JEE Advanced Maths Function Previous Year Questions with Solutions

Function JEE Advanced previous year questions with solutions are given on this page. These step-by-step solutions are prepared by our subject experts. The important topics in functions include domain, codomain, range of functions, relations, types of functions, etc. Practising these solutions will definitely boost the confidence of the students. Students are recommended to practise these solutions so that they can learn the problems easily. These solutions help students to get an idea of what type of questions are asked for the JEE Advanced exam.

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Question 1: If R = {(x, y) : x, y ∈ Z, x² + 3y² ≤ 8} is a relation on the set of integers Z, then the domain of R^-1 is

(a) {-2, -1, 1, 2}

(b) {0, 1}

(c) {-2, -1, 0, 1, 2}

(d) {-1, 0, 1}

Solution:

Given R = {(x, y) : x, y ∈ Z, x² + 3y² ≤ 8}

when x = 0, 3y² ≤ 8

Domain of R^-1 = value of y

= {-1, 0, 1}

Hence option d is the answer.

Question 2: Let the function f: [0, 1] → R be defined by f(x) = 4^x/(4^x + 2). Then the value of f(1/40) + f(2/40) + f(3/40) + … + f(39/40) – f(1/2)

Solution:

Given f(x) = 4^x/(4^x + 2)

f(1-x) = 4^1-x/(4^1-x + 2)

f(x) + f(1-x) = 4^x/(4^x + 2) + 4^1-x/(4^1-x + 2)

= 4^x/(4^x + 2) + (4/4^x)/((4/4^x) + 2)

= 4^x/(4^x + 2) + 2/(2 + 4^x)

= (4^x + 2)/(4^x + 2)

= 1

f(1/40) + f(2/40) + f(3/40) + … + f(39/40) – f(1/2) = (f(1/40) + f(39/40)) + f(2/40) + f(38/40) + …+ f(20/40) – f(½)

= (1 + 1 + …19 times) + f(½) – f(½)

= 19

Question 3: 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:

Given f(x) = 2x³ – 15x² + 36x + 1

f’(x) = 6x² – 30x + 36

= 6(x – 2)(x – 3)

f’(x) > 0

x ∈ (-∞, 2) ⋃ (3, ∞)

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

So f(x) is not one-one.

It is many one.

f(0) = 1, f (2) = 29, f (3) = 28

Range = [1, 29]

So f(x) is onto.

Hence option b is the answer.

Question 4: Let the function, f:[-7, 0] →R be continuous on [-7, 0] and differentiable on (-7, 0). If f(-7)= -3 and f’(x) ≤ 2, for all x∈ (-7, 0), then for all such functions f, f(-1) + f(0) lies in the interval:

(a) [-6, 20]

(b) (-∞, 20]

(c) (-∞, 11]

(d) [-3, 11]

Solution:

Given f(-7) = -3 and f’(x) ≤ 2

Applying LMVT in [-7,0], we get

(f(-7) – f(0))/-7 = f’(c) ≤ 2

(-3-f(0))/-7 ≤ 2

f(0) + 3 ≤ 14

f(0) ≤ 11

Applying LMVT in [-7, -1], we get

(f(-7) – f(-1))/(-7 +1) = f’(c) ≤ 2

-3 – f(-1))/-6 = f’(c) ≤ 2

f(-1) + 3 = ≤ 12

f(-1) ≤ 9

Therefore f(-1)+ f(0) ≤ 20

Hence option b is the answer.

Question 5: Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If α is the number of one-one functions from X to Y and β is the number of onto functions from Y to X, then the value of (1/5!)(β – α) is

Solution:

Given n(X) = 5

n(Y) = 7

α = number of one-one functions from X to Y = ⁷C₅ ×5! = 2520

β = number of onto functions from Y to X = (⁷C₃ + 3. ⁷C₃)5!

= 4× ⁷C₃ × 5!

= 16800

(β – α )/5! = (16800 – 2520)/5!

= 119

Question 6: Let [ t ] denote the greatest integer ≤ t. Then the equation in x, [ x ]² + 2[ x+2 ] – 7 = 0 has

(a) exactly four integral solutions.

(b) infinitely many solutions.

(c) no integral solution.

(d) exactly two solutions.

Solution:

Given [x]²+2[x+2]-7 = 0

let [x] = y

y²+3y-y -3 = 0

(y -1) (y+3) = 0

=> x ∈ [-3, -2) ⋃ [1, 2)

Hence option b is the answer.

Question 7: Let f(x) be a quadratic polynomial such that f(-1) + f(2) = 0. If one of the roots of f(x) = 0 is 3, then its other root lies in

(a) (-3, -1)

(b) (1, 3)

(c) (-1, 0)

(d) (0, 1)

Solution:

Let f(x) = ax² + bx + c

f(-1) = a – b + c

f(2) = 4a + 2b + c

Given f(-1) + f(2) = 0

=> a – b + c + 4a + 2b + c = 0

=> 5a + b + 2c = 0 ..(i)

Given that one root is 3.

f(3) = 0 => 9a + 3b + c = 0 ..(ii)

Multiply (i) by 3 and subtract (ii) from it, we get a/-5 = c/6

=> c/a = -6/5

Product of roots, αβ = c/a = -6/5

α = 3

So β = -2/5 ∈ (-1,0)

Hence option c is the answer.

Question 8: If f(x + y) = f(x) f(y) and ∑_x=1^∞ f(x) = 2. x, y∈ N, where N is the set of all natural number, then the value of f(4)/f(2) is:

(a) 2/3

(b) 1/9

(c) 1/3

(d) 4/9

Solution:

f(x + y) = f(x) f(y)

Put x = 1, y = 1

f(2) = (f(1))²

Put x = 2, y = 1

f(3) = f(2). f(1) = (f(1))³

Put x = 2, y = 2

f(4) = (f(2))² = (f(1))⁴

So f(n) = (f(1))ⁿ

∑_x=1^∞ f(x) = f(1) + f(2) + f(3) +…= 2

f(1)+(f(1))²+(f(1))³+… = 2

f(1)/(1-f(1)) = 2

f(1) = 2/3

f(2) = (2/3)²

f(4) = (2/3)⁴

f(4)/f(2) = (2/3)⁴/(2/3)² = 4/9

Hence option d is the answer.

Question 9: If f(x) = log_e((1 – x)/(1+x)), |x| < 1, then f(2x/(1+x²)) is equal to:

(a) 2f(x)

(b) 2f(x²)

(c) (f(x))²

(d) -2f(x)

Solution:

Given f(x) = log_e((1 – x)/(1+x))

f(2x/(1+x²)) = log_e((1 – 2x/(1+x²))/(1+2x/(1+x²)))

= log_e((1+x² – 2x)/(1+x²+2x))

= log_e((1-x)²/(1+x)²)

= log_e ((1-x)/(1+x))²

= 2 log ((1-x)/(1+x)

= 2 f(x)

Hence option a is the answer.

Question 10: Let f(n) = [⅓ + 3n/100]n, where [n] denotes the greatest integer less than or equal to n. Then ∑_n=1⁵⁶f(n) is equal to

(a) 56

(b) 689

(c) 1287

(d) 1399

Solution:

f(x) = [(⅓) + 3x/100]x

f(1) to f(22) = 0

f(23) = [(⅓) + (69/100)]23

f(24) = [(⅓) + (72/100)]24

f(23) + f(24) + …f(56) = 23 + 24 + …+55 + 2×56

= 33×39 + 112

= 1399

Hence option d is the answer.

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