JEE Advanced Previous Year Questions (2018 - 2025): Sets, Relations and Functions

2024

Now, put y = -x in eq. (1)
f(x) + f(-x) = f(0) { f(0) = 0 }
⇒ f(-x) = -f(x)
⇒ f is an odd function
From eq. (2)
f(-nx) = n f(-x)
⇒ f(-nx) = -n f(x)
⇒ f(mx) = m f(x) ∀ m ∈ ℤ⁻ ... (3)
From eq. (2) and eq. (3)
f(nx) = n f(x) ∀ n ∈ ℤ ... (4)
Now, put x = pq where p, q ∈ ℤ, q ≠ 0
f (npq) = n f (pq) ∀ n ∈ ℤ
Put n = q
f(p) = q f pq
⇒ p f(1) = q f (pq) {from eq. (4)}
Let f(1) = a
Then, p a = q f (pq)
⇒ f (pq) = apq
⇒ f(x) = a x ∀ x ∈ ℚ
Now, f (-35) = a (-35) = 12 ⇒ a = -20
⇒ f(x) = -20x ∀ x ∈ ℚ ... (5)
From the given functional equation, it is not possible to find a unique function for irrational values of 'x'. There are infinitely many such functions satisfying the given functional equation for irrational values of x, but in this problem, we finally need the function at rational values of 'x' only. So, for rational values of x, we get a unique function mentioned in (5).
Now, g(x + y) = g(x) ⋅ g(y)
⇒ ln(g(x + y)) = ln(g(x)) + ln(g(y))
Let ln(g(x)) = h(x)
⇒ h(x + y) = h(x) + h(y)
⇒ h(x) = k x ∀ x ∈ ℚ
⇒ g(x) = e^kx ∀ x ∈ ℚ ... (6)
And g (-13) = e ^-k3 = 2 ⇒ K = -3 ln(2)
⇒ K = ln (18)
⇒ g(x) = e^{(ln(1/8) . x)} = 18^x = 2^-3x∀ x ∈ ℚ
Now, f (14) = -5, g(-2) = 2⁶ = 64
g(0) = 1
So, (f 14 + g(-2) - 8 g(0))
= (-5 + 64 - 8)(1) = 51

Q2: Let the function f : ℝ → ℝ be defined by

.
Then the number of solutions of f(x) = 0 in ℝ is ______.  [JEE Advanced 2024 Paper 2]
Ans: 1
f(x) m= x2023 + 2024x + 2025e^πx (x² - x + 3) (sin x + 2)
∴ (sin x + 2) is never zero.
∴ For x2023 + 2024x + 20251 = 0
Let ϕ(x) = x2023 + 2024x + 2025.
ϕ'(x) = 2023x2022 + 2024 > 0 ∀ x ∈ ℝ
∴ ϕ(x) is a strictly increasing function.
∴ ϕ(x) = 0 for exactly one value of x.
∴ f(x) = 0 has one solution.

2023

Let domain and co-domain of a function y = f(x) are S and T respectively.
(A) There are infinitely many elements in domain and four elements in co-domain.
⇒ There are infinitely many functions from S to T.
⇒ Option (A) is correct
(B) If number of elements in domain is greater than number of elements in co-domain, then number of strictly increasing function is zero.
⇒ Option (B) is incorrect
(C) Maximum number of continuous functions = 4 × 4 × 4 = 64
(Every subset (0, 1),(1, 2),(3, 4) has four choices)
∵ 64<120⇒ option (C) is correct.
(D) For every point at which f(x) is continuous, f(x) = 0
⇒ Every continuous function from S to T is differentiable.
Option (D) is correct. 

Q2: Let f : [0, 1] → [0, 1] be the function defined by . Consider the square region S = [0, 1] × [0, 1]. Let  be called the green region and  be called the red region. Let  be the horizontal line drawn at a height ℎ ∈ [0, 1]. Then which of the following statements is(are) true?
(a) There exists an ℎ ∈ [1/4, 2/3] such that the area of the green region above the line L_ℎ equals the area of the green region below the line L_ℎ
(b) There exists an ℎ ∈ [1/4, 2/3] such that the area of the red region above the line L_ℎ equals the area of the red region below the line L_ℎ
(c) There exists an ℎ ∈ [1/4, 2/3] such that the area of the green region above the line L_ℎ equals the area of the red region below the line L_ℎ
(d) There exists an ℎ ∈ [1/4, 2/3] such that the area of the red region above the line L_ℎ equals the area of the green region below the line [JEE Advanced 2023 Paper 1]
Ans: (b), (c) & (d)
Given,

 option (A) is incorrect

 option (B) is correct.

⇒ h= 1/2 ⇒ option (C) is correct.
(D) ∵ Option (C) is correct ⇒ option (D) is also correct.

2022

Q1: Let |M| denote the determinant of a square matrix M. Let  be the function defined by 

where

Let p(x) be a quadratic polynomial whose roots are the maximum and minimum values of the function g(θ), and p(2)  =2 - √2. Then, which of the following is/are TRUE ?
(a)
(b)
(c)
(d)             [JEE Advanced 2022 Paper 1]Ans: (a) & (c)
Given,

Here,

and

and

Also,

  (skew symmetric)  

For option (A)  Correct.
For option (B)  Incorrect.
For option (C)  Correct.
For option (D)  Incorrect. 

2020

Then the value of  is ..........
Ans: 19
The given function f : [0, 1] → R be define by

So, f(x) + f(1 - x) = 1 .....(i)

=

= (19 times) {from Eq. (i)}
= 19. 

Q2: Let the function be defined by Suppose the function f has a local minimum at θ precisely when , where . Then the value of  is ............. [JEE Advanced 2020 Paper 2]
Ans: 0.5
The given function f : R → R be defined by

The local minimum of function 'f' occurs when

but

Where,

So,  = 0.50

Q3: Let f : [0, 2] → R be the function defined by

If  are such that , then the value of β - α is .......... [JEE Advanced 2020 Paper 1]
Ans: 1
The given function f : [0, 2] → R defined by

As, so,

Therefore the value of (β - α) = 1

Q3: For a polynomial g(x) with real coefficients, let m_g denote the number of distinct real roots of g(x). Suppose S is the set of polynomials with real coefficients defined by
For a polynomial f, let f' and f'' denote its first and second order derivatives, respectively. Then the minimum possible value of (m_f' + m_f''), where f ∈ S, is .............. [JEE Advanced 2020 Paper 1]
Ans: 5
Given set S of polynomials with real coefficients

and for a polynomial f ∈ S, Let

it have -1 and 1 as repeated roots twice, so graph of f(x) touches the X-axis at x = -1 and x = 1, so f'(x) having at least three roots x = -1, 1 and α. Where α ∈ (-1, 1) and f''(x) having at least two roots in interval (-1, 1)
So, m_f' = 3 and m_f'' = 2
∴ Minimum possible value of (m_f' + m_f'') = 5 

Q4: If the function f : R → R is defined by f(x) = |x| (x - sin x), then which of the following statements is TRUE?
(a) f is one-one, but NOT onto
(b) f is onto, but NOT one-one
(c) f is BOTH one-one and onto
(d) f is NEITHER one-one NOR onto             [JEE Advanced 2020 Paper 1]
Ans: (c)
The given function f : R → R is

The function 'f' is a odd and continuous function and as , so range is R, therefore, 'f' is a onto function.

⇒ f is strictly increasing function. ∀x ∈(0, ∞).
Similarly, for x < 0, -x + sin x > 0 and (- x) (1 - cos x) > 0, therefore, f′(x)> 0∀ x ∈(-∞, 0)
⇒ f is strictly increasing function, ∀x ∈ (0, ∞)
Therefore 'f' is a strictly increasing function for x ∈ R and it implies that f is one-one function. 

2018

Q1: Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If α is the number of one-one functions from X to Y and β is the number of onto functions from Y to X, then the value of 1 / 5!(β - α) is .................. [JEE Advanced 2018 Paper 2]
Ans: 119
Given, X has exactly 5 elements and Y has exactly 7 elements.
∴ n(X) = 5
and n(Y) = 7
Now, number of one-one functions from X to Y is
Number of onto functions from Y to X is  β

1, 1, 1, 1, 3 or 1, 1, 1, 2, 2

= 4 x 35 - 21= 140 - 21
= 119

Q2: Let  and (Here, the inverse trigonometric function sin^-1 x assumes values in [-π/2, π/2].).
Let f : E₁ → R be the function defined by  and g : E₂ → R be the function defined by  .      [JEE Advanced 2018 Paper 2]

The correct option is :
(a) P → 4; Q → 2; R → 1 ; S → 1
(b) P → 3; Q → 3; R → 6 ; S → 5
(c) P → 4; Q → 2; R → 1 ; S → 6
(d) P → 4; Q → 3; R → 6 ; S → 5                  [JEE Advanced 2018 Paper 2]
Ans: (a)
We have,

and

So,

The domain of f and g are 
and Range of

Range of f is R - {0} or (-∞, 0) ∪ (0, ∞)
Range of g is
Now, P → 4, Q → 2, R → 1, S → 1
Hence, option (a) is correct answer.