# JEE Main Relations and Functions Previous Year Questions With Solutions

Relations and Functions is one of the important concepts of JEE. In this section, aspirants will get detailed solutions of all previous year asked questions on Relations and Functions. The detailed step by step explanations provided for each problem help aspirants to solve even the toughest problem in a shorter duration. Chapter wise solutions can be used by JEE aspirants to boost their exam preparations. Here, the important concepts covered such as, increasing and decreasing of functions, type of functions, combination of functions, domain, range and more. Students are advised to download your pdf now and start practice.

### Important Points to Know:

 – Domain: A collection of the first values in the ordered pairs. (input Values) – Range: A collection of the second values in the ordered pairs (output values). – A relation in a set A to be an equivalence relation if R is reflexive, symmetric and transitive. Reflexive: If (x, x) Є R, for x Є A. Symmetric: If (x1, x2) Є R => for all x1, x2 Є R for all x1 , x2 Є A. Transitive: If (x1, x2) Є R and (x2, x3) Є R => for all x1, x3 Є R for all x1, x2, x3 Є A. – One-one and Many-one Functions A function f:X ->Y is one-function, if different elements of X have unique images in Y. A function f:X ->Y is Many-one function, if two or more elements of X have the same image in Y. – Types of Relations: Empty Relations, Inverse Relations, Universal Relations, Identity Relations, Reflexive Relations, Symmetric Relations, Transitive Relations. – Types of Functions: One to one function or Injective function, Many to one function, Onto Function or Surjective function, One-one and Onto function or Bijective function.

## JEE Main Past Year Questions With Solutions on Relations and Functions

Question 1: The function f : [0, 3] -> [1, 29] defined by f(x) = 2x3 – 15x2 + 36x + 1 is

(a) one-one and onto

(b) onto but not one-one

(c) one-one but not onto

(d) neither one-one nor onto

Solution:

The function f : [0, 3] -> [1, 29] defined by

f(x) = 2x3 – 15x2 + 36x + 1

f'(x) = 6x2 – 30x + 36

= − 6(x2 – 5x + 6)

= 6(x – 2)(x – 3)

f(x) is increasing in [0, 2] and decreasing in [2, 3].

We know, if function is strictly increasing or decreasing in its domain, then it is one-one. But given function is increasing as well as decreasing.

So, f(x) is many one.

f(0) = 1

f(2) = 29 and

f(3) = 28

Range is [1, 29].

The function is onto.

Question 2: Let f(x) = x2 and g(x) = sin x for all x Є R. Then the set of all x satisfying (f o g o g o f)(x) = (g o g o f)(x), where (f o g)(x) = f(g(x)) is.

Solution:

(g o f)(x) = g(f(x) = sin x2

g o (g o f)(x) = g(g(f(x)) = sin(sin x2)

(f o g o g o f) (x) = [sin (sin x2)]2 …(i)

Again,

(g o g o f) (x) = sin (sin2x) …(ii)

Now,

[sin (sin x2)]2 = sin (sin x2)

sin(sin x2)(sin(sin x2 – 1) = 0

=>sin x2 = 0, sin x2 = π/2

But, sin x2 = π/2 can not be a solution as sin x Є [-1, 1]

=>x2 = n π or x = ± √nπ where n Є {0, 1, 2, 3,…..}

Question 3: A function f from the set of natural numbers to integers defined by

f(n) = (n – 1)/2, when n is odd, and

f(n) = -n/2, when n is even

(a) one-one but not onto

(b) onto but not one-one

(c) one-one and onto both

(d) neither one-one nor onto

Solution:

If n = odd: clearly shown that function is one-one and onto.

For example, if f: N -> I

Let x, y Є N and both are even.

so, f(x) = f(y) => -x/2 = -y/2 => x = y

If n = even, values are set of all negative integers. Again, function is one-one and onto.

For example, if f: N -> I

Let x, y Є N and both are odd.

so, f(x) = f(y) => (x-1)/2 = (y-1)/2 => x = y

Therefore, option (c) is correct.

Question 4: The range of the function f(x) = 3|sin x| – 2|cos x| is

(a) [-2, √13]

(b) [-2, 3]

(c) [-3, 2]

(d) [3, √13]

Solution:

f(x) = 3|sin x| – 2|cos x|

If f(x) is continuous function and |sin x| and |cos x| are always positive.

Find Minimum and Maximum value of f(x):

f(x) is minimum when |sin x| = 0 and |cos x| = 1

The minimum value will be = 0 – 3 = -3

f(x) is max when |sin x| = 1 and |cos x| = 0

The max value will be = 2 – 0 = 2

The required range is [-3, 2]

Question 5: If f:R-> R satisfy f(x + y) = f(x) + f(y), for all x, y Є R and f(1) = 7 then $\sum_{r=1}^n f(r)$ is

(a) 7n(n + 1)

(b) 7n/2

(c) $\frac{7(n+1)}{2}$

(d) $\frac{7n(n+1)}{2}$

Solution:

f(x + y) = f(x) + f(y) for all x, y Є R

f(1 + 1) = 2 f(1) = 2(7)

f(2) = 2(7)

Therefore, f(3) = f(1) + f(2) = 7 + 2(7) = 3(7)

=> f(x) = ax

=> a(1) = 7 => a = 7

f(x) = 7x

=> $\sum_{r=1}^n f(r) = 7(1 +2r+…+n) = \frac{7n(n+1)}{2}$

Question 6: The range of the function f(x) = 7-xPx-3 is

(a) {1, 2, 3}

(b) {1, 2, 3, 4}

(c) {1, 2, 3, 4, 5}

(d) {1, 2, 3, 4, 5, 6}

Solution:

Given function is f(x) = 7-xPx-3

When 7 – x ≥ 0 => x ≤ 7

When x – 3 ≥ 0 => x ≥ 3

And 7 – x ≥ x – 3 => x ≤ 5

This implies, 3 ≤ x ≤ 5

=>x = 3, 4, 5

So, the range is {1, 2, 3}

Question 7: If fx(x) = 1/k (sinkx + coskx) where x Є R and k ≥ 1, then f4(x) – f6(x) is

(a) 1/6                (b) 1/4                  (c) 1/5                     (d) 1/12

Solution:

fx(x) = 1/k (sinkx + coskx)

At x = 4 and at x = 6

f4(x) = 1/4 (sin4x + cos4x) and f6(x) = 1/6 (sin6x + cos6x)

Now, f4(x) – f6(x) = 1/4 (sin4x + cos4x) – 1/6 (sin6x + cos6x) …(1)

Using identity: (a4 + b4) = (a2 + b2)2 – 2a2b2 and (a6 + b6) = (a2 + b2)3 – 3a2b2(a2 + b2) and also, we know, sin2 x + cos2x = 1

(1) => f4(x) – f6(x) = 1/4(1 – 2 sin2 x cos2x) – 1/6 (1 – 3 sin2 x cos2x)

= 1/4 -1/6

= 1/12

Question 8: The function f: N -> N defined by f(x) = x – 5[x/5], where N is the set of natural numbers and [x] denotes the greatest integer less than or equal to x, is

(a) one-one and onto

(b) one-one but not onto

(c) onto but not one-one

(d) neither one-one nor onto

Solution:

The function f: N -> N defined by f(x) = x – 5[x/5], where N is the set of natural numbers and [x] denotes the greatest integer.

Consider x in intervals of 5 natural numbers.

Where x Є N.

From the above graph, we can conclude that f(x) is neither one-one nor onto.

Question 9: Let f : (–1, 1) -> B, be a function defined by f(x) = tan-1(2x)/(1 – x2), then f is both one-one and onto when B is the interval

(a) [0, π/2]                        (b) (0, π/2)                           (c) (- π/2, π/2)                          (d) [-π/2, π/2]

Solution:

f(x) = tan-1(2x)/(1 – x2)

For x Є (-1, 1)

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

Replace x by tan A

$f(tan A) = tan^{-1}(\frac{2tan A}{1-tan^2 A})$

f(tan A) = tan-1 (tan 2A) = 2 tan-1 A

=> -π/2 < tan-1 (2x/(1-x2) < π/2

Question 10: The function f: R -> [-1/2, 1/2] defined as f(x) = x/(1+x2), is

(a) invertible

(b) Surjective but not injective

(c) Neither injective nor surjective

(d) injective but not surjective

Solution:

Given: f(x) = x/(1+x2)

Differentiate above function:

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

Let y = x/(1+x2)

or yx2 – x + y = 0

for y = 0 => x = 0

[for y ≠ 0]

And Range: [-1/2, 1/2]

Therefore, function is surjective but not injective.