Sets Relations And Functions Previous Year Questions With Solutions

In set theory, Sets, Relations and Functions are three different concepts but equally important for JEE preparation. The questions from the previous years of JEE question papers from this topic are provided on this page, along with a detailed solution for each question. These questions include all the important concepts and formulae. Students can expect two-three questions from this chapter in the JEE paper.

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Also Read:

• Functions and its Types
• Monotonicity and Extremum of Functions

JEE Main Maths Sets, Relations And Functions Previous Year Questions With Solutions

Question 1: If A = [(x, y) : x² + y² = 25] and B = [(x, y) : x² + 9y² = 144], then A ∩ B contains _______ points.

Solution:

A = Set of all values (x, y) : x² + y² = 25 = 5²

B = [x² / 144] + [y² / 16] = 1

i.e., [x² / (12)²] + [y² / (4)²] = 1.

Clearly, A ∩ B consists of four points.

Question 2: In a college of 300 students, every student reads 5 newspapers, and every newspaper is read by 60 students. The number of newspapers is ________.

Solution:

Let the number of newspapers be x.

If every student reads one newspaper, the number of students would be x (60) = 60x

Since every student reads 5 newspapers, the number of students = [x * 60] / [5] = 300

x = 25

Question 3: Let R be the relation on the set R of all real numbers defined by a R b if and only if |a − b| ≤ 1. Then R is __________.

Solution:

|a − a| = 0 < 1

Therefore, a R a ∀ a ∈ R

Therefore, R is reflexive.

Again a R b, |a − b| ≤ 1 ⇒ |b − a| ≤ 1 ⇒ b R a

Therefore, R is symmetric.

Again 1 R [½] and [½] R1 but [½] ≠ 1

Therefore, R is not anti-symmetric.

Further, 1 R 2 and 2 R 3, but 1 R 3 is not possible, [Because, |1 − 3| = 2 > 1]

Hence, R is not transitive.

Question 4: Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then R⁻¹ o R is ________.

Solution:

First, find R⁻¹.

R⁻¹ = {(5, 4) ; (4, 1) ; (6, 4) ; (6, 7) ; (7, 3)}.

Obtain the elements of R⁻¹ o R.

Pick the element of R and then of R⁻¹.

Since (4, 5) ∈ R and (5, 4) ∈ R⁻¹, we have (4, 4) ∈ R⁻¹ o R

Similarly, (1, 4) ∈ R, (4, 1) ∈ R⁻¹ ⇒ (1, 1) ∈ R⁻¹ o R

(4, 6) ∈ R, (6, 4) ∈ R⁻¹ ⇒ (4, 4) ∈ R⁻¹ o R,

(4, 6) ∈ R, (6, 7) ∈ R⁻¹ ⇒ (4, 7) ∈ R⁻¹ o R

(7, 6) ∈ R, (6, 4) ∈ R⁻¹ ⇒ (7, 4) ∈ R⁻¹ o R,

(7, 6) ∈ R, (6, 7) ∈ R⁻¹ ⇒ (7, 7) ∈ R⁻¹ o R

(3, 7) ∈ R, (7, 3) ∈ R⁻¹ ⇒ (3, 3) ∈ R⁻¹ o R,

Hence, R⁻¹ o R = {(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}.

Question 5: If

Solution:

f [ f (x) ] = (f (x) − 3)/ ( f (x) + 1)

Now f [f { f (x) }] = f ([3 + x] / [1 − x])

= 4x/4

= x

Therefore, f [f { f (x) }] = x.

Question 6: If f (x) = cos (log x), then find the value of f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)].

Solution:

f (x) = cos (log x)

Let y = f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)]

y = cos (log x) * cos (log 4) − [1 / 2] * [cos log (x / 4) + cos (log 4x)]

y = cos (log x) cos (log 4) − [1 / 2] * [cos (log x −log 4) + cos (log x + log 4)]

y = cos (log x) cos (log 4) − [1 / 2] * [2 cos (log x) cos (log 4)]

y = 0

Question 7: Let f : R → R be defined by f (x) = 2x + |x|, then f (2x) + f (−x) − f (x) = _______.

Solution:

f(x) = 2x + |x|

f(2x) = 2(2x) + |2x| = 4x + 2|x|

f(-x) = -2x + |-x| = -2x + |x|

-f(x) = -2x + (-|x|) = -2x – |x|

Hence, f(2x) + f(-x) – f(x) = 4x + 2|x| – 2x + |x| – 2x – |x|

= 2|x|

= 2x; x ≥ 0 and -2x; x < 0

Question 8: If f (x) = cos [π²] x + cos[−π²] x, where [x] stands for the greatest integer function, then find the function of the right angle.

Solution:

f (x) = cos [π²] x + cos[−π²] x

f (x) = cos (9x) + cos (−10x) {since π = 3.14}

= cos (9x) + cos (10x)

= 2 cos (19x / 2) cos (x / 2)

Now, right angle = π/2

So, f (π / 2) = 2 cos (19π / 4) cos (π / 4)

f (π / 2) = 2 *(−1 / √2) * (1/ √2)

= −1

Question 9: If 

Solution:

Let

= 1 − (2 / [x² + 1])          [Because [x² + 1] > 1 also (2 / [x² + 1]) ≤ 2]

So 1 − [2 / [x² + 1]] ≥ 1 − 2;

−1 ≤ f (x) < 1

Thus, f (x) has a minimum value equal to -1.

Question 10: The function f : R → R defined by f (x) = e^x is ________.

Solution:

Function f: R → R is defined by f (x) = e^x.

Let x₁, x₂ ∈ R and f (x₁) = f (x₂) or e^x₁ = e^x₂ or x₁ = x₂.

Therefore, f is one-one.

Let f (x) = e^x = y.

Taking log on both sides, we get x = log y.

As we know, negative real numbers have no pre-image, or the function is not onto, and zero is not the image of any real number.

Therefore, function f is one-one and into.

Question 11: If f: R → S defined by f (x) = sin x − √3 cos x + 1 is onto, then what is the interval of S?

Solution:

Given,

f (x) = sin x − √3 cos x + 1

As we know, the range of the function f(x) = a cos x + b sin x + c is given by:

c – √(a² + b²) ≤ f(x) ≤ c + √(a² + b²)

− √[1 + (√−3)²] ≤ (sin x − √3 cos x) ≤ √[1 + (√−3)²]

−2 ≤ (sin x − √3 cos x) ≤ 2

−2 + 1 ≤ (sin x − √3 cos x + 1) ≤ 2 + 1

−1 ≤ (sin x − √3 cos x + 1) ≤ 3

i.e., range = [−1, 3]

For f to be onto, the interval of S = [−1, 3].

Question 12: What is the domain of the function

Solution:

Given,

Let g (x) = sin⁻¹ (3 − x)

−1 ≤ 3 −x ≤ 1

Domain of g(x) is [2, 4] and let h (x) = log [|x| − 2]

|x|− 2 > 0

|x| > 2

x < −2 or x > 2

(−∞, −2) ∪ (2, ∞)

Also, log(|x| – 2) ≠ 0

|x| – 2 ≠ 1

|x| ≠ 3

We know that (f / g) (x) = f(x) / g(x) ∀ x ∈ D1 ∩ D2 − {x ∈ R : g (x) = 0}

Domain of f (x) = (2, 4] − {3} = (2, 3) ∪ (3, 4].

Question 13: If f (x) = a cos (bx + c) + d, then what is the range of f (x)?

Solution:

f (x) = a cos (bx + c) + d ..(i)

As we know, -1 ≤ cos θ ≤ 1

For minimum, cos (bx + c) = −1

From (i), f (x) = −a + d = (d − a)

For maximum, cos (bx + c) = 1

From (i), f (x) = a + d = (d + a)

Range of f (x) = [d − a, d + a]

Alternatively,

-1 ≤ cos(bx + c) ≤1

-a ≤ a cos(bx + c) ≤ a

-a + d ≤ a cos(bx + c) + d ≤ a + d

Range of f(x) = [d – a, a + d]

Question 14: The function f: R → R is defined by f (x) = cos² x + sin⁴x for x ∈ R, then what is f (R)?

Solution:

f (x) = cos² x + sin⁴x

y = f (x) = cos² x + sin²x (1 − cos²x)

y = cos² x + sin²x − sin²x cos²x

y = 1 − sin²x cos²x

y = 1 − [1 / 4] * [sin²2x]

3 / 4 ≤ f (x) ≤ 1, (Because 0 ≤ sin²2x ≤ 1)

f (R) ∈ [3/4, 1]

Question 15: If f (x) = 3x − 5, then f⁻¹(x) is _____________.

Solution:

Let f (x) = y ⇒ x = f⁻¹ (y).

Hence, f (x) = y = 3x − 5

⇒f⁻¹ (y) = x

f is one-one and onto, so f⁻¹ exists and is given by f⁻¹ (x) = [x + 5] / [3].

Also, Read:

JEE Advanced Maths Function Previous Year Questions with Solutions

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