# 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 Main preparation. The questions from the previous years of JEE Main from this topic are present 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 examination. Read More.

 Let us have a brief look on 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 ⊆ 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).

## 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 / R3], [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 f (x) = $\frac{x − 3}{x + 1}$, then f [f { f (x) }] equals ________.

Solution:

f [ f (x) ] = f (x) − 3 f (x) + 1

$\frac{\frac{x-3}{x+1} – 3}{\frac{x-3}{x+1} +1}$

= $\frac{x − 3 − 3x − 3}{x − 3 + x + 1}$

$\frac{3 + x}{1 − x}$

f (x) = (2, 4] − {3}.

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

Now 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 (2x) = 2 (2x) + |2x| = 4x + 2 |x|,

y = x2 + 1,

f (x) = 2x + |x|

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

= 2|x|

Question 8: If f (x) = cos [π2] x + cos[−π2] x, then find the function of the angle.

Solution:

f (x) = cos [π2] x + cos[−π2] x

f (x) = cos (9x) + cos (−10x)

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

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

f (π / 2) = 2 cos (19π / 4) cos (π / 4);

f (π / 2) = 2 * −1 / √2 * 1/ √2

= −1

Question 9: If f (x) = $\frac{x^2-1}{x^2+1}$, for every real number. Then what is the minimum value of f?

Solution:

Let f (x) = $\frac{x^2-1}{x^2+1}$

= $\frac{x^2+1-2}{x^2+1}$

= 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 the minimum value equal to 1.

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

Solution:

Function f: R → 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 = logy.

We know that 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 into.

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

Solution:

− √[1 + (√−3)2] ≤ (sinx − √3 cosx) ≤ √[1 + (√−3)2]

−2 ≤ (sinx − √3cosx) ≤ 2

−2 + 1 ≤ (sinx − √3 cosx + 1) ≤ 2 + 1

−1 ≤ (sinx − √3 cosx + 1) ≤ 3 i.e.,

range = [−1, 3]

For f to be onto S = [−1, 3].

Question 12: What is the domain of the function f (x) = $\frac{sin^{-1} (3-x)}{log(|x|-2)}$?

Solution:

f (x) = $\frac{sin^{-1} (3-x)}{log(|x|-2)}$

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, ∞)

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)

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]

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

⇒ x = $\frac{y + 5}{3}$

⇒f−1 (y) = x = $\frac{y + 5}{3}$

f−1 (x) = $\frac{x+ 5}{3}$

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