tn= -1 (n+ 12 )- (n- 12 )1+ (n+ 12 ) (n- 12 )

Tn= -1 (n+ 12 )- -1 (n- 12 )

Sn= r=1nTr= -1 (n+ 12 )- -1 ( 12 ) (telescoping)

As n , -1 (n+ 12 ) 2

S = 2- -1 ( 12 )

Answer: (3)

Keep practising — Inverse Trigonometric Functions

Ques. 2JEE Main 2025, 2 April (Evening Shift)

If y = ( 3 + -1 x2 ), then (x - y)2 + 3y2 is equal to .

Ques. 3JEE Main 2025, 29 January (Morning Shift)

Let S = \x : -1x = + -1x + -1(2x + 1) \. Then x S(2x - 1)2 is equal to .

Ques. 4JEE Main 2025, 28 January (Morning Shift)

( -1 35 + -1 513 + -1 3365 ) is equal to:

3365
1
3265
0

Ques. 5JEE Main 2025, 28 January (Evening Shift)

Let [x] denote the greatest integer less than or equal to x. Then the domain of f(x) = -1(2[x] + 1) is:

(- , )
(- , ) - \0\
(- , -1] [0, )
(- , -1] [1, )

Ques. 6JEE Main 2025, 24 January (Morning Shift)

If for some , ; , + =8 and 2 ( -1 )+cosec2 ( -1 )=36, then 2+ is

Ques. 7JEE Main 2025, 24 January (Evening Shift)

If > > >0, then the expression -1 \ + (1+ 2)( - ) \+ -1 \ + (1+ 2)( - ) \ + -1 \ + (1+ 2)( - ) \ is equal to :

3
2-( + + )
0

Ques. 8JEE Main 2025, 23 January (Morning Shift)

If 2 x 3 4, then -1 ( 1213 x+ 513 x ) is equal to

x+ -1 512
x- -1 43
x+ -1 45
x- -1 512

Ques. 9JEE Main 2025, 22 January (Morning Shift)

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of 16 ( ( -1x )2+ ( cosec-1x )2 ) is :

24 2
18 2
22 2
31 2

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