JEE Advanced Previous Year Questions (2018 - 2024): Trigonometric Functions & Equations | Mathematics (Maths) for JEE Main & Advanced PDF Download

JEE ADVANCED PYQs 2024

Q1: Let π2 < x < π be such that cot x = -5√11. Then ( sin 11x2 ) (sin 6x - cos 6x) + ( cos 11x2 ) (sin 6x + cos 6x) is equal to:
(a) √11 - 12√3
(b) √11 + 12√3
(c) √11 + 13√2
(d) √11 - 13√2 [JEE Advanced 2024 Paper 1]
Ans: (b)
Given the information, let's start by analyzing the trigonometric relationships involving:
cot x = -5√11 where π2 < x < π.
We know that cot x is the reciprocal of tan x. Hence,
cot x = cos xsin x = -5√11
From the above, it follows that:
cos x = -5k and sin x = √11k
For some constant k. Using the Pythagorean identity:
cos² x + sin² x = 1
Substituting the values of cos x and sin x into this identity:
(-5k)² + (√11k)² = 1
25k² + 11k² = 1
36k² = 1
k² = 136
k = 16 or k = -16
Therefore, we have two sets of values:
cos x = -5/6 and sin x = √11/6
Given that x is in the interval ( π2 , π), where sine is positive and cosine is negative, we take:
cos x = -5/6, sin x = √11/6
Now consider the expression:
(sin 11x2 ) (sin 6x - cos 6x) + (cos 11x2 ) (sin 6x + cos 6x)
Using trigonometric identities, we simplify the expression step by step.
Thus, the correct option is:
Option B
√11 + 12√3

JEE ADVANCED PYQs 2023

Q1: Consider an obtuse-angled triangle ABC in which the difference between the largest and the smallest angle is π/2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Then the inradius of the triangle ABC is : _______. [JEE Advanced 2023 Paper 2]
Ans: 0.25
In radius r = ΔS = a2R(sin A + sin B + sin C)
r = asin (π/2 - 2C) sin (π/2 + C) + sin C
= acos 2C + cos C + sin C
= acos 2C + √1 + sin 2C
= 3√7 / 16√(7/4) + √(7/2) = 14
⇒ r = 14 = 0.25
⇒ r = 0.25

JEE ADVANCED PYQs 2022

Q1: Consider the following lists :

The correct option is
(a) (I) →(P); (II) →(S); (III) →(P); (IV) →(S)
(b) (I) → (P); (II) → (P); (III) → (T); (IV) → (R)
(c) (I) → (Q); (II) →(P); (III) → (T); (IV) → (S)
(d) (I) →(Q); (II) →(S); (III) →(P); (IV) →(R)              [JEE Advanced 2022 Paper 1]
Ans: (b)
Solving all question one by one we get,

So,
∴ x has 2 elements → P

So,
∴ x has 2 elements → P

So,

∴ x has 6 elements →T

So,

∴ x has 4 elements →R

Q2: Let α and β be real numbers such that  .
If  and , then the greatest integer less than or equal to is: [JEE Advanced 2022 Paper 2]
Ans: 1
Given,
and

Let,

= 4/3

 JEE ADVANCED PYQs 2019

Q1: For non-negative integers n, let

Assuming cos⁻¹ x takes values in [0, π], which of the following options is/are correct?
(a) If α = tan(cos⁻¹ f(6)), then α2 + 2α −1 = 0
(b)
(c) sin(7 cos⁻¹ f(5)) = 0
(d)  [JEE Advanced 2019 Paper 2]
Ans: (a), (b) & (c)
It is given, that for non-negative integers 'n',

Now,

Now,
Now,

and Now,
Hence, options (a), (b) and (c) are correct.

Q2: Let f(x) = sin(π cos x) and g(x) = cos(2π sin x) be two functions defined for x > 0. Define the following sets whose elements are written in the increasing order:
X = {x : f(x) = 0}, Y = {x : f'(x) = 0}
Z = {x : g(x) = 0}, W = {x : g'(x) = 0}
List - I contains the sets X, Y, Z and W. List - II contains some information regarding these sets. 

Which of the following is the only CORRECT combination?
(a) (IV), (P), (R), (S)
(b) (III), (P), (Q), (U)
(c) (III), (R), (U)
(d) (IV), (Q), (T)       [JEE Advanced 2019 Paper 2]
Ans: (a)
For Z = {x : g(x) = 0}, x > 0
∵ g(x) = cos(2π sin x) = 0

here values of sin x, are in an A.P. but corresponding values of x are not in an AP so, (iii) → R.
For W = {x : g'(x) = 0}, x > 0
So, g'(x) = −2 π cos x sin(2π sin x) = 0
⇒ either cos x = 0 or sin(2π sin x) = 0 

⇒ (iv) → P, R, S
Hence, option (a) is correct. 

JEE ADVANCED PYQs 2018

Q1: In a ΔPQR = 30^∘ and the sides PQ and QR have lengths 10√3 and 10, respectively. Then, which of the following statement(s) is(are) TRUE?
(a) ∠QPR=45∘
(b) The area of the ΔPQR is 25√3 and ∠QRP=120∘
(c) The radius of the incircle of the ΔPQR is 103 − 15
(d) The area of the circumcircle of the ΔPQR is 100                    [JEE Advanced 2018 Paper 1]
Ans: (b), (c) & (d)
We have,
In ΔPQR

By cosine rule

Since, PR = QR = 10

Radius of incircle of

and radius of circumcircle

∴ Area of circumcircle of

Hence, option (b), (c) and (d) are correct answer.

Q2: Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the X-axis, Y-axis and Z-axis, respectively, where O(0, 0, 0) is the origin. Let  be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If p = SP, q = SQ, r = SR and t = ST, then the value of |(p × q) × (r × t)| is _________ [JEE Advanced 2018 Paper 2]
Ans: 0.5
Here, P(1, 0, 0), Q(0, 1, 0), R(0, 0, 1), T = (1, 1, 1) and

Now,

and

Now,

∴

= 1/2
= 0.5