JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

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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Let us have a brief look at the definition of Set, Relation and Function:

Set: A collection of well-defined objects which are distinct from each other.

Relation: If M and N are two non-empty sets, then a relation R from M to N is a subset of M x N. If R βŠ† M X N and (m,n) ∈ R, it indicates that “m” is related to “n” by the relation R, and written as mRn.

Function: Function is a special class of relation. A function f takes an input x, and returns a single output, say f(x).

Also Read:

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 βˆ’ b| ≀ 1. Then R is __________.

Solution:

|a βˆ’ a| = 0 < 1

Therefore, a R a βˆ€ a ∈ R

Therefore, R is reflexive.

Again a R b, |a βˆ’ b| ≀ 1 β‡’ |b βˆ’ a| ≀ 1 β‡’ b R a

Therefore, R is symmetric.

Again 1 R [Β½] and [Β½] R1 but [Β½] β‰  1

Therefore, R is not anti-symmetric.

Further, 1 R 2 and 2 R 3, but 1 R 3 is not possible, [Because, |1 βˆ’ 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 Rβˆ’1 o R is ________.

Solution:

First, find Rβˆ’1.

Rβˆ’1 = {(5, 4) ; (4, 1) ; (6, 4) ; (6, 7) ; (7, 3)}.

Obtain the elements of Rβˆ’1 o R.

Pick the element of R and then of Rβˆ’1.

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

Similarly, (1, 4) ∈ R, (4, 1) ∈ Rβˆ’1 β‡’ (1, 1) ∈ Rβˆ’1 o R

(4, 6) ∈ R, (6, 4) ∈ Rβˆ’1 β‡’ (4, 4) ∈ Rβˆ’1 o R,

(4, 6) ∈ R, (6, 7) ∈ Rβˆ’1 β‡’ (4, 7) ∈ Rβˆ’1 o R

(7, 6) ∈ R, (6, 4) ∈ Rβˆ’1 β‡’ (7, 4) ∈ Rβˆ’1 o R,

(7, 6) ∈ R, (6, 7) ∈ Rβˆ’1 β‡’ (7, 7) ∈ Rβˆ’1 o R

(3, 7) ∈ R, (7, 3) ∈ Rβˆ’1 β‡’ (3, 3) ∈ Rβˆ’1 o R,

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

Question 5: If

\(\begin{array}{l}f (x) =\frac{x βˆ’ 3}{x + 1}\end{array} \)
, then f [f { f (x) }] equals ________.

Solution:

f [ f (x) ] = (f (x) βˆ’ 3)/ ( f (x) + 1)

\(\begin{array}{l}=\frac{\frac{x-3}{x+1} – 3}{\frac{x-3}{x+1} +1}\end{array} \)
\(\begin{array}{l}=\frac{x βˆ’ 3 βˆ’ 3x βˆ’ 3}{x βˆ’ 3 + x + 1}\end{array} \)
\(\begin{array}{l}= \frac{3 + x}{1 βˆ’ x}\end{array} \)

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

\(\begin{array}{l}=\frac{\frac{3+x}{1-x}-3}{\frac{3+x}{1-x}+1}\\=\frac{3+x-3+3x}{3+x+1-x}\end{array} \)

= 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) βˆ’ [1 / 2] * [f (x / 4) + f (4x)].

Solution:

f (x) = cos (log x)

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

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

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

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

y = 0

Question 7: Let f : R β†’ R be defined by f (x) = 2x + |x|, then f (2x) + f (βˆ’x) βˆ’ 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 β‰₯ 0 and -2x; x < 0

Question 8: If f (x) = cos [Ο€2] x + cos[βˆ’Ο€2] x, where [x] stands for the greatest integer function, then find the function of the right angle.

Solution:

f (x) = cos [Ο€2] x + cos[βˆ’Ο€2] x

f (x) = cos (9x) + cos (βˆ’10x) {since Ο€ = 3.14}

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

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

Now, right angle = Ο€/2

So, f (Ο€ / 2) = 2 cos (19Ο€ / 4) cos (Ο€ / 4)

f (Ο€ / 2) = 2 *(βˆ’1 / √2) * (1/ √2)

= βˆ’1

Question 9: IfΒ 

\(\begin{array}{l}f (x) =\frac{x^2-1}{x^2+1}\end{array} \)
, for every real number, then what is the minimum value of f?

Solution:

Let

\(\begin{array}{l}f (x) = \frac{x^2-1}{x^2+1}\end{array} \)
\(\begin{array}{l}=\frac{x^2+1-2}{x^2+1}\end{array} \)

= 1 βˆ’ (2 / [x2 + 1])Β  Β  Β  Β  Β  [Because [x2 + 1] > 1 also (2 / [x2 + 1]) ≀ 2]

So 1 βˆ’ [2 / [x2 + 1]] β‰₯ 1 βˆ’ 2;

βˆ’1 ≀ f (x) < 1

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

Question 10: The function f : R β†’ R defined by f (x) = ex is ________.

Solution:

Function f: R β†’ R is defined by f (x) = ex.

Let x1, x2 ∈ R and f (x1) = f (x2) or ex1 = ex2 or x1 = x2.

Therefore, f is one-one.

Let f (x) = ex = 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 β†’ S defined by f (x) = sin x βˆ’ √3 cos x + 1 is onto, then what is the interval of S?

Solution:

Given,

f (x) = sin x βˆ’ √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)

βˆ’ √[1 + (βˆšβˆ’3)2] ≀ (sin x βˆ’ √3 cos x) ≀ √[1 + (βˆšβˆ’3)2]

βˆ’2 ≀ (sin x βˆ’ √3 cos x) ≀ 2

βˆ’2 + 1 ≀ (sin x βˆ’ √3 cos x + 1) ≀ 2 + 1

βˆ’1 ≀ (sin x βˆ’ √3 cos x + 1) ≀ 3

i.e., range = [βˆ’1, 3]

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

Question 12: What is the domain of the function

\(\begin{array}{l}f(x) =\frac{sin^{-1} (3-x)}{log(|x|-2)}\end{array} \)
?

Solution:

Given,

\(\begin{array}{l}f(x) = \frac{sin^{-1} (3-x)}{log(|x|-2)}\end{array} \)

Let g (x) = sinβˆ’1 (3 βˆ’ x)

βˆ’1 ≀ 3 βˆ’x ≀ 1

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

|x|βˆ’ 2 > 0

|x| > 2

x < βˆ’2 or x > 2

(βˆ’βˆž, βˆ’2) βˆͺ (2, ∞)

Also, log(|x| – 2) β‰  0

|x| – 2 β‰  1

|x| β‰  3

We know that (f / g) (x) = f(x) / g(x) βˆ€ x ∈ D1 ∩ D2 βˆ’ {x ∈ R : g (x) = 0}

Domain of f (x) = (2, 4] βˆ’ {3} = (2, 3) βˆͺ (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 ΞΈ ≀ 1

For minimum, cos (bx + c) = βˆ’1

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

For maximum, cos (bx + c) = 1

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

Range of f (x) = [d βˆ’ 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 β†’ 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 βˆ’ cos2x)

y = cos2 x + sin2x βˆ’ sin2x cos2x

y = 1 βˆ’ sin2x cos2x

y = 1 βˆ’ [1 / 4] * [sin22x]

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

f (R) ∈ [3/4, 1]

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

Solution:

Let f (x) = y β‡’ x = fβˆ’1 (y).

Hence, f (x) = y = 3x βˆ’ 5

\(\begin{array}{l}\Rightarrow x=\frac{y + 5}{3}\end{array} \)

β‡’fβˆ’1 (y) = x

\(\begin{array}{l}=\frac{y + 5}{3}\end{array} \)
\(\begin{array}{l}f^{-1}(x)=\frac{x+ 5}{3}\end{array} \)

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

Also, Read:

JEE Advanced Maths Function Previous Year Questions with Solutions

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