JEE Main Previous Year Questions (2016- 2024): Functions | Mathematics (Maths) Class 12 PDF Download

Q.1. For three positive integers  and r = pq + 1 such that 3,3log_y ⁡x, 3log_z ⁡y, 7log_x ⁡z are in A.P. with common difference 1/2. Then r − p − q is equal to     (JEE Main 2023)
(a) -6
(b) 12
(c) 6
(d) 2

Ans. d

& r = pq + 1

Q.2. Let |M| denote the determinant of a square matrix M. Let g: [0, π/2] → R be the function defined by     (JEE Advanced 2022)

Solution:  

where

Let p(x) be a quadratic polynomial whose roots are the maximum and minimum values of the function g(θ), and p(2) = 2 − √2. Then, which of the following is/are TRUE?     (JEE Advanced 2022)
(a)
(b)
(c)
(d)

Correct Answe r is Option (a, c)

Q.3. The domain of the function     (JEE Main 2022)
(a) [1, ∞)
(c) [−1, 2]
(c) [−1, ∞)
(d) (−∞, 2]

Correct Answer is Option (c)
Solution:  


x² + 2x + 7
5x ≥ − 5
x ≥ −1

x² − 3x + 2 ≥ − x² − 2x − 7
2x² − x + 9 ≥ 0
x ∈ R
(i) ∩ (ii)
Domain ∈ [−1, ∞)

Q.4. The function f(x) = xe^x(1−x), x ∈ R, is:      (JEE Main 2022)
(a) increasing in
(b) decreasing in
(c) increasing in
(d) decreasing in

Correct Answer is Option (a)
Solution:  
f(x) = xe^x(1−x), x ∈ R
f′(x) = xe^x(1−x) . (1 − 2x) + e^x(1−x) 
= e^x(1−x)[x − 2x² + 1]
= −e^x(1−x)[2x² − x − 1]
= −e^x(1−x)(2x + 1)(x − 1)
∴ f(x) is increasing inand decreasing in

Q.5. Let be such that and Let f(x) = ax² + bx + c be such that f(1) = 3, f(−2) = λ and  f(3) = 4. If f(0) + f(1) + f(−2) + f(3) = 14, then λ is equal to:      (JEE Main 2022)
(a) −4
(b) 13/2
(c) 23/2
(d) 4

Correct Answer is Option (d)
Solution:  
f(1) = a + b + c = 3 ..... (i)
f(3) = 9a + 3b + c = 4 .... (ii)
f(0) + f(1) + f(−2) + f(3) = 14
OR c + 3 + (4a − 2b + c) + 4 = 14
OR 4a − 2b + 2c = 7 ..... (iii)
From (i) and (ii) 8a + 2b = 1 ..... (iv)
From (iii) −(2) × (i)
⇒ 2a − 4b = 1 ..... (v)
From (iv) and (v) a = 1/6, b = −1/6 and c = 3
f(−2) = 4a − 2b + c
4/6 + 2/6 + 3 = 4

Q.6. Let α, β and γ be three positive real numbers. Let f(x) = αx⁵ + βx³ + γx, x ∈ R and g : R → R be such that g(f(x)) = x for all x ∈ R. If a₁, a₂, a₃,…, a_n be in arithmetic progression with mean zero, then the value ofis equal to:      (JEE Main 2022)
(a) 0
(b) 3
(c) 9
(d) 27

Correct Answer is Option (a)
Solution:  


∴ First and last term, second and second last and so on are equal in magnitude but opposite in sign.
f(x) = αx⁵ + βx³ + γx  = 0α + 0β + 0γ
= 0

Q.7. Considering only the principal values of the inverse trigonometric functions, the domain of the function
      (JEE Main 2022)
(a) (−∞, 1/4]
(b)
(c) (−1/3, ∞)
(d) (−∞, 1/3]

Correct Answer is Option (b)
Solution:  

⇒ −x² − 3 ≤ x² − 4x + 2 ≤ x² + 3
⇒ 2x² − 4x + 5 ≥ 0 & −4x ≤ 1
x ∈ R & x ≥
So domain is 

Q.8. The domain of the functionwhere [t] is the greatest integer function, is:      (JEE Main 2022)
(a)
(b)
(c)
(d)

Correct Answer is Option (c)
Solution: 
−1 ≤ 2x² − 3 < 2
or 2 ≤ 2x² < 5
or 1 ≤ x² < 5/2
 
log_1/2(x² − 5x + 5) > 0
0 < x² − 5x + 5 < 1
x² − 5x + 5 > 0 & x² − 5x + 4 < 0

& x ∈ (−∞, 1) ∪ (4, ∞)
Taking intersection

Q.9. Let f, g : N − {1} → N be functions defined by f(a) = α, where α is the maximum of the powers of those primes p such that pα divides a, and g(a) = a + 1, for all a ∈ N − {1}. Then, the function f + g is     (JEE Main 2022)
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto

Correct Answer is Option (d)
Solution:  
f, g : N − {1} → N defined as
f(a) = α, where α is the maximum power of those primes p such that pα divides a.
g(a) = a + 1,
Now,
f(2) = 1, g(2) = 3 ⇒ (f + g)(2) = 4
f(3) = 1, g(3) = 4 ⇒ (f + g)(3) = 5
f(4) = 2, g(4) = 5 ⇒ (f + g)(4) =7
f(5) = 1, g(5) = 6 ⇒ (f + g)(5) = 7
∵ (f + g)(5) = (f + g)(4)
∴ f+g is not one-one
Now, ∵ f_min = 1, g_min = 3
So, there does not exist any x ∈ N − {1} such that (f + g)(x) = 1, 2, 3
∴ f + g is not onto

Q.10. If the maximum value of a, for which the function fa(x) = tan−1⁡2x − 3ax + 7 is non-decreasing in  is equal to     (JEE Main 2022)
(a)
(b)
(c)
(d)

Correct Answer is Option (a)
Solution:  
f_a(x) = tan⁻¹2x − 3ax + 7
∵ f_a(x) is non-decreasing in

Q.11. 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:     (JEE Main 2022)
(a) 4
(b) 10
(c) 11
(d) 16

Correct Answer is Option (b)
Solution:  
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
 
Adding all these and applying n → ∞
 
f(3x) − f(0) = 3x/2
Putting x = 8/3
f(8) − f(0) = 4
⇒ f(0) = 3
Putting x = 14/3
f(14) − 3 = 7 ⇒ f(14) = 0

Q.12. 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 ____________.     (JEE Main 2022)
(a) ⁵⁰P₁₇
(b) ⁵⁰P₃₃
(c) 33! × 17!
(d) 50!/2 

Correct Answer is Option (b)
Solution:  
As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction f(3) > f(9) > f(15) ....... > f(99)
So number of ways = ⁵⁰C₁₇ . 1 . 33!
= 50!/2 

Q.13. If the absolute maximum value of the functionin the interval [−3, 0] is f(α), then:     (JEE Main 2022)
(a) α = 0
(b) α = −3
(c) α ∈ (−1, 0)
(d) α ∈ (−3, −1)

Correct Answer is Option (b)

Q.14. 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:     (JEE Main 2022)
(a) 60
(b) 90
(c) 108
(d) 126

Correct Answer is Option (b)
Solution:  
Given, f(1) + f(2) = f(3)
It means f(1), f(2) and f(3) are dependent on each other. But there is no condition on f(4), so f(4) can be f(4) = 1, 2, 3, 4, 5, 6.
For f(1), f(2) and we have to find how many functions possible which will satisfy the condition f(1) + f(2) = f(3)
Case 1: 
When f(3) = 2 then possible values of f(1) and f(2) which satisfy f(1) + f(2) = f(3) is f(1) = 1 and f(2) = 1.
And f(4) can be = 1, 2, 3, 4, 5, 6
∴ Total possible functions = 1 × 6 = 6
Case 2: 
When f(3) = 3 then possible values
(1) f(1) = 1 and f(2) = 2 (2)
f(1) = 2 and f(2) = 1
And f(4) can be = 1, 2, 3, 4, 5, 6.
∴ Total functions = 2 × 6 = 12
Case 3: 
When f(3) = 4 then
(1) f(1) = 1 and f(2) = 3
(2) f(1) = 2 and f(2) = 2
(3) f(1) = 3 and f(2) = 1
And f(4) can be = 1, 2, 3, 4, 5, 6
∴ Total functions = 3 × 6 = 18
Case 4: 
When f(3) = 5 then
(1) f(1) = 1 and f(4) = 4
(2) f(1) = 2 and f(4) = 3
(3) f(1) = 3 and f(4) = 2
(4) f(1) = 4 and f(4) = 1
And f(4) can be = 1, 2, 3, 4, 5 and 6
∴ Total functions = 4 × 6 = 24
Case 5: 
When f(3)=6 then
(1) f(1) = 1 and f(2) = 5
(2) f(1) = 2 and f(2) = 4
(3) f(1) = 3 and f(2) = 3
(4) f(1) = 4 and f(2) = 2
(5) f(1) = 5 and f(2) = 1
And f(4) can be = 1, 2, 3, 4, 5 and 6
∴ Total possible functions = 5 × 6 = 30
∴ Total functions from those 5 cases we get
= 6 + 12 + 18 + 24 + 30 = 90

Q.15. Let  and S2 = {x ∈ R : 3^2x − 3^x+1 − 3^x+2 + 27 ≤ 0}. Then, S₁ ∪ S₂ is equal to:     (JEE Main 2022)
(a) (−∞, −2] ∪ (1, 2)
(b) (−∞, −2] ∪ [1, 2]
(c) (−2, 1] ∪ [2, ∞)
(d) (−∞, 2]

Correct Answer is Option (b)
Solution:  
Given,

x² + 3x + 5 is a quadratic equation
a = 1 > 0 and D = (−3)² − 4 . 1 . 5 = −11 < 0
∴ x² + 3x + 5 > 0 (always)
So, we can ignore this quadratic term



∴ x ∈ (−α, −2] ∪ (1, 2)
∴ S₁ = (−α, −2] ∪ (1, 2)
Now,
3^2x − 3^x+1 − 3^x+2 + 27 ≤ 0
⇒ (3^x)² − 3 . 3^x − 3² . 3^x + 27 ≤ 0
Let 3^x = t
⇒ t² − 3 . t − 3² . t + 27 ≤ 0
⇒ t(t − 3) − 9(t − 3) ≤ 0
⇒ (t − 3)(t − 9) ≤ 0

∴ 3 ≤ t ≤ 9
⇒ 3¹ ≤ 3^x ≤ 3² 
⇒ 1 ≤ x ≤ 2
∴ x ∈ [1, 2]
∴ S₂ = [1, 2]
∴ S₁ ∪ S₂ = (−α, 2] ∪ (1, 2) ∪ [1, 2]

Q.16. The domain of the function     (JEE Main 2022)
(a)
(b) (−∞, −1] ∪ [1, ∞) ∪ {0}
(c)
(d)

Correct Answer is Option (d)
Solution:  





From (3) and (4), we get

Q.17. Let a function f : N → N be defined by     (JEE Main 2022)

then, f is
(a) one-one but not onto
(b) onto but not one-one
(c) neither one-one nor onto
(d) one-one and onto

Correct Answer is Option (d)
Solution:  
When n = 1, 5, 9, 13 thenwill give all odd numbers.
When n = 3, 7, 11, 15 .....
n − 1 will be even but not divisible by 4
When n = 2, 4, 6, 8 .....
Then 2n will give all multiples of 4
So range will be N.
And no two values of n give same y, so function is one-one and onto.

Q.18. Let f : R → R be defined as f (x) = x − 1 and g : R − {1, −1} → R be defined as g(x)=Then the function fog is:     (JEE Main 2022)
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto

Correct Answer is Option (d)
Solution:  
f : R → R defined as
f(x) = x − 1 and g : R → {1, −1} → R, g(x) = 
 
∴ Domain of fog(x)=R−{−1,1}
And range of fog(x) = (−∞, −1] ∪ (0, ∞)
 
 
 
∴ fog(x) is neither one-one nor onto.

Q.19. Let f(x) = 2cos⁻¹x + 4cot⁻¹x − 3x² − 2x + 10, x ∈ [−1, 1]. If [a, b] is the range of the function f, then 4a − b is equal to:    (JEE Main 2022)
(a) 11
(b) 11 − π
(c) 11 + π
(d) 15 − π

Correct Answer is Option (b)
Solution:  
f(x) = 2cos⁻¹x + 4cot⁻¹x − 3x² − 2x + 10 ∀ x ∈ [−1, 1]

 So f(x) is decreasing function and range of f(x) is [f(1), f(−1)], which is [π + 5, 5π + 9]
Now 4a − b = 4(π + 5) − (5π + 9)
= 11 − π

Q.20. Let f(x) =, x ∈ R − {0, −1, 1}. If fⁿ⁺¹(x) = f(fⁿ(x)) for all n ∈ N, then f⁶(6) + f⁷(7) is equal to:    (JEE Main 2022)
(a) 7/6
(b)
(c) 7/12
(d)

Correct Answer is Option (b)
Solution:  
Given,

Also given, fⁿ⁺¹(x) = f(fⁿ(x)) ..... (1)
∴ For n = 1
f¹⁺¹(x) = f(f¹(x))
⇒ f²(x) = f(f(x))

From equation (1), when n = 2
f²⁺¹(x) = f(f²(x))
⇒ f³(x) = f(f²(x))
 
Similarly,
f⁴(x) = f(f³(x))

 ∴ f⁵(x) = f(f⁴(x))
= f(x)
 
f⁶(x) = f(f⁵(x))
 
= −1x (Already calculated earlier)
f⁷(x) = f(f⁶(x))

 ∴ f⁶(6) =
 
So, f⁶(6) + f⁷(7)

Q.21. Let f : R → R and g : R → R be two functions defined by f(x) = log_e(x² + 1) − e^−x + 1 and  Then, for which of the following range of α, the inequality  holds?    (JEE Main 2022)
(a) (2, 3)
(b) (−2, −1)
(c) (1, 2)
(d) (−1, 1)

Correct Answer is Option (a)
Solution:  
f(x) = log_e(x² + 1) − e^−x + 1
 

g(x) = e^−x − 2e^x 
g′(x)−−e^−x − 2e^x < 0 ∀x ∈ R
⇒ f(x) is increasing and g(x) is decreasing function.

 
= α² − 5α + 6 < 0
= (α − 2)(α − 3) < 0
= α ∈ (2, 3)

Q.22. Let f(x) be a polynomial function such that f(x) + f′(x) + f″(x) = x⁵ + 64. Then, the value of is equal to:    (JEE Main 2022)
(a) −15
(b) −60
(c) 60
(d) 15

Correct Answer is Option (a)
Solution:  

f(x) + f′(x) + f″(x) = x⁵ + 64
Let f(x) = x⁵ + ax⁴ + bx³ + cx² + dx + e
f′(x) = 5x⁴ + 4ax³ + 3bx² + 2cx + d
f″(x) = 20x³ + 12ax² + 6bx + 2c
x5(a + 5)x⁴ + (b + 4a + 20)x³ + (c + 3b + 12a)x² + (d + 2c + 6b)x + e + d + 2c = x⁵ + 64
⇒ a + 5 = 0
b + 4a + 20 = 0
c + 3b + 12a = 0
d + 2c + 6b = 0
e + d + 2c = 64
∴ a = −5, b = 0, c = 60, d = −120, e = 64
∴ f(x) = x⁵ − 5x⁴ + 60x² − 120x + 64
 
By L' Hospital rule

= -15

Q.23. Let f : R → R be defined as f(x) = x³ + x − 5. If g(x) is a function such that f(g(x)) = x, ∀′x′ ∈ R, then g'(63) is equal to ______________.    (JEE Main 2022)
(a) 1/49
(b) 3/49
(c) 43/49
(d) 91/49

Correct Answer is Option (a)
Solution:  
f(x) = 3x² + 1
f'(x) is bijective function
and f(g(x)) = x ⇒ g(x) is inverse of f(x)
g(f(x)) = x
g′(f(x)) . f′(x) = 1
g′(f(x)) =
Put x = 4 we get
g′(63) = 1/49

Q.24. Let f : N → R be a function such that f(x + y) = 2f(x)f(y) for natural numbers x and y. If f(1) = 2, then the value of α for whichholds, is:    (JEE Main 2022)
(a) 2
(b) 3
(c) 4
(d) 6

Correct Answer is Option (c)
Solution:  
Given,
f(x + y) = 2f(x)f(y)
and f(1) = 2
For x = 1 and y = 1,
f(1 + 1) = 2f(1)f(1)
⇒ f(2) = 2(f(1))² = 2(2)² = 2³ 
For x = 1, y = 2,
f(1 + 2) = 2f(1)y(2)
⇒ f(3) = 2 . 2 . 2³ = 2⁵ 
For x = 1, y = 3,
f(1 + 3) = 2f(1)f(3)
⇒ f(4) = 2 . 2 . 2⁵ = 2⁷ 
For x = 1, y = 4,
f(1 + 4) = 2f(1)f(4)
⇒ f(5) = 2 . 2 . 27 = 29 ..... (1)
Also given

⇒ f(α + 1) + f(α + 2) + f(α + 3) + ... + f(α + 10) = 512/3(2²⁰ − 1)
⇒ f(α + 1) + f(α + 2) + f(α + 3) + .... + f(α + 10) =
This represent a G.P with first term = 2⁹ and common ratio = 2² 
∴ First term = f(α + 1) = 29 ..... (2)
From equation (1), f(5) = 29
∴ From (1) and (2), we get f(α + 1) = 2⁹ = f(5)
⇒ f(α + 1) = f(5)
⇒ f(α + 1) = f(4 + 1)
Comparing both sides we get,
α = 4

Q.25. The domain of the function
    (JEE Main 2022)
(a) (−∞, 1) ∪ (2, ∞)
(b) (2, ∞)
(c)
(d)

Correct Answer is Option (d)
Solution:  


The solution to this inequality is

for x² − 3x + 2 > 0 and ≠ 1

Combining the two solution sets (taking intersection)

Q.26. The sum of absolute maximum and absolute minimum values of the function f(x) = |2x² + 3x − 2| + sin⁡x cos⁡x in the interval [0, 1] is:    (JEE Main 2022)
(a)
(b)
(c)
(d)

Correct Answer is Option b
Solution:  

 f′(x) = −4x − 3 + cos⁡2x < 0
For x ≥ 1/2: f′(x) = 4x + 3 + cos⁡2x > 0
So, minima occurs at x = 1/2


So, maxima is possible at x = 0 or x = 1
Now checking for x = 0 and x = 1, we can see it attains its maximum value at x = 1

 
Sum of absolute maximum and minimum value

Q.27. For the function f(x) = 4log_e(x − 1) − 2x² + 4x + 5, x > 1, which one of the following is NOT correct?   (JEE Main 2022)
(a) f is increasing in (1, 2) and decreasing in (2, ∞)
(b) f(x) = −1 has exactly two solutions
(c) f′(e) − f″(2) < 0
(d) f(x) = 0 has a root in the interval (e, e + 1)

Correct Answer is Option (c)
Solution:  
f(x) = 4log_e⁡(x − 1) − 2x² + 4x + 5, x > 1

For 1 < x < 2 ⇒ f′(x) > 0
For x > 2 ⇒ f′(x) < 0 (option A is correct)
f(x) = −1 has two solution (option B is correct)
f(e) > 0
f(e + 1) < 0
f(e) ⋅ f(e + 1) < 0 (option D is correct)

(option C is incorrect)

Q.28. For p, q ∈ R, consider the real valued function f(x) = (x − p)² − q, x ∈ R and q > 0. Let a₁, a₂,  a₃ and a₄ be in an arithmetic progression with mean p and positive common difference. If |f(a_i)| = 500 for all i = 1, 2, 3, 4, then the absolute difference between the roots of f(x) = 0 is ___________.    (JEE Main 2022)

Solution:  
∵ a₁, a₂, a₃, a₄ 
∴ a₂ = p − 3d, a₂ = p − d, a₃ = p + d and a₄ = p + 3d
Where d > 0
∵ |f(a_i)| = 500
⇒ |9d² − q| = 500
and |d² − q| = 500 ..... (i)
either 9d² − q = d2 − q
⇒ d = 0 not acceptable
∴ 9d² − q = q − d² 
∴ 5d² − q = 0 ..... (ii)
Roots of f(x) = 0 are p + √q and p − √q
∴ absolute difference between roots = |2√q| = 50

Q.29. The number of functions f, from the set A = {x ∈ N : x² − 10x + 9 ≤ 0} to the set B = {n² : n ∈ N} such that f(x) ≤ (x − 3)² + 1, for every x ∈ A, is ___________.    (JEE Main 2022)

Solution:  
A = {x ∈ N, x² − 10x + 9 ≤ 0}
= {1, 2, 3, ...., 9}
B = {1, 4, 9, 16, .....}
f(x) ≤ (x − 3)² + 1
f(1) ≤ 5, f(2) ≤ 2, .......... f(9) ≤ 37
x = 1 has 2 choices
x = 2 has 1 choice
x = 3 has 1 choice
x = 4 has 1 choice
x = 5 has 2 choices
x = 6 has 3 choices
x = 7 has 4 choices
x = 8 has 5 choices
x = 9 has 6 choices
∴ Total functions = 2 x 1 x 1 x 1 x 2 x 3 x 4 x 5 x 6 = 1140

Q.30. Let f(x) = 2x2 − x − 1 and S = {n ∈ Z : |f(n)| ≤ 800}. Then, the value ofis equal to ___________.    (JEE Main 2022)

Solution:  
∵ |f(n)| ≤ 800
⇒ −800 ≤ 2n² − n − 1 ≤ 800
⇒ 2n² − n − 801 ≤ 0
 
∴ n = −19, −18, −17, .........., 19, 20.

= 2 . 2 . (12 + 22 + ... + 192) + 2 . 20² − 20 − 40
= 10620

Q.31. 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 ______________.    (JEE Main 2022)

Solution:  
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
∴ Maximum or minimum value of f(x) occur at critical points or boundary points
 
f(7/5) = 0 + 4 = 4
as both |5x − 7| and x² + 2x are increasing in nature after x = 7/5
∴ f(2) = 3 + 8 = 11
∴ f(7/5)_min = 4 and f(2)_max = 11
Sum is 4 + 11 = 15

Q.32. Let f(x) be a quadratic polynomial with leading coefficient 1 such that f(0) = p, p ≠ 0, and f(1) = 13. If the equations f(x) = 0 and f ∘ f ∘ f ∘ f(x) = 0 have a common real root, then f(−3) is equal to ________________.    (JEE Main 2022)

Solution:  
Let f(x) = (x − α)(x − β)
It is given that f(0) = p ⇒ αβ = p
and f(1) = 1/3 ⇒ (1 − α)(1 − β) = 1/3
Now, let us assume that, α is the common root of f(x) = 0 and f ∘ f ∘ f ∘ f(x) = 0
f ∘ f ∘ f ∘ f(x) = 0
⇒ f ∘ f ∘ f(0) = 0
⇒ f ∘ f(p) = 0
So, f(p) is either α or β.
(p − α)(p − β) = α
(αβ − α)(αβ − β) = α ⇒ (β − 1)(α − 1)β = 1 (∵ α ≠ 0)
So, β = 3
(1 − α)(1 − 3) = 1/3
α = 7/6

Q.33. Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If f(g(x)) = 8x² − 2x and g(f(x)) = 4x² + 6x + 1, then the value of f(2) + g(2) is _________.    (JEE Main 2022)

Ans. 18

Q.34. Let c, k ∈ R. If f(x) = (c + 1)x² + (1 − c²)x + 2k and f(x + y) = f(x) + f(y) − xy, for all x, y ∈ R, then the value of |2(f(1) + f(2) + f(3) + ...... + f(20))| is equal to ____________.    (JEE Main 2022)

Solution:  
f(x) is polynomial
Put y = 1/x in given functional equation we get



⇒ 2(c + 1) = 2K − 1 ..... (1)
and put x = y = 0 we get
f(0) = 2 + f(0) − 0 ⇒ f(0) = 0 ⇒ k = 0
∴ k = 0 and 2c = −3 ⇒ c = −3/2

Q.35. Let [t] denote the greatest integer ≤ t and {t} denote the fractional part of t. The integral value of α for which the left hand limit of the functionat x = 0 is equal tois _____________.    (JEE Main 2022)

Solution:  



⇒ 3α2 − 10α + 3 = 0
∴ α = 3 or 1/3
∵ α in integer, hence α = 3

Q.36. Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
 
Let g : S → S be a function such that  
Then g(10)g(1) + g(2) + g(3) + g(4) + g(5)) is equal to _____________.     (JEE Main 2022)

Solution:  

∴ f(1) = 2, f(2) = 4, ......, f(5) = 10
and f(6) = 1, f(7) = 3, f(8) = 5, ......, f(10) = 9

∴ f(g(10)) = 9 ⇒ g(10) = 10
f(g(1)) = 2 ⇒ g(1) = 1
f(g(2)) = 1 ⇒ g(2) = 6
f(g(3)) = 4 ⇒ g(3) = 2
f(g(4)) = 3 ⇒ g(4) = 7
f(g(5)) = 6 ⇒ g(5) = 3
∴ g(10)g(1) + g(2) + g(3) + g(4) + g(5)) = 190

Q.37. Let f : R → R be a function defined by Then is equal to ______________.    (JEE Main 2022)

Solution:  


i.e. f(x) + f(1 − x) = 2


= 49 x 2 + 1 = 99

Q.38. Let f : R → R satisfy f(x + y) = 2^xf(y) + 4^yf(x), ∀x, y ∈ R. If f(2) = 3, then 14. f′(4)/f′(2) is equal to ____________.    (JEE Main 2022)

Solution:  
∵ f(x + y) = 2^xf(y) + 4^yf(x) ....... (1)
Now, f(y + x)2^yf(x) + 4^xf(y) ...... (2)
∴ 2^xf(y) + 4^yf(x) = 2yf(x) + 4^xf(y)
(4y − 2y)f(x) = (4x − 2x)f(y)

∴ f(x) = k(4^x − 2^x)
∵ f(2) = 3 then k = 1/4

Q.39. Let f : R → R be a function defined byIf the function g(x) = f(f(f(x)) + f(f(x)), then the greatest integer less than or equal to g(1) is ____________.    (JEE Main 2022)

Solution:  
Given,

and g(x) = f(f(f(x))) + f(f(x))
∴ g(1) = f(f(f(1))) + f(f(1))







Now, f(f(f(1))) = f(1) = 3^1/50 
∴ g(1) = f(f(f(1))) + f(f(1)) = 3^1/50 + 1
Now, greatest integer less than or equal to g(1)
= [g(1)]
= [3^1/50 + 1]
= [3^1/50] + [1]
= [1.02] + 1
= 1 + 1 = 2

Q.40. The number of points where the function f(x)=
[t] denotes the greatest integer ≤ t, is discontinuous is _____________.    (JEE Main 2022)

Solution:  
∵ f(−1) = 2 and f(1) = 3
For x ∈ (−1, 1), (4x² − 1) ∈ [−1, 3)
hence f(x) will be discontinuous at x = 1 and also
whenever 4x² − 1 = 0, 1 or 2

So there are total 7 points of discontinuity.

Q.41. The number of one-one functions f : {a, b, c, d} → {0, 1, 2, ......, 10} such that 2f(a) − f(b) + 3f(c) + f(d) = 0 is ___________.                 (JEE Main 2022)

Solution:  
Given one-one function
f : {a, b, c, d} → {0, 1, 2, .... 10}
and 2f(a) − f(b) + 3f(c) + f(d) = 0
⇒ 3f(c) + 2f(a) + f(d) = f(b)
Case I: 
(1) Now let f(c) = 0 and f(a) = 1 then
3 × 0 + 2 × 1 + f(d) = f(b)
⇒ 2 + f(d) = f(b)
Now possible value of f(d) = 2, 3, 4, 5, 6, 7, and 8.
f(d) can't be 9 and 10 as if f(d) = 9 or 10 then f(b) = 2 + 9 = 11 or f(b) = 2 + 10 = 12, which is not possible as here any function's maximum value can be 10.
∴ Total possible functions when f(c) = 0 and f(a) = 1 are = 7
(2) When f(c) = 0 and f(a) = 2 then
3 × 0 + 2 × 2 + f(d) = f(b)
⇒ 4 + f(d) = f(b)
∴ possible value of f(d) = 1, 3, 4, 5, 6
∴ Total possible functions in this case = 5
(3) When f(c) = 0 and f(a) = 3 then
3 × 0 + 2 × 3 + f(d) = f(b)
⇒ 6 + f(d) = f(b)
∴ Possible value of f(d) = 1, 2, 4
∴ Total possible functions in this case = 3
(4) When f(c) = 0 and f(a) = 4 then
3 × 0 + 2 × 4 + f(d) = f(b)
⇒ 8 + f(d) = f(b)
∴ Possible value of f(d) = 1, 2
∴ Total possible functions in this case = 2
(5) When f(c) = 0 and f(a) = 5 then
3 × 0 + 2 × 5 + f(d) = f(b)
⇒ 10 + f(d) = f(b)
Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.
∴ No function is possible in this case.
∴ Total possible functions when f(c) = 0 and f(a) = 1, 2, 3 and 4 are = 7 + 5 + 3 + 2 = 17
Case II: 
(1) When f(c) = 1 and f(a) = 0 then
3 × 1 + 2 × 0 + f(d) = f(b)
⇒ 3 + f(d) = f(b)
∴ Possible value of f(d) = 2, 3, 4, 5, 6, 7
∴ Total possible functions in this case = 6
(2) When f(c) = 1 and f(a) = 2 then
3 × 1 + 2 × 2 + f(d) = f(b)
⇒ 7 + f(d) = f(b)
∴ Possible value of f(d) = 0, 3
∴ Total possible functions in this case = 2
(3) When f(c) = 1 and f(a) = 3 then
3 × 1 + 2 × 3 + f(d) = f(b)
⇒9 + f(d) = f(b)
∴ Possible value of f(d) = 0
∴ Total possible functions in this case = 1
∴ Total possible functions when f(c) = 1 and f(a) = 0, 2 and 3 are
= 6 + 2 + 1 = 9
Case III: 
(1) When f(c) = 2 and f(a) = 0 then
3 × 2 + 2 × 0 + f(d) = f(b)
⇒ 6 + f(d) = f(b)
∴ Possible values of f(d) = 1, 3, 4
∴ Total possible functions in this case = 3
(2) When f(c) = 2 and f(a) = 1 then,
3 × 2 + 2 × 1 + f(d) = f(b)
⇒ 8 + f(d) = f(b)
∴ Possible values of f(d) = 0
∴ Total possible function in this case = 1
∴ Total possible functions when f(c) = 2 and f(a) = 0, 1 are = 3 + 1 = 4
Case IV: 
(1) When f(c) = 3 and f(a) = 0 then
3 × 3 + 2 × 0 + f(d) = f(b)
⇒ 9 + f(d) = f(b)
∴ Possible values of f(d) = 1
∴ Total one-one functions from four cases
= 17 + 9 + 4 + 1 = 31

Q.42. The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is ____________.                  (JEE Main 2021)

Solution:  
A = 4-digit numbers divisible by 3
A = 1002, 1005, ....., 9999.
9999 = 1002 + (n − 1)3
⇒ (n − 1)3 = 8997 ⇒ n = 3000
B = 4-digit numbers divisible by 7
B = 1001, 1008, ......., 9996
⇒ 9996 = 1001 + (n − 1)7
⇒ n = 1286
A ∩ B = 1008, 1029, ....., 9996
9996 = 1008 + (n − 1)21
⇒ n = 429
So, no divisible by either 3 or 7
= 3000 + 1286 − 429 = 3857
total 4-digits numbers = 9000
required numbers = 9000 − 3857 = 5143 

Q.43. If A = {x ∈ R : |x − 2| > 1},

C = {x ∈ R : |x − 4| ≥ 2} and Z is the set of all integers, then the number of subsets of the
set (A ∩ B ∩ C)^c ∩ Z is ________________.                 (JEE Main 2021)

Solution:  
A = (−∞, 1) ∪ (3, ∞)
B = (−∞, −2) ∪ (2, ∞)
C = (−∞, 2] ∪ [6, ∞)
So, A ∩ B ∩ C = (−∞, −2) ∪ [6, ∞)
z ∩ (A ∩ B ∩ C)' = {−2, −1, 0, −1, 2, 3, 4, 5}
Hence, no. of its subsets = 2⁸ = 256. 

Q.44. Let S = {1, 2, 3, 4, 5, 6, 7}. Then the number of possible functions f : S → S
such that f(m . n) = f(m) . f(n) for every m, n ∈ S and m . n ∈ S is equal to _____________.            (JEE Main 2021)

Solution:  
F(mn) = f(m) . f(n)
Put m = 1 f(n) = f(1) . f(n) ⇒ f(1) = 1
Put m = n = 2

Put m = 2, n = 3

f(5), f(7) can take any value
Total = (1 × 1 × 7 × 1 × 7 × 1 × 7) + (1 × 1 × 3 × 1 × 7 × 1 × 7)
= 490 

Q.45. Let A = {n ∈ N | n² ≤ n + 10,000}, B = {3k + 1 | k∈ N} an dC = {2k | k ∈ N}, then the sum of all the elements of the set A ∩(B − C) is equal to _____________.             (JEE Main 2021)

Solution:  
B − C ≡ {7, 13, 19, ......, 97, .......}
Now, n² − n ≤ 100 × 100
⇒ n(n − 1) ≤ 100 × 100
⇒ A = {1, 2, ......., 100}.
So, A∩(B − C) = {7, 13, 19, ......., 97}
Hence, sum =

Q.46. Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f : A → A such that f(1) + f(2) = 3 − f(3) is equal to              (JEE Main 2021)

Solution:  
f(1) + f(2) = 3 − f(3)
⇒ f(1) + f(2) = 3 + f(3) = 3
The only possibility is: 0 + 1 + 2 = 3
⇒ Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.
So number of bijective functions.

Q.47. If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x³) + x g(x³) is divisible by x² + x + 1, then P(1) is equal to ___________.            (JEE Main 2021) 

Solution:  
Given, p(x) = f(x³) + xg(x³)
We know, x² + x + 1 = (x − ω) (x − ω²)
Given, p(x) is divisible by x² + x + 1. So, roots of p(x) is ω and ω².
As root satisfy the equation,
So, put x = ω
p(ω) = f(ω³) + ωg(ω³) = 0
= f(1) + ωg(1) = 0 [ω³ = 1]


Comparing both sides, we get


So, f(1) = 0
Now, p(1) = f(1) + 1 . g(1) = 0 + 0 = 0

Q.48. If a + α = 1, b + β = 2 andthen the value of the expressionis __________.            (JEE Main 2021)  

Solution:  

Replace x with 1/x

(i) + (ii)

Q.49. Let  A = {n∈N: n is a 3-digit number}
B = {9k + 2: k ∈ N}
and C = {9k + l: k ∈ N} for some l(0 < l < 9)
If the sum of all the elements of the set A ∩ (B ∪ C) is 274 × 400, then l is equal to ________.             (JEE Main 2021)

Solution:  
3 digit number of the form 9K + 2 are {101, 109, .............992}
⇒ Sum equal to 100/2 (1093) = s₁ = 54650
274 × 400 = s₁ + s₂
274 × 400 = 100/2 [101 + 992] + s₂
274 × 400 = 50 × 1093 + s₂
s₂ = 109600 − 54650
s₂ = 54950
s₂ = 54950 = 100/2[(99 + l) + (990 + l)]
1099 = 2l + 1089
l = 5 

Q.50. The range of the function,
           (JEE Main 2021)
(a) (0, √5)
(b) [-2, 2]
(c)
(d) [0, 2]

Correct Answer is Option (d)
Solution:  





So, Range of f(x) is [0, 2] 

Q.51. Let f : N → N be a function such that f(m + n) = f(m) + f(n) for every m, n ∈ N. If f(6) = 18, then f(2) . f(3) is equal to:           (JEE Main 2021)
(a) 6
(b) 54
(c) 18
(d) 36

Correct Answer is Option (b)
Solution:  
f(m + n) = f(m) + f(n)
Put m = 1, n = 1
f(2) = 2f(1)
Put m = 2, n = 1
f(3) = f(2) + f(1) = 3f(1)
Put m = 3, n = 3
f(6) = 2f(3) ⇒ f(3) = 9
⇒ f(1) = 3, f(2) = 6
f(2) . f(3) = 6 × 9 = 54 

Q.52. The domain of the function
           (JEE Main 2021)

(a)
(b)
(c)
(d)

Correct Answer is Option (c)
Solution:  



(1) & (2) 

Q.53. Which of the following is not correct for relation R on the set of real numbers?           (JEE Main 2021)
(a) (x, y) ∈ R ⇔ 0 < |x| − |y| ≤ 1 is neither transitive nor symmetric.
(b) (x, y) ∈ R ⇔ 0 < |x − y| ≤ 1 is symmetric and transitive.
(c) (x, y) ∈ R ⇔ |x| − |y| ≤ 1 is reflexive but not symmetric.
(d) (x, y) ∈ R ⇔ |x − y| ≤ 1 is reflexive and symmetric. 

Correct Answer is Option (b)
Solution:  
Note that (a, b) and (b, c) satisfy 0 < |x − y| ≤ 1 but (a, c) does not satisfy it so 0 ≤ |x − y| ≤ 1 is symmetric but not transitive.
For example,
x = 0.2, y = 0.9, z = 1.5
0 ≤ |x – y| = 0.7 ≤ 1
0 ≤ |y – z| = 0.6 ≤ 1
But |x – z| = 1.3 > 1
So, (b) is correct.

Q.54. The domain of the functionis:            (JEE Main 2021)
(a)
(b)
(c)
(d)

Correct Answer is Option (d)
Solution: 

Q.55. Let [t] denote the greatest integer less than or equal to t. Let
f(x) = x − [x], g(x) = 1 − x + [x], and h(x) = min{f(x), g(x)}, x ∈ [−2, 2]. Then h is:           (JEE Main 2021)
(a) A continuous in [−2, 2] but not differentiable at more than four points in (−2, 2)
(b) not continuous at exactly three points in [−2, 2]
(c) continuous in [−2, 2] but not differentiable at exactly three points in (−2, 2)
(d) not continuous at exactly four points in [−2, 2] 

Correct Answer is Option (a)
Solution: 

min{x − [x], 1 − x + [x]}

h(x) = min{x − [x], 1 − [x − [x])}

⇒ always continuous in [−2, 2] but not differentiable at 7 points. 

Q.56. Out of all patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If K% of them are suffering from both ailments, then K can not belong to the set:           (JEE Main 2021)
(a) {80, 83, 86, 89}
(b) {84, 86, 88, 90}
(c) {79, 81, 83, 85}
(d) {84, 87, 90, 93}

Correct Answer is Option (c)
Solution: 
n(A ∪ B) ≥ n(A) + n(B) − n(A ∩ B)
100 ≥ 89 + 98 − n(A ∪ B)
n(A ∩ B) ≥ 87
87 ≤ n(A ∩ B) ≤ 89

Q.57. Let N be the set of natural numbers and a relation R on N be defined by R = {(x, y) ∈ N × N : x³ − 3x²y − xy² + 3y³ = 0}. Then the relation R is:            (JEE Main 2021)
(a) symmetric but neither reflexive nor transitive
(b) reflexive but neither symmetric nor transitive
(c) reflexive and symmetric, but not transitive
(d) an equivalence relation

Correct Answer is Option (b)
Solution: 
x³ − 3x²y − xy² + 3y³ = 0
⇒ x(x² − y²) − 3y(x² − y²) = 0
⇒ (x − 3y)(x − y)(x + y) = 0
Now, x = y ∀(x, y) ∈N × N so reflexive but not symmetric & transitive.
See, (3, 1) satisfies but (1, 3) does not. Also (3, 1) & (1, −1) satisfies but (3, −1) does not. 

Q.58. Let f : R → R be defined as f(x + y) + f(x − y) = 2f(x)f(y), f(1/2) = −1. Then, the value ofis equal to:             (JEE Main 2021)
(a) cosec²(21) cos(20) cos(2)
(b) sec²(1) sec(21) cos(20)
(c) cosec²(1) cosec(21) sin(20)
(d) sec²(21) sin(20) sin(2)

Correct Answer is Option (a)
Solution: 
f(x) = cosλx


⇒ λ = 2π
Thus f(x) = cos2πx
Now k is natural number
Thus f(k) = 1

Q.59. Consider function f : A → B and g : B → C (A, B, C ⊆ R) such that (gof)⁻¹ exists, then:                           (JEE Main 2021)
(a) f and g both are one-one
(b) f and g both are onto
(c) f is one-one and g is onto
(d) f is onto and g is one-one

Correct Answer is Option (c)
Solution: 
∴ (gof)⁻¹ exist ⇒ gof is bijective
⇒ 'f' must be one-one and 'g' must be ONTO. 

Q.60. If [x] be the greatest integer less than or equal to x, thenis equal to:                 (JEE Main 2021)
(a) 0
(b) 4
(c) -2
(d) 2

Correct Answer is Option (b)
Solution: 

= [4] + [-4.5] + [5] + [-5.5] + [6] +..... + [-49.5] + [50]
= 4 - 5 + 5 - 6 + 6 ......-50 + 50
= 4 

Q.61. Let g : N → N be defined as
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1, for all n ≥ 0.
Then which of the following statements is true?         (JEE Main 2021)
(a) There exists an onto function f : N → N such that fog = f
(b) There exists a one-one function f : N → N such that fog = f
(c) gogog = g
(d) There exists a function : f : N → N such that gof = f

Correct Answer is Option (a)
Solution: 
g : N → N
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1



If f : N → N, if is a one-one function such that f(g(x)) = f(x) ⇒ g(x) = x, which is not the case
If f : N → N f is an onto function
such that f(g(x)) = f(x),
one possibility is

Here f(x) is onto, also f(g(x)) = f(x) ∀ x∈N 

Q.62. If the domain of the function is the interval (α, β], then α + β is equal to:          (JEE Main 2021)
(a) 3/2
(b) 2
(c) 1/2
(d) 1

Correct Answer is Option (a)
Solution: 
O ≤ x² − x + 1 ≤ 1
⇒ x² − x ≤ 0
⇒ x ∈ [0, 1]


⇒ 0 < 2x − 1 ≤ 2
1 < 2x ≤ 3
1/2 < x ≤ 3/2
Taking intersection
x ∈ (1/2, 1]
⇒ α = 1/2, β = 1
⇒ α + β = 3/2

Q.63. The number of solutions of sin⁷x + cos⁷x = 1, x∈ [0, 4π] is equal to          (JEE Main 2021)
(a) 11
(b) 7
(c) 5
(d) 9

Correct Answer is Option (c)
Solution: 
sin⁷x ≤ sin²x ≤ 1 ...... (1)
and cos⁷x ≤ cos²x ≤ 1 ..... (2)
also sin²x + cos²x = 1
⇒ equality must hold for (1) & (2)
⇒ sin⁷x = sin²x & cos⁷x = cos²x
⇒ sin x = 0 & cos x = 1
or
cos x = 0 & sin x = 1
⇒ x = 0, 2π, 4π, π/2, 5π/2
⇒ 5 solutions

Q.64. Let [x] denote the greatest integer less than or equal to x. Then, the values of x∈R satisfying the equationlie in the interval:          (JEE Main 2021)
(a) [0, 1/e)
(b) [log_e2, log_e3)
(c) [1, e)
(d) [0, log_e2)

Correct Answer is Option (d)
Solution: 


Let [e^x] = t
⇒ t² + t − 2 = 0
⇒ t = −2, 1
[e^x] = −2 (Not possible)
or [e^x] = 1 ∴ 1 ≤ e^x < 2
⇒ ln⁡(1) ≤ x < ln⁡(2)
⇒ 0 ≤ x < ln⁡(2)
⇒ x ∈ [0, In 2)

Q.65. Let f : R − {α/6} → R be defined byThen the value of α for which (fof)(x) = x, for all x ∈ R − {α/6}, is:          (JEE Main 2021)
(a) No such α exists
(b) 5
(c) 8
(d) 6

Correct Answer is Option (b)
Solution: 

5x + 3 = 6xy − αy
x(6y − 5) = αy + 3


fo f(x) = x
f(x) = f⁻¹(x)
From eqⁿ (i) & (ii)
Clearly (α = 5)

Q.66. Let [ x ] denote the greatest integer ≤ x, where x ∈ R. If the domain of the real valued function f(x)=is (− ∞, a) ]∪ [b, c) ∪ [4, ∞), a < b < c, then the value of a + b + c is:          (JEE Main 2021) 
(a) 8
(b) 1
(c) -2
(d) -3

Correct Answer is Option (c)
Solution: 
For domain, 

Case I:
When |[x]|−2 ≥ 0
and |[x]|−3 > 0
∴ x ∈ (− ∞, −3) ∪ [4, ∞) ...... (1)
Case II:
When |[x]|−2 ≤ 0
and |[x]|−3 < 0
∴ x ∈ [−2, 3) ..... (2)
So, from (1) and (2) we get
Domain of function
= (− ∞, −3) ∪ [−2, 3) ∪ [4, ∞)
∴ (a + b + c) = −3 + (−2) + 3 = −2 (a < b < c)
⇒ Option (c) is correct.

Q.67. Let f : R − {3} → R − {1} be defined by
Let g : R → R be given as g(x) = 2x − 3. Then, the sum of all the values of x for which f⁻¹(x) + g⁻¹(x) = 13/2 is equal to:           (JEE Main 2021) 
(a) 3
(b) 5
(c) 2
(d) 7

Correct Answer is Option (b)
Solution: 
Finding inverse of f(x)


Similarly for g⁻¹(x)


⇒ 6x − 4 + x² + 2x − 3 = 13x − 13
⇒ x² − 5x + 6 = 0
⇒ (x − 2)(x − 3) = 0
⇒ x = 2 or 3

Q.68. If the functions are defined asthen what is the common domain of the following functions:
           (JEE Main 2021)
(a) 0 ≤ x ≤ 1
(b) 0 ≤ x < 1
(c) 0<x<1
(d) 0 < x ≤ 1 

Correct Answer is Option (c)
Solution: 

⇒ x ≥ 0 & 1 − x ≥ 0 ⇒ x ∈ [0, 1]

⇒ x ≥ 0 & 1 − x ≥ 0 ⇒ x ∈[0, 1]

⇒ x ≥ 0 & 1 − x > 0 ⇒ x ∈ [0, 1)

⇒ 1 − x ≥ 0 & x > 0 ⇒ x ∈ (0, 1]

⇒ 1 − x ≥ 0 & x ≥ 0 ⇒ x ∈ [0, 1]
⇒ x ∈ (0, 1)

Q.69. The real valued functionwhere [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to:           (JEE Main 2021)
(a) all real except integers
(b) all non-integers except the interval [ −1, 1 ]
(c) all integers except 0, −1, 1
(d) all real except the interval [ −1, 1 ]

Correct Answer is Option (b)
Solution: 
Domain of cos⁡ec⁻¹x:
x ∈ (−∞, −1] ∪ [1, ∞)
and, x − [x] > 0
⇒ {x} > 0
⇒ x ≠ I
∴ Required domain = (−∞, −1] ∪ [1, ∞)− I

Q.70. Consider the function f : R → R defined by
Then f is:           (JEE Main 2021)
(a) not monotonic on (−∞, 0) and (0, ∞) 
(b) monotonic on (0, ∞) only 
(c) monotonic on (−∞, 0) only
(d) monotonic on (−∞, 0) ∪ (0, ∞)

Correct Answer is Option (a)
Solution: 



∴ f'(x) is an oscillating function which is non-monotonic on (−∞, 0) and (0, ∞).

Q.71. In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement?           (JEE Main 2021)
(a) Q and R
(b) None of these
(c) P and R
(d) P and Q

Correct Answer is Option (b)
Solution: 
As none play all three games the intersection of all three circles must be zero.
Hence none of P, Q, R justify the given statement.

Q.72. The inverse of y = 5^log⁡x is:           (JEE Main 2021) 
(a) x = 5^log⁡y 

(b)
(c)
(d) x = y^log⁡y5

Correct Answer is Option (b)
Solution: 
y = 5^log⁡x

⇒ log⁡y = log⁡x . log5

Q.73. Let A = {2, 3, 4, 5, ....., 30} and '≃' be an equivalence relation on A × A, defined by (a, b) ≃ (c, d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to:            (JEE Main 2021) 
(a) 5
(b) 6
(c) 8
(d) 7

Correct Answer is Option (d)
Solution: 

ad = bc
(a, b) R (4, 3) ⇒ 3a = 4b

b must be multiple of 3
b = {3, 6, 9 ..... 30}
(a, b) = {(4, 3), (8, 16), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21)}
⇒ 7 ordered pair

Q.74. Let f be a real valued function, defined on R − {−1, 1} and given by

Then in which of the following intervals, function f(x) is increasing?           (JEE Main 2021) 
(a) (−∞, −1) ∪ ([1/2, ∞) − {1})
(b) (−∞, ∞) − {−1, 1)
(c) (−∞, 1/2] − {−1}
(d) (−1, 1/2]

Correct Answer is Option (a)
Solution: 

Q.75. The range of a ∈ R for which the function f(x) = (4a − 3)(x + log_e 5) + 2(a − 7) cot(x/2) sin²(x/2), x ≠ 2nπ, n∈N has critical points, is:           (JEE Main 2021)
(a) [1, ∞)
(b) (−3, 1)
(c)
(d) (−∞, −1]  

Correct Answer is Option (c)
Solution: 

f(x) = (4a − 3)(x + ln ⁡5) + (a − 7)sin⁡ x
f′(x) = (4a − 3) + (a − 7)cos⁡ x = 0

Q.76. Let [ x ] denote greatest integer less than or equal to x. If for n ∈ N,

           (JEE Main 2021)
(a) 2^n-1
(b) n
(c) 2
(d) 1

Correct Answer is Option (d)
Solution: 

(1 − x + x³) = a₀ + a₁x + a₂^x2 + ...... + a_3n x ³ⁿ
Put x = 1
1 = a₀ + a₁ + a₂ + a₃ + a₄ + ........ + a_3n ...... (1)
Put x = −1
1 = a₀ − a₁ + a₂ − a₃ + a₄ + ........(−1)³ⁿa_3n ..... (2)
Add (1) + (2)
⇒ a₀ + a₂ + a₄ + a₆ + ...... = 1
Sub (1) − (2)
⇒ a₁ + a₃ + a₅ + a₇ + ...... = 0

= (a0 + a2 + a4 + ......) + 4(a1 + a3 + .....)
= 1 + 4 × 0
= 1

Q.77. The number of elements in the set {x ∈ R : (|x| − 3) |x + 4| = 6} is equal to:         (JEE Main 2021)
(a) 4
(b) 2
(c) 3
(d) 1

Correct Answer is Option (b)
Solution:  
Case 1:
x ≤ −4
(−x − 3)(−x − 4) = 6
⇒ (x + 3)(x + 4) = 6
⇒ x² + 7x + 6 = 0
⇒ x = −1 or −6
but x ≤ −4
x = −6
Case 2:
x ∈ (−4, 0)
(−x − 3)(x + 4) = 6
⇒ −x² − 7x − 12 − 6 = 0
⇒ x² + 7x + 18 = 0
D < 0 No solution
Case 3:
x ≥ 0
(x − 3)(x + 4) = 6
⇒ x² + x − 12 − 6 = 0
⇒ x² + x − 18 = 0

Q.78. Let A = {1, 2, 3, ...., 10} and f : A → A be defined as

Then the number of possible functions g : A → A such that gof = f is:         (JEE Main 2021)
(a) 5⁵
(b) 10⁵
(c) 5!
(d) ¹⁰C₅

Correct Answer is Option (b)
Solution: 

f(1) = 2
f(2) = 2
f(3) = 4
f(4) = 4
f(5) = 6
f(6) = 6
f(7) = 8
f(8) = 8
f(9) = 10
f(10) = 10
∴ f(1) = f(2) = 2
f(3) = f(4) = 4
f(5) = f(6) = 6
f(7) = f(8) = 8
f(9) = f(10) = 10
Given, g(f(x)) = f(x)
when x = 1, g(f(1)) = f(1) ⇒ g(2) = 2
when, x = 2, g(f(2)) = f(2) ⇒ g(2) = 2
∴ x = 1, 2, g(2) = 2
Similarly, at x = 3, 4, g(4) = 4
at x = 5, 6, g(6) = 6
at x = 7, 8, g(8) = 8
at x = 9, 10, g(10) = 10
 
Here, you can see for even terms mapping is fixed. But far odd terms 1, 3, 5, 7, 9 we can map to any one of the 10 elements.
∴ For 1, number of functions = 10
For 3, number of functions = 10
For 9, number of functions = 10
∴ Total number of functions = 10 × 10 × 10 × 10 × 10 = 10⁵ 

Q.79. Let R = {(P, Q) | P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1, −1) is the set:         (JEE Main 2021)
(a) S = {(x, y)|x² + y² = √2}
(b) S = {(x, y)|x² + y² = 2}
(c) S = {(x, y)|x² + y² = 1}
(d) S = {(x, y)|x² + y² = 4}

Correct Answer is Option (b)
Solution: 
Given R = {(P, Q) | P and Q are at the same distance from the origin}.
Then equivalence class of (1, −1) will contain al such points which lies on circumference of the circle of centre at origin and passing through point (1, −1).
i.e., radius of circle = 
∴ Required equivalence class of (S)
{(x, y)|x² + y² = 2}

Q.80. 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 form the set A to the set A × B. Then:        (JEE Main 2021)
(a) 2y = 273x
(b) y = 91x
(c) 2y = 91x
(d) y = 273x

Correct Answer is Option (c)
Solution: 
Number of elements in A = 3
Number of elements in B = 5
Number of elements in A × B = 15 

Number of one-one function
x = 5 × 4 × 3
x = 60
 
Number of one-one function
y = 15 × 14 × 13
y = 15 × 4 × 14/4 × 13
y = 60 × 7/2 × 13
2y = (13)(7x)
2y = 91x 

Q.81. A function f(x) is given bythen the sum of the seriesis equal to:        (JEE Main 2021)
(a) 39/2
(b) 19/2
(c) 49/2
(d) 29/2

Correct Answer is Option (a)
Solution: 



Adding equation (i) and (ii)
f(x) + f(2 − x) = 1

Q.82. Let f, g : N → N such that f(n + 1) = f(n) + f(1) ∀ n ∈ N and g be any arbitrary function. Which of the following statements is NOT true?       (JEE Main 2021)
(a) If g is onto, then fog is one-one
(b) f is one-one
(c) If f is onto, then f(n) = n ∀ n ∈ N
(d) If fog is one-one, then g is one-one

Correct Answer is Option (a)
Solution: 
f(n + 1) = f(n) + 1
f(2) = 2f(1)
f(3) = 3f(1)
f(4) = 4f(1)
f(n) = nf(1)
f(x) is one-one 

Q.83. Let f : R → R be defined as f (x) = 2x – 1 and g : R - {1} → R be defined as g(x) =Then the composition function f(g(x)) is:       (JEE Main 2021)
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto

Correct Answer is Option (a)
Solution: 
Given, f(x) = 2x − 1; f : R → R

f[g(x)] = 2g(x) − 1



Now, draw the graph of 
∵ Any horizontal line does not cut the graph at more than one points, so it is one-one and here, co-domain and range are not equal, so it is into.
Hence, the required function is one-one into. 

Q.84. If g(x) = x² + x - 1 and (gof) (x) = 4x² - 10x + 5, then f (5/4) is equal to    (2020)
(a) 3/2
(b) - 1/2
(c) 1/2
(d) - 3/2

Correct Answer is Option (b)
Solution: 
We have
g(x) = x² + x - 1 and (gof) (x) = 4x² - 10x + 5
Now, g(f(x)) = 4x² - 10x + 5 = 4x² - 8x + 4 - 2x +1
⇒ g(f(x)) = (4 - 8x + 4x²) + (2 - 2x) - 1
⇒ g(f(x)) = (2-2x)² + (2-2x) - 1 ⇒ f(x) = 2 - 2x
Hence,

Q.85. The inverse function of f(x)
 

Correct Answer is Option (d)
Solution: 
Let,

Q.86. Let f: (1, 3) → R be a function defined by

where [x] denotes the greatest integer ≤ x . Then, the range of f is    (2020)

Correct Answer is Option (b)
Solution: 
We have,

Since f(x) is a decreasing function, then

Q.87. Let f and g be differentiable functions on R such that fog is the identity function. If for some a, b ∈ R g'(a) = 5 and g(a) = b then f'(b) is equal to    (2020)
(a) 1/5
(b) 1
(c) 5
(d) 2/5

Correct Answer is Option (a)
Solution: 
f and g are differentiable functions on R and fog is the identity function. So,
f(g(x)) = x
⇒ f'(g(x)) . g'(x) = 1     (1)
Substituting x = a in Eq. (1), we get
f'(g(a)). g'(a) = 1
⇒ f'(b) × 5 = 1 ⇒ f'(b) = 1/5

Q.88. For x ∈ R - {0, 1}, let f₁(x) = 1/x, f₂ (x) = 1 - x and f₃(x) = 1/1-x be three given functions. If a function, J(x) satisfies (f₂oJof₁) (x) = f₃(x) then J(x) is equal to:    (2019)
(a) f₃(x)
(b) 1/x f₃(x)
(c) f₂(x)
(d) f₁(x)

Correct Answer is Option (a)
Solution: 
The given relation is

Q.89. If the fractional part of the number 2⁴⁰³/15 is k/15, then k is equal to:    (2019)
(a) 6
(b) 8
(c) 4
(d) 14

Correct Answer is Option (b)
Solution: 
2⁴⁰³ = 2⁴⁰⁰ · 2³
= 2^{4 × 100} · 2³
= (2⁴)¹⁰⁰· 8
= 8(2⁴)¹⁰⁰ = 8(16)¹⁰⁰
= 8(1 + 15)¹⁰⁰
= 8 + 15μ
When 2⁴⁰³ is divided by 15, then remainder is 8.
Hence, fractional part of the number is 8/15
Therefore, value of k is 8

Q.90. Let A = {x ∈ R: x is not a positive integer}. Define a function f: A → R as f(x) = 2x/x - 1, then f is:    (2019)
(a) Not injective
(b) Neither injective nor surjective
(c) Surjective but not injective
(d) Injective but not surjective

Correct Answer is Option (d)
Solution: 
As A = {x ∈ R: x is not a positive integer}
A function f: A → R given by f(x) = 2x/x-1  
f(x₁) = f(x₂) ⇔ x₁ = x₂
So, f is one-one.
As f(x) ≠ 2 for any x ∈ A ⇒ f is not onto.
Hence f is injective but not subjective.

Q.91. Let N be the set of natural numbers and two functions f and g be defined as f, g : N → N such that    (2019) 

and g(n) = n - (- 1)ⁿ. Then fog is:
(a) Onto but not one-one.
(b) One-one but not onto.
(c) Both one-one and onto.
(d) Neither one-one nor onto.

Correct Answer is Option (a)
Solution: 




⇒ fog is onto but not one - one

Q.92. Let f: R → R be defined by
 
Then the range of f is:    (2019)
(a) [- 1/2, 1/2]
(b) R - [-1,1]
(c) R - [- 1/2, 1/2]
(d) (-1,1) - {0}

Correct Answer is Option (a)
Solution: 

Q.93. Let a function f: (0, ∞) → (0, ∞) be defined by

(a) Not injective but it is surjective
(b) Injective only
(c) Neither injective nor surjective
(d) Both injective as well as surjective

Correct Answer is Option (a)
Solution: 
f: (0, ∞) → (0, ∞)

∵ f(1) = 0 and 1 ∈ domain but 0 ∉ co-domain
Hence, f(x) is not a function.

Q.94. If f(x) = log_e(1 - x)/(1 + x) , |x| < 1, then f(2x/1 + x²) is equal to :     (2019)
(a) 2f(x)
(b) 2f(x²)
(c) (f(x))² 
(d) -2f(x)

Correct Answer is Option (a)
Solution: 

Q.95. Let f(x) = a^x (a > 0) be written as f(x) = f₁(x) + f₂(x), where f₁(x) is an even function and f₂(x) is an odd function. Then f₁(x + y) + f₁(x - y) equals:     (2019)
(a) 2 f₁(x) f₁(y)
(b) 2 f₁(x + y) f₁(x - y)
(c) 2 f₁(x) f₂(y)
(d) 2 f₁(x + y) f₂(x - y)

Correct Answer is Option (a)
Solution: 
Given function can be written as

Q.96. If the function f: R - {1, -1} → A defined by f(x) = x²/1 - x², is surjective, then A is equal to:     (2019)
(a) R - {-1}
(b) [0, ∞]
(c) R- [-1, 0]
(d) R - (-1, 0]

Correct Answer is Option (c)
Solution: 

∴ f(x) increases in x ∈ (0, ∞)
Also f(0) = 0 and

∴ Set A = R - [-1, 0)
And the graph of function f(x) is

Q.97. 

where the function f satisfies f(x + y) = f(x) f(y) for all natural numbers x, y and f(1) = 2. Then the natural number 'a' is:     (2019)
(a) 2
(b) 16
(c) 4
(d) 3

Correct Answer is Option (d)
Solution: 
∵ f(x + y) = f(x) x f(y)
⇒ Let f(x) = t^x
∵ f(1) = 2
∵ t = 2
⇒ f(x) = 2^x


⇒ a = 3

Q.98. The domain of the definition of the function

(a) (-1, 0) ∪ (1, 2) ∪ (3, ∞)
(b) (-2, -1) ∪ (-1, 0) ∪ (2, ∞)
(c) (-1, 0) ∪ (1, 2) ∪ (2, ∞)
(d) (1, 2) ∪ (2, ∞)

Correct Answer is Option (c)
Solution: 
To determine domain, denominator ≠ 0 and x³ - x > 0
i.e., 4 - x² ≠ 0 x ≠ ±2       ...(1)
and x (x - 1) (x + 1) > 0

x∈(-1, 0) ∪ (1, ∞)    ...(2)
Hence domain is intersection of (1) & (2).
i.e.,x ∈ (-1, 0) ∪ (1, 2) ∪ (2, ∞)

Q.99. Let f(x) = x², x ∈ R. For any A ⊆ R, define g(A) = {x∈R: f(x) ∈ A}. If S = [0, 4], then which one of the following statements is not true?     (2019)
(a) g(f(S)) ≠ S
(b) f(g(S)) = S
(c) g(f(S)) = g(S)
(d) f (g (S)) ≠ f (S)

Correct Answer is Option (c)
Solution: 
f(x) = x² ; x ∈ R
g(A) = {x ∈ R: f(x) ∈ A} S = [ 0, 4]
g(S) = {x ∈ R: f(x) ∈ S}
= {x ∈ R: 0 ≤ x² ≤ 4} = { x ∈ R : -2 ≤ x ≤ 2}
∴ g(S) ≠ S
∴ f(g (S)) ≠ f(S)
g(f(S)) = {x ∈ R: f(x) ∈ f(S)}
= { x ∈ R : x² ∈ S²} = { x ∈ R : 0 ≤ x² ≤ 16}
= {x ∈ R : -4 ≤ x ≤ 4}
∴ g(f(S)) ≠ g(S)
∴ g(f(S)) = g (S) is incorrect.

Q.100. Let f(x) = log_e (sinx), (0 < x < π) and g(x) = sin^-1 (e^-x), (x > 0). If α is a positive real number such that a = (fog)' (α) and b = (fog) (α), then:     (2019)
(a) aα² + bα + a = 0
(b) aα² - bα - a = 1
(c) aα² - bα - a = 0 
(d) aα² + bα - a = - 2a²

Correct Answer is Option (b)
Solution: 
f(x) = ln (sin x), g (x) = sin^-1 (e^-x)
⇒ f(g(x)) = ln (sin (sin^-1 e^-x)) = - x
⇒ f(g(α)) = - α
But given that (fog) (α) = b
∴ - α = b and f' (g(α)) = a, i.e., a = - 1
∴ aα² - bα - a = - α² + a² - (- 1)
⇒ aα² - bα - a = 1

Q.101. For x ∈ (0, 3/2), let f(x) = √x, g(x) = tan x and h(x) = (1-x²)/1+x²). If φ(x) = ((hof)og), (x), then φ (π/3) is equal to:     (2019)
(a) tan π/12
(b) tan 11π/12
(c) tan 7π/12
(d) tan 5π/12 

Correct Answer is Option (b)
Solution: 
∵ φ(x) = ((hof)og)(x)
∵ φ(π/3) = h(f(g(π/3))) = h(f(√3)) = h(3^1/4)
= (1 - √3)/(1+√3) = - 1/2 (1 + 3 - 2√3) = √3 - 2 = -(-√3 + 2)
= -tan 15º = tan(180º - 15º) = tan(π - π/12) = tan 11π/12

Q.102. Let f(x) = 2¹⁰dx + 1 and g(x) = 3¹⁰x - 1. If (fog)(x) = x, then x is equal to:    (2017)
(a) (2¹⁰ - 1)/(2¹⁰ - 3^-10)
(b) (1 - 2^-10)/(3¹⁰ - 2^-10)
(c) (3¹⁰ - 1)/(3¹⁰ - 2^-10)
(d) (1 - 3^-10)/(2¹⁰ - 3^-10)

Correct Answer is Option (b)
Solution: 
f(g(x)) = x
f(3¹⁰x - 1) = 2¹⁰(3¹⁰.x - 1) = x
= 1/(3¹⁰ - 2^-110)
2¹⁰(3¹⁰x - 1) + 1 = x
x(6¹⁰ - 1) = 2¹⁰ - 1
x = (2¹⁰ - 1)/(6¹⁰ - 1)    =  (1 - 2^-10)/(3¹⁰ - 2^-10)

Q.103. 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:    (2017)
(a) One-one but not onto
(b) One-one and onto
(c) Neither one-one nor onto
(d) Onto but not one-one

Correct Answer is Option (c)
Solution: 

f(10) = 10 – 5(2) = 0 which is not in codomain
So, the function is many one + into

Q.104. If f(x) + 2f(1/x) = 3x, x ≠ 0, and S = {x ∈ R: f(x) = f(-x)}; then S:    (2016)
(a) Is an empty set
(b) Contains exactly one element
(c) Contains exactly two elements.
(d) Contains more than two elements.

Correct Answer is Option (c)
Solution: