JEE Main Previous Year Questions (2016- 2025): Functions | Mathematics (Maths) Class 12 PDF Download

Ans. d

Ans. a




∴ Intersection of (1), (2) and (3)

Ans. a

Ans. b

Ans. d

Ans. a



Taking intersection of (i), (ii) and (iii)

Ans. d

Similarly,

Ans. a

Ans. c

Ans. d

Ans. b

Ans. a


so function is one-one but not onto

Ans. c

Ans. c

Clearly x < 0 satisfies which are included in option (1) only.

Ans. a
Step 1: Total Functions with 1 ∈ f (A)
Any function f : A → B where A = {1, 2, 3, 4} and B = {1,4, 9, 16} is defined by choosing one of the four elements of B for each element of A. 
Thus, the total number of functions is 4⁴ = 256.
To count those functions where 1 appears at least once in the set f(A), we can use the complementary counting method: subtract the functions that never use 1. 
If 1 is excluded, each element of A has only 3 choices (namely, {4, 9, 16}), so the number of such functions is 3⁴ = 81.
Thus, the number of functions such that 1 ∈ f(A) is 256 - 81 = 175.
Step 2: Counting Many-One Functions
In this context, "many-one functions" are understood to be non-injective functions. Since an injective (one-to-one) function from A to B must be a permutation (because both sets have 4 elements), the number of one-to-one functions is 4! = 24.
It is important to note that every injective function f : A → B has f(A) = B (a full permutation) which automatically means 1 ∈ f(A).
Thus, the number of many-one (non-injective) functions f : A → B with 1 ∈ f(A) is found by subtracting the one-to-one functions from the total functions that include 1: 175 - 24 = 151.
This detailed explanation shows that the number of many-one functions f : A → B such that 1 ∈ f(A) is indeed 151.

Ans. 96

Ans. c

Ans. d

Ans. a

Neither one-one nor onto.

Ans. d

Ans. c
The function can be simplified to
For x = 3 and x = -5, f(x) equals 0. 
Therefore, f(x) is not one-one as it yields the same output for different input values.
The range of f(x) is [-2, 1.6], indicating that f(x) does not cover all possible real values. 
Consequently, f(x) is not onto.
Thus, the function is neither one-one nor onto.

Ans. a

Ans. c
f(1)+f(3)=14
Case I
f(1)=2,f(3)=12
f(1)=12,f(3)=2
Total one-one function
=2 × 5 × 4 × 3
=120
Case II
f(1)=4,f(3)=10
f(1)=10,f(3)=4
Total one-one function
= 2 × 5 × 4 × 3
=120
Total cases = 120 + 120 = 240

Ans. d
To find the domain of the function we need to consider the domain conditions for both the square root function and the logarithmic function.
The square root function requires that the argument of the square root be non-negative, so x² - 25 ≥ 0. 
This inequality is satisfied when x ≤ -5 or x ≥ 5.
The denominator of the rational part of f(x), (4 - x²), cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have 4 - x² ≠ 0. This inequality is violated when x = ±2.
Combining these conditions gives us the domain for the rational part of the function: x ∈ (-∞,-5] U (5,∞) and x ≠ 2,-2.
Moving on to the logarithmic function, log₁₀(x² +2x - 15), the argument must be positive: x² + 2x - 15 > 0. 
This is a quadratic inequality, which we can factor to find the solution: (x +5)(x - 3) > 0. 
From this, we see that the inequality is satisfied for x < -5 or x > 3.
The overall domain of f(x) is the intersection of the domains for each piece. 
Taking the intersection of the two sets gives us: x € (-∞,-5) U (5,∞),
Since the question states that the domain is of the form (-∞, α) U [β, ∞), we can infer that α =- 5 and β = 5.
We calculate α² + β³ as follows: α² + β³ = (-5)² +5³ =25 + 125 = 150.
So the correct answer, representing the sum of α² and β³, is: Option D 150.

Ans. b


From the graph of g (f(x)), we can say
g (f(x)) ⇒ Many one into

Ans. d
To find we first need to understand the composition of f with itself, i.e.,
We can then repeatedly apply g to get the given expression.
First, let's calculate

To evaluate this expression, we substitute for x in the function f (x):

Now, we simplify the expression:

So, g(x) = x for all x in the domain of g, which is R - {2/3}.
It's important to note that the domain restriction is preserved through the composition because f{x} has a vertical asymptote at x = 2/3 which doesn't intersect the graph.
So, g (x) is the identity function on its domain, which means that applying g any number of times will result in the same input for x in the given domain. 
Hence, we have: 

This corresponds to option D, which is 4 .

Ans. b

Ans. a

Ans. d

Range of f (g(x)) is [0, 1]

Ans. d

Ans. b
f : N - {1} → N
f(n) = The highest prime factor of n.
f(2) = 2
f(4) = 2 
⇒ many one
4 is not image of any element
⇒  into
Hence many one and into
Neither one-one nor onto.

Ans. 24

Ans. 1010

Ans. 96
To determine the range of the function  let's start by simplifying the expression. 
Let sin²θ = x, so cos²θ = 1 − x . 
The function then transforms into:

Simplify the numerator and denominator separately:

Thus, the function becomes:

Next, we need to find the range of this function. Let's analyze the function by testing specific values of x in the interval [0 , 1] (since sin²θ ranges from 0 to 1):

It appears that f (x) achieves values within [1, 3]. To confirm this, we need to solve the quadratic inequality:

By solving the inequalities, it can be confirmed that the function indeed ranges from 1 to 3 on the interval [0,1]. Hence, we have:

The common ratio of the infinite geometric progression is:

Given the first term a = 64, the sum S of the infinite geometric progression can be given as:

Substituting the values a = 64 and r = 1/3, we get:

Therefore, the sum of the infinite geometric progression is 96.

Ans. 18
​

Ans. 1024

Ans. 44

Ans. d

& r = pq + 1

Solution: 

where

Correct Answe r is Option (a, c)

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, ∞)

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

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

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

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 

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

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

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

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

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 

Correct Answer is Option (b)

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

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]

Correct Answer is Option (d)
Solution:  





From (3) and (4), we get

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.

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.

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

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)

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)

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

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

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

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)

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

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)

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

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

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

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

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

Ans. 18

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

Solution: 



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

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

Solution: 


i.e. f(x) + f(1 − x) = 2


= 49 x 2 + 1 = 99

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

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

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.

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

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 

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. 

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 

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 =

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.

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

Solution: 

Replace x with 1/x

(i) + (ii)

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 

Correct Answer is Option (d)
Solution: 





So, Range of f(x) is [0, 2] 

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 

Correct Answer is Option (c)
Solution: 



(1) & (2)

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.

Correct Answer is Option (d)
Solution: 

Correct Answer is Option (a)
Solution: 

min{x − [x], 1 − x + [x]}

h(x) = min{x − [x], 1 − [x − [x])}

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

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. 

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


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

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

Correct Answer is Option (b)
Solution: 

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

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 

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

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

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)

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)

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.

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

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)

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

Correct Answer is Option (a)
Solution: 



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

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.

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

⇒ log⁡y = log⁡x . log5

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

Correct Answer is Option (a)
Solution: 

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

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

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

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 

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}

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 

Correct Answer is Option (a)
Solution: 



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

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 

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. 

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,

Correct Answer is Option (d)
Solution: 
Let,

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

Since f(x) is a decreasing function, then

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

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

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

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.

Correct Answer is Option (a)
Solution: 




⇒ fog is onto but not one - one

Correct Answer is Option (a)
Solution: 

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.

Correct Answer is Option (a)
Solution: 

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

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

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

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, ∞)

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.

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

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

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)

Correct Answer is Option (c)
Solution: 

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

Correct Answer is Option (c)
Solution: