JEE  >  JEE Main Previous year questions (2016-22): Functions

# JEE Main Previous year questions (2016-22): Functions Notes | Study Mathematics (Maths) Class 12 - JEE

## Document Description: JEE Main Previous year questions (2016-22): Functions for JEE 2022 is part of Mathematics (Maths) Class 12 preparation. The notes and questions for JEE Main Previous year questions (2016-22): Functions have been prepared according to the JEE exam syllabus. Information about JEE Main Previous year questions (2016-22): Functions covers topics like and JEE Main Previous year questions (2016-22): Functions Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main Previous year questions (2016-22): Functions.

Introduction of JEE Main Previous year questions (2016-22): Functions in English is available as part of our Mathematics (Maths) Class 12 for JEE & JEE Main Previous year questions (2016-22): Functions in Hindi for Mathematics (Maths) Class 12 course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: JEE Main Previous year questions (2016-22): Functions Notes | Study Mathematics (Maths) Class 12 - JEE
 1 Crore+ students have signed up on EduRev. Have you?

Q.1. Let |M| denote the determinant of a square matrix M. Let g: [0, π/2] → R be the function defined by

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)

Ans. a, c

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

Ans. c

x2 + 2x + 7
5x ≥ − 5
x ≥ −1

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

Q.3. The function f(x) = xex(1−x), x ∈ R, is:      (JEE Main 2022)
(a) increasing in
(b) decreasing in
(c) increasing in
(d) decreasing in

Ans. a
f(x) = xex(1−x), x ∈ R
f′(x) = xex(1−x) . (1 − 2x) + ex(1−x)
= ex(1−x)[x − 2x2 + 1]
= −ex(1−x)[2x2 − x − 1]
= −ex(1−x)(2x + 1)(x − 1)
∴ f(x) is increasing inand decreasing in

Q.4. 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

Ans. d
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.5. 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,…, an be in arithmetic progression with mean zero, then the value ofis equal to:      (JEE Main 2022)
(a) 0
(b) 3
(c) 9
(d) 27

Ans. a

∴ 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.6. 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]

Ans. b

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

Q.7. The domain of the functionwhere [t] is the greatest integer function, is:      (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans. c
−1 ≤ 2x− 3 < 2
or 2 ≤ 2x2 < 5
or 1 ≤ x2 < 5/2

log1/2(x2 − 5x + 5) > 0
0 < x2 − 5x + 5 < 1
x2 − 5x + 5 > 0 & x2 − 5x + 4 < 0

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

Q.8. 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

Ans. d
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, ∵ fmin = 1, gmin = 3
So, there does not exist any x ∈ N − {1} such that (f + g)(x) = 1, 2, 3
∴ f + g is not onto

Q.9. 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)

Ans. a
fa(x) = tan−12x − 3ax + 7
∵ fa(x) is non-decreasing in

Q.10. 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

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

Ans. b

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.12. 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)

Ans. b

Q.13. 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

Ans. b
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.14. Let  and S2 = {x ∈ R : 32x − 3x+1 − 3x+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]

Ans. b
Given,

x+ 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,
32x − 3x+1 − 3x+2 + 27 ≤ 0
⇒ (3x)2 − 3 . 3x − 32 . 3x + 27 ≤ 0
Let 3x = t
⇒ t2 − 3 . t − 32 . t + 27 ≤ 0
⇒ t(t − 3) − 9(t − 3) ≤ 0
⇒ (t − 3)(t − 9) ≤ 0

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

Q.15. The domain of the function     (JEE Main 2022)
(a)
(b) (−∞, −1] ∪ [1, ∞) ∪ {0}
(c)
(d)

Ans. d

From (3) and (4), we get

Q.16. 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

Ans. d
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.17. 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

Ans. d
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.18. 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

Ans. b
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.19. 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)

Ans. b
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))

f7(x) = f(f6(x))

∴ f6(6) =

So, f6(6) + f7(7)

Q.20. Let f : R → R and g : R → R be two functions defined by f(x) = loge(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)

Ans. a
f(x) = loge(x2 + 1) − e−x + 1

g(x) = e−x − 2ex
g′(x)−−e−x − 2ex < 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.21. 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

Ans. a

f(x) + f′(x) + f″(x) = x5 + 64
Let f(x) = x5 + ax4 + bx3 + cx+ 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 = x+ 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.22. 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

Ans. a
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.23. 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

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

Ans. d

The solution to this inequality is

for x2 − 3x + 2 > 0 and ≠ 1

Combining the two solution sets (taking intersection)

Q.25. 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)

Ans. b

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.26. For the function f(x) = 4loge(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)

Ans. c
f(x) = 4loge⁡(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.27. For p, q ∈ R, consider the real valued function f(x) = (x − p)2 − q, x ∈ R and q > 0. Let a1, a2,  aand a4 be in an arithmetic progression with mean p and positive common difference. If |f(ai)| = 500 for all i = 1, 2, 3, 4, then the absolute difference between the roots of f(x) = 0 is ___________.    (JEE Main 2022)

Ans. 50
∵ a1, a2, a3, a4
∴ a2 = p − 3d, a2 = p − d, a3 = p + d and a4 = p + 3d
Where d > 0
∵ |f(ai)| = 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.28. 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)

Ans. 1140
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.29. Let f(x) = 2x2 − x − 1 and S = {n ∈ Z : |f(n)| ≤ 800}. Then, the value ofis equal to ___________.    (JEE Main 2022)

Ans. 10620
∵ |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.30. 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)

Ans. 15
f(x) = |5x − 7| + [x+ 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.31. 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)

Ans. 25
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.32. 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)) = 4x+ 6x + 1, then the value of f(2) + g(2) is _________.    (JEE Main 2022)

Ans. 18

Q.33. 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)

Ans. 3395
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.34. 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)

Ans. 3

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

Q.35. 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)

Ans. 180

∴ 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.36. Let f : R → R be a function defined by Then is equal to ______________.    (JEE Main 2022)

Ans. 99

i.e. f(x) + f(1 − x) = 2

= 49 x 2 + 1 = 99

Q.37. Let f : R → R satisfy f(x + y) = 2xf(y) + 4yf(x), ∀x, y ∈ R. If f(2) = 3, then 14. f′(4)/f′(2) is equal to ____________.    (JEE Main 2022)

Ans. 248
∵ f(x + y) = 2xf(y) + 4yf(x) ....... (1)
Now, f(y + x)2yf(x) + 4xf(y) ...... (2)
∴ 2xf(y) + 4yf(x) = 2yf(x) + 4xf(y)
(4y − 2y)f(x) = (4x − 2x)f(y)

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

Q.38. 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)

Ans. 2
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) = 31/50
∴ g(1) = f(f(f(1))) + f(f(1)) = 31/50 + 1
Now, greatest integer less than or equal to g(1)
= [g(1)]
= [31/50 + 1]
= [31/50] + [1]
= [1.02] + 1
= 1 + 1 = 2

Q.39. The number of points where the function f(x)=
[t] denotes the greatest integer ≤ t, is discontinuous is _____________.    (JEE Main 2022)

Ans. 7
∵ f(−1) = 2 and f(1) = 3
For x ∈ (−1, 1), (4x− 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.40. 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)

Ans. 31
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.41. The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is ____________.                  (JEE Main 2021)

Ans. 5143
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
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.42. 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   C)c  Z is ________________.                 (JEE Main 2021)

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

Q.43. 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)

Ans. 490
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.44. 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)

Ans. 832
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.45. 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)

Ans. 720
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.46. 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)

Ans. 0
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.47. If a + α = 1, b + β = 2 andthen the value of the expressionis __________.            (JEE Main 2021)

Ans. 2

Replace x with 1/x

(i) + (ii)

Q.48. Let  A = {nN: n is a 3-digit number}
B = {9k + 2: k  N}
and C = {9k + l 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)

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

Ans. d

So, Range of f(x) is [0, 2]

Q.50. 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

Ans. b
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.51. The domain of the function
(JEE Main 2021)

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

Ans. c

(1) & (2)

Q.52. 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.

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

Ans. d

Q.54. 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]

Ans. a
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.55. 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}

Ans. c
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.56. Let N be the set of natural numbers and a relation R on N be defined by R = {(x, y) ∈ N × N : x− 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

Ans. b
x− 3x2y − xy2 + 3y3 = 0
⇒ x(x2 − y2) − 3y(x2 − y2) = 0
⇒ (x − 3y)(x − y)(x + y) = 0
Now, x = y (x, y) × 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.57. Let f : R  R be defined as f(x + y) + f(x − y) = 2f(x)f(y), f(1/2) = −1Then, 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)

Ans. a
f(x) = cosλx

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

Q.58. 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

Ans. c
(gof)−1 exist  gof is bijective
'f' must be one-one and 'g' must be ONTO.

Q.59. 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

Ans. b

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

Q.60. 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

Ans. a
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)  xN

Q.61. If the domain of the function is the interval (αβ], then α + β is equal to:          (JEE Main 2021)
(a) 3/2
(b) 2
(c) 1/2
(d) 1

Ans. a
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.62. 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

Ans. c
sin7 sin2 1 ...... (1)
and cos7 cos2 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, 4ππ/2, 5π/2
5 solutions

Q.63. 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) [loge2, loge3)
(c) [1, e)
(d) [0, loge2)

Ans. d

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

Q.64. 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

Ans. b

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.65. 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

Ans. c
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.66. 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

Ans. b
Finding inverse of f(x)

Similarly for g−1(x)

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

Q.67. 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

Ans. c

⇒ 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.68. 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 ]

Ans. b
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, ∞)

Ans. a

∴ 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

Ans. b
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 = 5log⁡x is:           (JEE Main 2021)
(a) x = 5log⁡y

(b)
(c)
(d) x = ylog⁡y5

Ans. b
y = 5log⁡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

Ans. d
(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]

Ans. a

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

Ans. c

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) 2n-1
(b) n
(c) 2
(d) 1

Ans. d

(1 − x + x3) = a+ a1x + a2x2 + ...... + a3n x 3n
Put x = 1
1 = a0 + a1 + a2 + a3 + a4 + ........ + a3n ...... (1)
Put x = −1
1 = a0 − a+ a− a3 + a4 + ........(−1)3na3n ..... (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

Ans. b
Case 1:
4
( 3)( 4) = 6
(x + 3)(x + 4) = 6
x2 + 7x + 6 = 0
x = 1 or 6
but x  4
x = 6
Case 2:
(4, 0)
( 3)(x + 4) = 6
x2  7x  12  6 = 0
x2 + 7x + 18 = 0
D < 0 No solution
Case 3:
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!