Most Important PYQs Functions

Answer Keys and Solutions JEE Main Crash Course

ANSWER KEYS
1. (3) 2. (4) 3. (3) 4. (1) 5. (4) 6. (15) 7. (2) 8. (3125)
9. (190) 10. (4) 11. (4) 12. (3) 13. (1) 14. (4) 15. (360) 16. (432)
17. (1) 18. (4) 19. (2) 20. (3) 21. (3) 22. (26) 23. (3) 24. (3)
25. (1) 26. (10) 27. (4) 28. (4) 29. (2039) 30. (4)

1. (3)
The domain of the function
2
−1 3x +x−1 x−1
−1
f (x)= sin ( )+ cos ( )
2
x+1
(x−1)

For, cos −1
(
x−1

x+1
)

x−1
−1 ≤ ≤ 1
x+1

2
⇒ −1 ≤ 1 − ≤ 1
x+1

−2
⇒ −2 ≤ ≤ 0
x+1

1
⇒ 0 ≤ ≤ 1
x+1

⇒ x + 1 ∈ [1, ∞)

⇒ x ∈ [0, ∞) . . . .(i)

For, sin −1 3x +x−1

( )
2
(x−1)

2
3x +x−1
−1 ≤ ≤ 1
2
( x−1 )

2 2 2
⇒ −(x − 1) ≤ 3x + x − 1 ≤ (x − 1) , x ≠ 1

Now, −(x − 1)
2
2
≤ 3x + x − 1, x ≠ 1

2
⇒ 4x − x ≥ 0 , x ≠ 1

⇒ x(4x − 1)≥ 0, x ≠ 1

1
⇒ x ∈ (−∞, 0] ∪[ , ∞) −{1} . . . .(ii)
4

And 3x 2
+ x − 1 ≤ (x − 1) , x ≠ 1
2

2
⇒ 2x + 3x − 2 ≤ 0, x ≠ 1

⇒(x + 2)(2x − 1)≤ 0, x ≠ 1

1
⇒ x ∈[−2, ] . . . .(iii)
2

Domain of the function sin from the equations (ii) & (iii) is
−1 3x +x−1
( )
2
(x−1)

1 1
⇒ x ∈[−2, 0]∪[ , ] . . . .(iv)
4 2

Now the domain of the given function will be the intersection of the equation (i) & (iv)
Hence, domain is x ∈[ 1

4
,
1

2
]∪{0}

2. (4)
2 −1
f (x) = ln(4x + 11x + 6) + sin (4x + 3)

10x + 6
−1
+ cos ( )
3

2
(i) 4x + 11x + 6 > 0

2
4x + 8x + 3x + 6 > 0

(4x + 3)(x + 2) > 0

3
x ∈ (−∞, −2) ∪ (− , ∞)
4

(ii) 4x + 3 ∈ [−1, 1]
x ∈ [−1, −1/2]

10x + 6
(iii) ∈ [−1, 1]
3

9 3
x ∈ [− ,− ]
10 10

3 1 3 1
x ∈ (− ,− ] α = − ,β = −
4 2 4 2

5
α + β = −
4

36|α + β| = 45

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Most Important PYQs Functions
Answer Keys and Solutions JEE Main Crash Course

3. (3) We have,
| [ x ] | −2
f (x)= √
| [ x ] | −3

For domain,
| [ x ] | −2
( )≥ 0
| [ x ] | −3

Case I : When |[x]| − 2 ≥ 0 and |[x]|−3 > 0, then

∴ x ∈(−∞, −3)∪[4, ∞) . . .(1)

Case II : When |[x]|−2 ≤ 0 and |[x]|−3 < 0, then

∴ x ∈[−2, 3) . . .(2)

So, from (1) and (2), we get

Domain of function is
(−∞, −3)∪[−2, 3)∪[4, ∞)

∴ (a + b + c)= −3 +(−2)+3 = −2

(∵ a < b < c)

4. (1)
Given,
f : R → R be a function defined by f (x)= log √m
{√2(sin x − cos x)+m − 2} , for some m,
Also given the range of f is [0, 2],
Now we know that,
−√2 ≤ sin x − cos x ≤ √2

⇒ −2 ≤ √2(sin x − cos x)≤ 2

(Assuming √2(sin x − cos x)= k)

⇒ −2 ≤ k ≤ 2 …(1)

Now taking function f (x)= log √m

(k + m − 2)

Given, 0 ≤ f (x)≤ 2
⇒ 0 ≤ log (k + m − 2)≤ 2
√m

⇒ 1 ≤ k + m − 2 ≤ m

⇒ −m + 3 ≤ k ≤ 2 …(2)

Now from equations (1) & (2), we get

−m + 3 = −2

⇒ m = 5

5. (4)
Given:
[x]
f (x)=
2
1+x

So,
2
f (x)= ; x ∈ [2, 3)
2
1+x

3
f (x)= ; x ∈ [3, 4)
2
1+x

4
f (x)= ; x ∈ [4, 5)
2
1+x

5
f (x)= ; x ∈ [5, 6)
2
1+x

Hence,
5 2
f (x)∈( , ]
37 5

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6. (15)
Given,
2
f (x)=|5x − 7|+[x + 2x]

2
=|5x − 7|+[(x + 1) − 1]

2
=|5x − 7|+[(x + 1) ]−1

(as [x − 1]=[x]-1 where [.] is greatest integer function)

Now critical points of

7
f (x)= , √5 − 1, √6 − 1, √7 − 1, √8 − 1, 2
5

∴ Maximum or minimum value of f (x) occur at critical points or boundary points.

So, f(
5

4
)=
3

4
+ 4 =
19

4
and f ( 7

5
)= 0 + 4 = 4

As both |5x − 7| and x 2

+ 2x are increasing in nature after x = 7

So, f (2)=|10 − 7|+[4 + 4]= 3 + 8 = 11

∴ f(
7

5
) = 4 and f (2) max
= 11
min

So sum of minimum and maximum value is 4 + 11 = 15

7. (2)
1
25

Given f (x)= [2(1 − x 50

25
)(2 + x )]
2

25 25 50
f (x)= [(2 − x )(2 + x )]
1
50 50
= (4 − x )
1

50 50
1

i.e. f (f (x))= (4 − ((4 − x 50

)
50
) ) = x

Now g(x)= f (f (f (x)))+f (f (x))

= f (x)+x
1

So g(1)= f (1)+1 = 3 50
+ 1
1

Hence [g(1)]=[3 50 + 1]= 2

8. (3125)
Given:
3x+2
1
f (x)=
2x+3

Now,
2 1 1
f (x)= f of (x)
3x+2
3( ) +2
2x+3
2
⇒ f (x)=
3x+2
2( ) +3
2x+3

9x+6+4x+6
2
⇒ f (x)=
6x+4+6x+9

13x+12
2
⇒ f (x) =
12x+13

13x+12
3( ) +2
12x+13
3
⇒ f (x)=
13x+12
2( ) +3
12x+13

3 63x+62
⇒ f (x)=
62x+63
63x+62
3( ) +2
62x+63
4
⇒ f (x)=
63x+62
2( ) +3
62x+63

4 313x+312
⇒ f (x)=
312x+313

5 1563x+1562 ax+b
∴ f (x)= =
1562x+1563 bx+a

∴ a + b = 1563 + 1562 = 3125

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Most Important PYQs Functions
Answer Keys and Solutions JEE Main Crash Course

9. (190)
2n, if n = 1, 2, 3, 4, 5
Given, f (n)={
2n − 11 if n = 6, 7, 8, 9, 10

So, f (1)= 2, f (2)= 4, .....f (5)= 10

And f (6)= 1, f (7)= 3, f (8)= 5 ......f (10)= 9
n + 1 ; n ∈ odd
f (g(n))={
n − 1 ; n ∈ even

So, 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

So, g(10)⋅[g(1)+g(2)+g(3)+g(4)+g(5)]
= 10 ⋅[1 + 6 + 2 + 7 + 3]= 190

10. (4)
2
x
f og(x)= f (g(x)) = f( )
2
x +1

2 2 2
x x −x −1 −1
= − 1 = =
2 2 2
x +1 x +1 x +1

We know that, 0 ≤ x 2
< ∞, ∀x ∈ R

1 −1
2
⇒ 1 ≤ x + 1 < ∞, ∀x ∈ R ⇒ 1 ≥ > 0, ∀x ∈ R ⇒ −1 ≤ < 0, ∀x ∈ R
2 2
x +1 x +1

So, range of f og(x) is [−1, 0)⊂ R .

Hence, the function f og(x) is into function and f og(−x)= f (g(−x))= −1

2
=
−1

2
= f (g(x))
( −x ) +1 x +1

∴ f og(x) is an even function. So, it is a many one function.

Hence, f og(x) is neither one-one nor onto function.
11. (4) f : (0, ∞) → (0, ∞)

f (x) = ∣
∣1 −
1

x
∣
∣ is not a function
∵ f (1) = 0 and 1 ∈ domain but 0 ∉ co-domain
Hence, f (x) is not a function.
12. (3)
Given,
2
x +2x+1
f (x)=
2
x +1
2 2
( x +1 ) (2x+2)−(x +2x+1)2x
′
f (x) =
2
2
( x +1 )

3 2 3 2
2x +2x + 2x +2− 2x −4x − 2x
′
⇒ f (x) =
2
2
( x +1 )

2
2(1−x ) 2(x+1)(x−1)
′
⇒ f (x) = = −
2 2
2 2
( x +1 ) ( x +1 )

Clearly f (x) is one-one in (−∞, −1) and also in (1, ∞) but f (x) is not one-one in (−∞, ∞)
13. (1)
For one value of n we will get only one corresponding value of f (n), so f (n) is one-one
Now, for n = 2, 4, 6 ⋯ ; f (n)= 4, 8, 12. . .

for n = 3, 7, 11 ⋯ ; f (n)= 2, 6, 10. . .

for n = 1, 5, 9 ⋯ ; f (n)= 1, 3, 5, 7. . .

So range of f (n) is N
Hence f (n) is onto
14. (4)
x + 1 if x is odd
f (x)={
x if x is even

∵ g : A → A such that g(f (x))= f (x)

⇒ If x is even then g(x)= x …(1)

If x is odd then g(x + 1)= x + 1 …(2)

from (1) and (2) we can say that g(x)= x if x is even

⇒ If x is odd then g(x) can take any value in set A
so number of g(x)= 10 5
× 1

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Most Important PYQs Functions
Answer Keys and Solutions JEE Main Crash Course

15. (360) f (1) + f (2) + 1 = f (4) ≤ 6

f (1) + f (2) ≤ 5

Case (i) f (1) = 1 ⇒ f (2) = 1, 2, 3, 4 ⇒ 4 mappings

Case (ii) f (1) = 2 ⇒ f (2) = 1, 2, 3 ⇒ 3 mappings
Case (iii) f (1) = 3 ⇒ f (2) = 1, 2 ⇒ 2 mappings
Case (iv) f (1) = 4 ⇒ f (2) = 1 ⇒ 1 mapping
f (5) & f (6) both have 6 mappings each
Number of functions = (4 + 3 + 2 + 1) × 6 × 6 = 360
16. (432)
Given,
f (m ⋅ n)= f (m)⋅f (n)

Taking m = 1
⇒ f (1 ⋅ n)= f (1)⋅f (n)

⇒ f (1)= 1

Now taking m = 3 & n = 3 we get,

f (9) = f (3) × f (3) ,
So, (f (3), f (9)) have two possibility (1, 1) & (3, 9)
Now taking m = 2 & n = 1 we get,
f (2 ⋅ 1)= f (2)⋅f (1)

Now here f (1)= 1, so f (2) can take all 6 numbers,

Similarly, for f (5) & f (8) there will be 6 ways each,
So, total possible function will be, = 1 × 6 × 2 × 6 × 6 × 1 = 432
17. (1)
Given, f (1, 3, 5, 7, ⋯ , 99)→(2, 4, 6, 8, ⋯ , 100)
The number of elements in domain and codomain is 99.
Now, let us assume f (1)= 100

Now as per diagram we have only 1 way for arranging f (3)> f (5)> f (7)⋯ > f (99),
Similarly if we choose f (1)= 98 then again we have only 1 way for arranging f (3)> f (5)> f (7)⋯ > f (99) ,
So we can see the arrangement f (3)> f (5)> f (7)⋯ > f (99) depends upon f (1),
So number of ways for choosing f (1) is 50
C
1

18. (4)
Total cases will be 5
C4 × 4!

Now favourable cases for 2f (b)= f (c)+f (d)−f (a) will be,
Case (I) if f (b)= 1 then f (c), f (d), f (a) can take the value 3, 4, 5 & 4, 3, 5 respectively
Case (II) if f (b)= 2 then f (c), f (d), f (a) can take value 3, 5, 4 & 5, 3, 4 respectively
Case (III) if f (b)= 3 then f (c), f (d), f (a) can take value 2, 5, 1 & 5, 2, 1 respectively
So total favourable case will be 6
So probability=
favourable cases 6 1
= =
total outcomes 5×4! 20

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Most Important PYQs Functions
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19. (2)
Given,
1
f (x)+f ( )= 1 + x
1−x

x = 2 ⇒ f (2)+f (−1)= 3 …(1)

1
x = −1 ⇒ f (−1)+f ( )= 0 …(2)
2

1 1 3
x = ⇒ f( )+f (2)= …(3)
2 2 2

On subtracting equation (3) and equation (2) we get,

3
f (2) − f (−1) = . . . . . . . . (4)
4

On adding equation (1) and equation (4) we get,

9
2f (2)=
2
9
⇒ f (2)=
4

20. (3)
x
5 5
f (x)= x and f (2 − x)= x
5 +5 5 +5

f (x)+f (2 − x)= 1

1 2 39
⇒ f( )+f ( )+ … + f ( )
20 20 20

1 39 19 21 20
=(f ( )+f ( ))+ … +(f ( )+f ( ))+f ( )
20 20 20 20 20

1 39
= 19 + =
2 2

21. (3)
Let
3 2
f (x)= x − ax + bx − c

′ 2
f (x)= 3x − 2ax + b

′′
f (x)= 6x − 2a

'''
f (x)= 6

Also,
′
f (1)= a

⇒ 3 − 2a + b = a

⇒ 3a = b + 3

Also,
′′
f (2)= 12 − 2a = b

⇒ 12 − 2a = 3a − 3

⇒ 5a = 15 ⇒ a = 3

⇒ b = 6

And,
'''
f (3)= c ⇒ c = 6

Hence,
3 2
f (x)= x − 3x + 6x − 6

Now,
f (3)+2f (0)= 27 − 27 + 18 − 6 − 12 = 0

f (2)+f (1)= 8 − 12 + 12 − 6 + 1 − 3 + 6 − 6 = 0

So,
f (3)+2f (0)= f (2)+f (1)

⇒2f (0)−f (1)+f (3)= f (2)

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22. (26)
Given that: kf (k) + 2 = 0 for k = 2, 3, 4, 5 which means (k − 2), (k − 3), (k − 4), (k − 5) are the factors of this expression.
Let kf (k) + 2 = a(k − 2)(k − 3)(k − 4)(k − 5) . . . . (i)
P ut k = 0

2 = a(−2)(−3)(−4)(−5)

1
a =
60

Put a = 1

60
in (i), we get
1
kf (k) + 2 = (k − 2)(k − 3)(k − 4)(k − 5)
60

Now, put k = 10
1
10f (10) + 2 = × 8 × 7 × 6 × 5
60

10 f (10) = 26

So, 52 − 10 f (10) = 26
23. (3)
Given,
f (x + y)= f (x)⋅f (y) and f (1)= 3
Now taking x = 1 & y = 1 we get,
2
f (1 + 1)= f (1)⋅f (1)⇒ f (2)= 3

Similarly f (2 + 1)= f (2)⋅f (1)⇒ f (3)= 3 2

× 3 = 3
3

And so on f (n)= 3 n

So, ∑ n

r=1
f (r)= 3279

⇒ f (1)+f (2)+f (3). . . . . . . f (n)= 3279

2 3 n
⇒ 3 + 3 + 3 . . . . . . . +3 = 3279
n
3 −1
⇒ 3 × = 3279
3−1
n
3 −1
⇒ = 1093
2

n
⇒ 3 − 1 = 2186

n
⇒ 3 = 2187

n 7
⇒ 3 = 3

⇒ n = 7

24. (3)
Given,
Functional equation f (x + y)= f (x)+f (y)−1
Now taking x = 0 & y = 0 in above equation we get,
f (0 + 0)= f (0)+f (0)−1 ⇒ f (0)= 1

Now we know that,

f ( x+h ) −f ( x )
′
f (x)= lim
h
h→0

f ( x ) +f ( h ) −1−f ( x )
′
⇒ f (x)= lim
h
h→0

f ( h ) −1
′
⇒ f (x)= lim
h
h→0

f ( h ) −1
Now let lim = k
h
h→0

So, the equation becomes f ′

(x)= k

Now putting x = 0 in above equation we get,

′ ′
f (0)= k ⇒ k = 2 {as given f (0)= 2}

So, f ′
(x)= 2

Now integrate both side we get,

f (x)= 2x + c

Now again taking x = 0 we get,

f (0)= 2 × 0 + c ⇒ c = 1 {as f (0)= 1}

So, f (x)= 2x + 1
And f (−2)= 2 ×(−2)+1 ⇒ f (−2)= −3
Hence, |f (−2)|= 3

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25. (1)
Given functional equation is
f (m + n)= f (m)+f (n); m, n ∈ N . . . .(1)

Put m = n = 3
f (3 + 3) = f (3) + f (3)

⇒ f (3) = 9[∵ f (6)= 18]

Put m = 2, n = 1 in equation (1)

∴ f (3) = f (2 + 1) = f (2) + f (1)

= f (1 + 1) + f (1)

= f (1) + f (1) + f (1)

9 = 3f (1)

⇒ f (1) = 3

∴ f (2) = f (1 + 1) = f (1) + f (1) = 6

Now,
f (2) ⋅ f (3) = (6)(9) = 54

26. (10)
f (x)
∵ lim = 1 ⇒ f '(0)= 1
x
x→0

2 2
f (x + y)= f (x)+f (y)+xy + x y

Differentiate w.r.t. x keeping y constant

2
f '(x + y)= f '(x)+0 + y + 2xy

put y = −x
2 2
f '(0)= f '(x)+x − 2x

2
1 = f '(x)−x

2
f '(x)= 1 + x

f '(3)= 10 .
27. (4)
Given,
1
f (n) + f (n + 1) = 1, ∀ n ∈ {1, 2, 3}
n

⇒ nf (n) + f (n + 1) = n

At n = 1,
f (1) + f (2) = 1 ....(1)
At n = 2,
2f (2) + f (3) = 2 ....(2)
At n = 3,
3f (3) + f (4) = 3 ....(3)
Put the value of f (2) from equation (1) in equation (2),
2(1 − f (1)) + f (3) = 2

⇒ f (3) = 2f (1) ....(4)

Put the value of f (3) from equation (4) in equation (3),
3(2f (1)) + f (4) = 3

⇒ f (4) = 3 − 6f (1)

∵ f : {1, 2, 3, 4} → {a ∈ Z : |a| ≤ 8}

⇒ −8 ≤ f (4) ≤ 8

⇒ −8 ≤ 3 − 6f (1) ≤ 8

⇒ −11 ≤ −6f (1) ≤ 5

−5 11
⇒ ≤ f (1) ≤
6 6

⇒ f (1) = 0, 1

Case I: f (1) = 0
⇒ f (2) = 1, f (3) = 0, f (4) = 3

Case II: f (1) = 1

⇒ f (2) = 0, f (3) = 2, f (4) = −3

Therefore, two such functions are possible.

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28. (4) Greatest integer function

⎧ 0 ; 0 ≤ x < 1
⎪

f (x)=[x]=⎨ −1 ; − 1 ≤ x < 0
⎩
⎪
−2 ; − 2 ≤ x < −1

Given series
1 1 1 1 2 1 3 1 99
S =[− ]+[− − ]+[− − ]+[− − ]+ … +[− − ]
3 3 100 3 100 3 100 3 100

−1 0 ≤ r ≤ 66
General term T
1 r
r
=[− − ]={
3 100
−2 r > 66
66 99
⇒ S = ∑ (−1)+ ∑ (−2)=(−67)+(−2)×33
r=0 r=67

= − 133

29. (2039)
Given:
f (x)= ax − 3
1

b −1 x−7 3

g(x)= x + c, x ∈ R (f og) (x)= ( )

2

Now, let
h(x)=(f ∘ g)(x)
1

−1 x−7 3
⇒ h (x)= ( )
2

Let
1

x−7 3

y = ( )
2

x−7
3
⇒ y =( )
2

3
⇒ x = 2y + 7

So, inverse of h −1
(x) is 2x 3
+ 7 i.e.,
3
h(x)= f og(x)= 2x + 7

Also,
b
f og(x)= a(x + c)−3

b 3
⇒ ax + ac − 3 = 2x + 7

On comparing, we get
a = 2, b = 3, c = 5

So,
f (x)= 2x − 3

3
g(x)= x + 5

Now,
f og (ac)= f og(10)= f (1005)= 2007

(gof )(b)= gof (3)= g(3)= 32

Therefore,
f og (ac)+(gof )(b)= 2007 + 32 = 2039

30. (4)
Given:
2
x − 4x +[x]+3 = x[x]

2
⇒ x − 4x + 3 = x[x]−[x]

⇒(x − 1)(x − 3)=[x](x − 1)

⇒ x = 1 or x − 3 =[x]
If x − 3 =[x]
⇒ x–[x]= 3

⇒{x}= 3

which is not Possible, since {x}∈ [0, 1).

Hence, x = 1 is the only one solution in (−∞, ∞).

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