JEE Main Maths Functions previous year questions with solutions are provided here. Students are advised to revise the previous year JEE Main questions on functions so that they can understand the exam pattern and difficulty level. They can also download the PDF of JEE Main Maths functions previous year questions with solutions for offline use. Practice the previous year JEE Main questions on functions and score better ranks for the JEE Main exam.

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Question 1:

The total number of functions, f : {1, 2, 3, 4} → {1, 2, 3, 4, 5, 6} such that f(1) + f(2) = f(3), is equal to

(A) 60

(B) 90

(C) 108

(D) 126

Solution:

Case 1:

If f(3) = 3 then f(1) and f(2) take 1 OR 2

Number of ways = 2×6

= 12

Case 2:

If f(3) = 5 then f(1) and f(2) take 2 OR 3

OR 1 and 4

Number of ways = 2×6×2

= 24

Case 3:

If f(3) = 2

then f(1) = f(2) = 1

Number of ways = 6

Case 4:

If f(3) = 4 then f(1) = f(2) = 2

Number of ways = 6

OR f(1) and f(2) take 1 and 3

Number of ways = 12

Case 5:

If f(3) = 6 then f(1) = f(2) = 3 => 6 ways

OR f(1) and f(2) take 1 and 5 => 12 ways

OR f(2) and f(1) take 2 and 4 => 12 ways

Total number of functions = 12 + 24 + 6 + 6 + 12 + 6 + 12 + 12

= 90

Hence, option (B) is the answer.

Question 2:

Let f be any function defined on R and let it satisfy the condition:|f (x) – f (y)|≤ |(x – y)²|, ∀ x, y ∈ R. If f (0) = 1, then:

(A) f (x) < 0, ∀ x ∈ R

(B) f (x) can take any value in R

(C) f (x) = 0, ∀ x ∈ R

(D) f (x) > 0, ∀ x ∈ R

Solution:

Given that |f (x) – f (y)| ≤ |(x – y)²|, ∀ x, y ∈ R

|[f (x) – f (y)] / [(x – y)]| ≤ |x – y|

lim _x→y |[f (x) – f (y)] / [(x – y)]| ≤ 0

|f’ (y)| ≤ 0

f’ (y) = 0

f (y) = c

As f(0) = 1, f (y) = 1, for all y belongs to R.

⇒ f(x) = 1

Hence, option (D) is the answer.

Question 3:

The sum of the maximum and minimum values of the function f(x) = |5x – 7| + [x² + 2x] in the interval [5/4, 2], where [t] is the greatest integer ≤ t, is

Solution:

Given that f(x) = |5x – 7| + [x² + 2x]

= |5x – 7| + [(x + 1)²] – 1

Critical points of f(x) = 7/5, √5 – 1, √6 – 1, √7 – 1, √8 – 1, 2

The minimum or maximum value of f(x) occurs at critical points.

So f(5/4) = (¾) + 4

= 19/4

f(7/5) = 0 + 4 = 4

Since both |5x – 7| and x² + 2x are increasing in nature after x = 7/5.

So, f(2) = 3 + 8

= 11

f(7/5)_min = 4

f(2)_max = 11

Therefore, the sum of the maximum and minimum values of the function f(x) = 4 + 11

= 15

Question 4:

Let f : R → R be a continuous function such that f(3x) – f(x) = x. If f(8) = 7, then f(14) is equal to

(A) 4

(B) 10

(C) 11

(D) 16

Solution:

Given that f(3x) – f(x) = x ..(1)

x→ x/3

⇒ f(x) – f(x/3) = x/3 ..(2)

Again x → x/3

⇒ f(x/3) – f(x/9) = x/3² ..(3)

Similarly,

f(x/3^n-2) – f(x/3^n-1) = x/3^n-1 ..(n)

Add the above equations and put n→ ∞

lim_n→∞ [f(3x) – f(x/3^n-1)] = x(1 + ⅓ + 1/3² + …)

f(3x) – f(0) = 3x/2

Put x = 8/3

f(8) – f(0) = 4

Given that f(8) = 7

So f(0) = 3

Put x = 14/3

f(14) – 3 = 7

⇒ f(14) = 10

Hence, option (B) is the answer.

Question 5:

Let f :R→R be defined as f(x) = 2x-1 and g:R – {1} →R be defined as g(x) = (x-½)/(x-1). Then the composition function f(g(x)) is:

(A) both one-one and onto

(B) onto but not one-one

(C) neither one-one nor onto

(D) one-one but not onto

Solution:

Given that f(x) = 2x-1

g(x) = (x-½)/(x-1)

f(g(x)) = 2g(x) – 1

= 2[(x – 1/2)/(x – 1)] – 1

= x/(x – 1)

f(g(x)) = 1 + 1/(x – 1)

So the function is one-one but not onto.

Hence, option (D) is the answer.

Question 6:

If f: R→ R is a function defined by f(x) = [x – 1] cos ((2x – 1)/2)π, where [.] denotes the greatest integer function, then f is:

(A) discontinuous only at x = 1

(B) discontinuous at all integral values of x except at x = 1

(C) continuous only at x = 1

(D) continuous for every real x

Solution:

Given that f(x) = [x – 1] cos ((2x – 1)/2)π

Doubtful points are x = n, n∈I

We find the LHL and RHL.

LHL = lim _{x→ n-}[x – 1] cos ((2x – 1)/2)π

= (n-2) cos((2n – 1)/2)π

= 0

RHL = lim _{x→ n+}[x – 1] cos ((2x – 1)/2)π

= (n-1) cos((2n – 1)/2)π

= 0

⇒ f(n) = 0

⇒ continuous for every real x.

Hence, option (D) is the answer.

Question 7:

The function f(x) = ((4x³– 3x²)/6) – 2sin x + (2x – 1)cos x

(A) increases in [1/2, ∞)

(B) decreases (-∞, 1/2]

(C) increases in (-∞, 1/2]

(D) decreases [1/2, ∞)

Solution:

Given that f(x) = ((4x³– 3x²)/6) – 2sin x + (2x – 1)cos x

f’(x) = (2x² – x) – 2 cos x + 2 cos x – sin x(2x-1)

= (2x – 1) (x – sin x )

When x>0, x – sin x > 0

When x < 0, x – sin x < 0

f’(x) ≥ 0 in x∈(-∞, 0] ⋃ [1/2, ∞)

and f’(x) ≤ 0 in x∈ [0, ½]

⇒ f(x) increases in [1/2, ∞).

Hence, option (A) is the answer.

Question 8:

Let f: R → R be defined as

Let A = {x ∈ R ∶ f is increasing}. Then A is equal to:

(A) (-5, -4) ∪ (4,∞)

(B) (-5, ∞)

(C) (-∞, -5) ∪ (4, ∞)

(D) (-∞, -5) ∪ (-4, ∞)

Solution:

Given

f’(x) = {-55 ; x< -5; 6(x² – x – 20) ; -5 < x < 4; 6(x²– x – 6) ; x > 4

= {-55 ; x< -5; 6(x – 5)(x + 4) ; –5 < x < 4; 6(x – 3)(x + 2) ; x > 4

Hence, f(x) is monotonically increasing in (-5, -4) ∪ (4, ∞)

Hence, option (A) is the answer.

Question 9:

Let f, g: N→N such that f(n + 1)= f(n)+ f(1) for all n ∈ N and g be any arbitrary function. Which of the following statements is NOT true?

(A) f is one-one

(B) If fog is one-one, then g is one-one

(C) If g is onto, then fog is one-one

(D) If f is onto, then f(n) = n for all n ∈ N

Solution:

Given that f(n + 1) = f(n) +f(1)

=> f(n + 1) – f(n) = f(1)

This implies an A.P. with common difference = f(1)

Here, the general term = T_n = f(1) + (n-1)f(1)

= nf(1)

f(n) = nf(1)

Therefore, f(n) is one-one.

For fog to be one-one, g should be one-one.

For f to be onto, f(n) must take all the values of natural numbers.

Since f(x) is increasing, f(1) =1

f(n) = n

If g is many one, then fog is many one.

So the statement “if g is onto, then fog is one-one” is not true.

Hence, option (C) is the answer.

Question 10:

Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements, and y denote the total number of one-one functions from set A to set A × B. Then:

(A) y = 273x

(B) 2y = 91x

(C) y = 91x

(D) 2y = 273x

Solution:

Number of elements in A, n(A) = 3

Number of elements in B, n(B) = 5

Number of elements in A × B = 15

Number of one-one function is x = 5 × 4 × 3

x = 60

Number of one-one function is y = 15 × 14 × 13

y = 15 × 4 × (14/4) × 13

y = 60 × 7/2 × 13

2y = (13)(7x)

2y = 91x

Hence, option (B) is the answer.

Question 11:

The number of bijective functions f : {1, 3, 5, 7, …, 99} → {2, 4, 6, 8, ….., 100} such that f(3)≥ f(9) ≥ f(15) ≥ f(21) ≥…. ≥ f(99), is

(A) ⁵⁰P₁₇

(B) ⁵⁰P₃₃

(C) 33! ×17!

(D) 50!/2

Solution:

Since the function is one-one and onto, out of 50 elements of the domain set, 17 elements follow the restriction

f(3) > f(9) > f(15) ….. > f(99)

Therefore the number of ways = ⁵⁰C ₁₇ ×1 × 33!

= ⁵⁰P₃₃

Hence, option (B) is the answer.

Question 12:

If g(x) = x² + x – 1 and (gof)(x) = 4x² – 10x + 5, then f(5/4) is equal to

(A) -3/2

(B) -1/2

(C) 1/2

(D) 3/2

Solution:

Given that g(x) = x² + x – 1

gof(x) = 4x² – 10x + 5

g(f(5/4)) = 4×(5/4)² – 10×(5/4) + 5

= (25/4) – (25/2) + 5

= (-25/4) + 5

= -5/4

g(f(x)) = (f(x))² + f(x) – 1

g(f(5/4)) = (f(5/4))² + f(5/4) – 1

-5/4 = (f(5/4))² + f(5/4) – 1

(f(5/4))² + f(5/4) + (¼) = 0

=> [f(5/4) + (½)]² = 0

=> f(5/4) = -1/2

Hence, option (B) is the answer.

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