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Functions

Q.No: 1
Test Name : CAT Paper 2004

Let f(x) = ax2 – b|x|, where a and b are constants. Then at x = 0, f(x) is

A
maximized whenever a > 0, b > 0
B
maximized whenever a > 0, b < 0
C
minimized whenever a > 0, b > 0
D
minimized whenever a > 0, b< 0
Solution:
Q.No: 2
Test Name : CAT Paper 2004

Let u = (log2x)2 – 6log2 x + 12 where x is a real number. Then the equation xu = 256, has

A
no solution for x
B
exactly one solution for x
C
exactly tow distinct solutions for x
D
exactly three distinct solutions for x
Solution:
Q.No: 3
Test Name : CAT Paper 2004
Answer the questions on the basis of the information given below.

How many of the following products are necessarily zero for every x.

A
0
B
1
C
2
D
3
Solution:
Q.No: 4
Test Name : CAT Paper 2004
Answer the questions on the basis of the information given below.


A
Which of the following is necessarily true?
B
C
D
Solution:
Q.No: 5
Test Name : CAT Paper 2005

Let g(x) be a function such that g(x + 1) + g(x – 1) = g(x) for every real x. Then for what value of p is the relation g(x+p) = g(x) necessarily true for every real x?

A
5
B
3
C
2
D
6
Solution:
Q.No: 6
Test Name : CAT Paper 2005

A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs. 250 and Rs. 300 per day respectively. In addition, a male operator gets Rs. 15 per call he answers and female operator gets Rs. 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job?

A
15
B
14
C
12
D
10
Solution:
Q.No: 7
Test Name : CAT Paper 2005

Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

A
5
B
10
C
9
D
15
Solution:
Q.No: 8
Test Name : CAT Paper 2007

A function f(x) satisfies f(1) = 3600 and f(1) + f(2) + ... + f(n) = n2 f(n), for all positive integers n > 1. What is the value of f(9)?

A
80
B
240
C
200
D
100
E
120
Solution:
Q.No: 9
Test Name : CAT Paper 2008

Directions for Questions 5 and 6:


Let f(x) = ax2 + bx + c, where a, b and c are certain constants and a ≠ 0 It is known that

f(5) = – 3f(2). and that 3 is a root of f(x) = 0.




What is the other root of f(x) = 0?

A
-7
B
-4
C
2
D
6
E
cannot be determined
Solution:


Q.No: 10
Test Name : CAT Paper 2008



Let f(x) = ax2 + bx + c, where a, b and c are certain constants and a ≠ 0 It is known that

f(5) = – 3f(2). and that 3 is a root of f(x) = 0.




What is the value of a + b + c?

A
9
B
14
C
13
D
37
E
cannot be determined
Solution:
Q.No: 11
Test Name : CAT Paper 2008

Let f(x) be a function satisfying f(x).f(y) = f(x.y) for all real x, y. If f(2) = 4, then what is the value of f(1/2)?

A
0
B
1/4
C
1/2
D
1
E
cannot be determined
Solution:
Q.No: 12
Test Name : CAT Paper 1991
Q 61 to 100 : Choose the best answer choice from those provided

A function can sometimes reflect on itself, i.e. if y = f(x), then x = f(y). Both of them retain the same structure and form. Which of the following functions has this property?

A
B
C
D
None of the above.
E
NA
Solution:
Q.No: 13
Test Name : CAT Paper 1991
Q 61 to 100 : Choose the best answer choice from those provided

If y = f(x) and f(x) = (1−x) / (1 + x), which of the following is true?

A
f(2x) = f(x) − 1
B
x = f(2y) − 1
C
f(1/x) = f(x)
D
x = f(y)
E
NA
Solution:
Q.No: 14
Test Name : CAT Paper 1991
Q 61 to 100 : Choose the best answer choice from those provided

Three machines, A, B and C can be used to produce a product. Machine A will take 60 hours to produce a million units. Machine B is twice as fast as Machine A. Machine C will take the same amount of time to produce a million units as A and B running together. How much time will be required to produce a million units if all the three machines are used simultaneously?

A
12 hours
B
10 hours
C
8 hours
D
6 hour
E
NA
Solution:
Q.No: 15
Test Name : CAT Paper 1993
Q58 to 100 : Choose the appropriate answer choice.

Q74 and 75 : A function f(x) is said to be even if f(–x) = f(x), and odd if f(–x) = –f(x). Thus, for example, the function given by f(x) = x2 is even, while the function given by f(x) = x3 is odd. Using this definition, answer the following questions.

The function given by is

A
even
B
odd
C
neither
D
both
Solution:
Q.No: 16
Test Name : CAT Paper 1993
Q58 to 100 : Choose the appropriate answer choice.

Q74 and 75 : A function f(x) is said to be even if f(–x) = f(x), and odd if f(–x) = –f(x). Thus, for example, the function given by f(x) = x2 is even, while the function given by f(x) = x3 is odd. Using this definition, answer the following questions.

The sum of two odd functions

A
is always an even function
B
is always an odd function
C
is sometimes odd and sometimes even
D
may be neither odd nor even
Solution:
Q.No: 17
Test Name : CAT Paper 1993
Q86 to 88 : The following functions have been defined for three numbers A, B and C :
@ (A, B) = average of A and B
*(A, B) = product of A and B
/(A, B) = A divided by B
Answer these questions with the above data.

The sum of A and B is given by

A
*(@(A, B), 2)
B
/(@(A, B), 2)
C
@(*(A, B), 2)
D
@(/(A, B), 2)
Solution:
Q.No: 18
Test Name : CAT Paper 1993
Q86 to 88 : The following functions have been defined for three numbers A, B and C :
@ (A, B) = average of A and B
*(A, B) = product of A and B
/(A, B) = A divided by B
Answer these questions with the above data.

The sum of A, B, and C is given by

A
*(@(*(@(B, A), 2), C), 3)
B
/(@(*(@(B, A), 3), C), 2)
C
/(*(@(*(B, A), 2), C), 3)
D
None of these
Solution:
Q.No: 19
Test Name : CAT Paper 1994
Q51 – 90 : Choose the best alternative.

Q63 – 64 : are based on the following information:

A
2
B
6
C
8
D
–2
Solution:
Q.No: 20
Test Name : CAT Paper 1994
Q51 – 90 : Choose the best alternative.

Q63 – 64 : are based on the following information:

A
a < 0, b < 0
B
a > 0, b > 0
C
D
Solution:
Q.No: 21
Test Name : CAT Paper 1994
Q51 – 90 : Choose the best alternative.

Q87 – 90 : are based on the following information:

fog(x) is equal to

A
1
B
gof(x)
C
D
Solution:
Q.No: 22
Test Name : CAT Paper 1994
Q51 – 90 : Choose the best alternative.

Q87 – 90 : are based on the following information:

For what value of x; f (x) = g(x −3)?

A
–3
B
C
–4
D
None of these
Solution:
Q.No: 23
Test Name : CAT Paper 1994
Q51 – 90 : Choose the best alternative.

Q87 – 90 : are based on the following information:

What is the value of (gofofogogof) (x) (fogofog)(x)?

A
x
B
x2
C
D
Solution:
From Question 87, fog(x) = gof (x) = x, you will realise that if you were to form a chain of these functions for even number of times, you would still end up getting x.
E.g. fogofog(x) = fog(x) = x. Since both the brackets have the functions repeated for even number of times, each of their value will be x and their product will be x2.
Q.No: 24
Test Name : CAT Paper 1994
Q51 – 90 : Choose the best alternative.

Q87 – 90 : are based on the following information:

What is the value of fo(fog)o(gof)(x)?

A
x
B
x2
C
2x + 3
D
Solution:
From question 87, gof(x) = fog(x) = x.
fo(fog)o(gof)(x) = fo(fog)(x) = f(x) = 2x + 3.
Q.No: 25
Test Name : CAT Paper 1995

Direction for questions 88 to 91: Answer the questions based on the following information.

le(x, y) = Least of (x, y)
mo(x) = |x|
me(x, y) = Maximum of (x, y)

Find the value of me(a + mo(le(a, b)); mo(a + me(mo(a), mo(b))), at a = –2 and b = –3.

A
1
B
0
C
5
D
3
Solution:
If a = –2 and b = –3, then our expression will be me(–2 + mo(le(–2, –3)), mo(–2 + me(mo(–2), mo(–3)))
= me(–2 + mo(–3), mo(–2 + me(2, 3)))
= me(–2 + 3, mo(–2 + 3))
= me(1, mo(1)) = me(1, 1) = 1.
Q.No: 26
Test Name : CAT Paper 1995

Direction for questions 88 to 91: Answer the questions based on the following information.

Which of the following must always be correct for a, b > 0?

A
mo(le(a, b)) ≥ (me(mo(a), mo(b))
B
mo(le(a, b)) > (me(mo(a), mo(b))
C
mo(le(a, b)) < (le(mo(a), mo(b))
D
mo(le(a,b)) = le(mo(a), mo(b))
Solution:
Please note that the fastest way to solve these sums is the method of simulation, i.e., select any arbitrary values in the range given and verify whether the option holds good. E.g. a = 2, b = 3.
In this case, option (a) LHS = mo(le(2, 3)) = mo(2) = 2.
RHS = (me(mo(2), mo(3)) = (me(2, 3)) = 3.
Hence, LHS < RHS.
(b) LHS = mo(le(2, 3)) = mo(2) = 2. RHS = (me(mo(2), mo(3)) = (me(2, 3)) = 3.
Hence, LHS < RHS.
(c) LHS = mo(le(2, 3)) = mo(2) = 2. RHS = (le(mo(2), mo(3)) = le(2, 3) = 2.
Hence, LHS = RHS.
(d) LHS = mo(le(2, 3)) = mo(2) = 2. RHS = (le(mo(2), mo(3)) = le(2, 3) = 2.
Hence, LHS = RHS.
Q.No: 27
Test Name : CAT Paper 1995

Direction for questions 88 to 91: Answer the questions based on the following information.

For what values of ‘a’ is me(a2 – 3a, a – 3) < 0?

A
a > 3
B
0 < a < 3
C
a < 0
D
a = 3
Solution:
Q.No: 28
Test Name : CAT Paper 1995

Direction for questions 88 to 91: Answer the questions based on the following information.

For what values of ‘a’ is le(a2 – 3a, a – 3) < 0?

A
a > 3
B
0 < a < 3
C
a < 0
D
Both b and c
Solution:
Q.No: 29
Test Name : CAT Paper 1996

Direction for questions 115 and 116: Answer the questions based on the following information. A, S, M and D are functions of x and y, and they are defined as follows.



What is the value of M(M(A(M(x, y), S(y, x)), x), A(y, x)) for x = 2, y = 3?

A
60
B
140
C
25
D
70
Solution:
Q.No: 30
Test Name : CAT Paper 1996

Direction for questions 115 and 116: Answer the questions based on the following information. A, S, M and D are functions of x and y, and they are defined as follows.



What is the value of S[M(D(A(a, b), 2), D(A(a, b), 2)), M(D(S(a, b), 2), D(S(a, b), 2))]?

A
a2 + b2
B
ab
C
a2 - b2
D
Solution:
Q.No: 31
Test Name : CAT Paper 2003 (L)
DIRECTIONS for Questions 126 to 150: Answer the questions independently of each other.

A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition

A
11 ≤ e ≤ 66
B
10 ≤ e ≤ 66
C
11 ≤ e ≤ 65
D
0 ≤ e ≤ 11
Solution:
The least number of edges will be when one point is
connected to each of the other 11 points, giving a total
of 11 lines. One can move from any point to any other
point via the common point.
The maximum edges will be when a line exists between
any two points. Two points can be selected from 12
points in 12C2 i.e. 66 lines.
Q.No: 32
Test Name : CAT Paper 1990

A
p(–p)k–1
B
p(1–p)k–1
C
{p(1–p)}k–1
D
None of these
Solution:
Q.No: 33
Test Name : CAT 2017 Actual Paper Slot 1

If f1 (x) = x2 + 11x + n and f2 (x) = x, then the largest positive integer n for which the equation f1 (x) = f2 (x) has two distinct real roots, is

Solution:
Q.No: 34
Test Name : CAT 2017 Actual Paper Slot 1


A
2
B
1/3
C
6
D
2/3
Solution:
Q.No: 35
Test Name : CAT 2017 Actual Paper Slot 2

Let f(x) = x2 and g(x) = 2x, for all real x. Then the value of f(f(g(x)) + g(f(x))) at x = 1 is

A
16
B
18
C
36
D
40
Solution:
Q.No: 36
Test Name : CAT 2017 Actual Paper Slot 2

If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(1) is

Solution:
Q.No: 37
Test Name : CAT 2018 Actual Paper Slot 1

If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals

Solution:
Q.No: 38
Test Name : CAT 2019 Actual Paper Slot 1

Consider a function f satisfying f(x + y) = f(x) f(y) where x, y are positive integers, and f(1) = 2. If f(a + 1) + f(a + 2) + ... + f(a + n) = 16(2n – 1) then a is equal to

Solution:
Q.No: 39
Test Name : CAT 2019 Actual Paper Slot 1

For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd.

If m is a positive integer such that 8f(m + 1) – f(m) = 2, then m equals

Solution:
Q.No: 40
Test Name : CAT 2019 Actual Paper Slot 2

Let f be a function such that f (mn) = f (m) f (n) for every positive integers m and n. If f (1), f (2) and f (3) are positive integers, f (1) < f (2), and f (24) = 54, then f (18) equals

Solution:
Q.No: 41
Test Name : CAT Actual Paper 2020 Slot-1

If f(5 + x) = f(5 – x) for every real x, and f(x) = 0 has four distinct real roots, then the sum of these roots is

A
40
B
10
C
20
D
0
Solution:
Q.No: 42
Test Name : CAT Actual Paper 2020 Slot-2


A
1
B
4
C
0
D
16
Solution:
Q.No: 43
Test Name : CAT Actual Paper 2020 Slot-3

If f(x + y) = f(x)f(y) and f(5) = 4, then f(10) – f(–10) is equal to

A
15.9375
B
0
C
3
D
14.0625
Solution:
Q.No: 44
Test Name : CAT Actual Paper 2021 Slot-2


A
B
C
D
Solution:
Q.No: 45
Test Name : CAT Actual Paper 2021 Slot-3


A
-15
B
-20
C
-16
D
-12
Solution:
Q.No: 46
Test Name : CAT Actual Paper 2022 Slot-1

Let a, b, c be non-zero real numbers such that b2 < 4ac, and f(x) = ax2 + bx + c. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be

A
the empty set
B
the set of all integers
C
either the empty set or the set of all integers
D
the set of all positive integers
Solution:
Q.No: 47
Test Name : CAT Actual Paper 2022 Slot-1


A
0
B
50
C
100
D
25
Solution:
Q.No: 48
Test Name : CAT Actual Paper 2022 Slot-2

Suppose for all integers x, there are two functions f and g such that f(x) + f(x – 1) – 1 = 0 and g(x) = x2. If f(x2 – x) = 5, then the value of the sum f(g(5)) + g(f(5)) is

Solution:
Q.No: 49
Test Name : CAT Actual Paper 2022 Slot-2

Let f(x) be a quadratic polynomial in x such that f(x) ≥ 0 for all real numbers x. If f(2) = 0 and f(4) = 6, then f(–2) is equal to

A
12
B
6
C
36
D
24
Solution:
Q.No: 50
Test Name : CAT Actual Paper 2022 Slot-3


A
x ≤ r
B
x > r
C
x ≥ r
D
x ≠ r
Solution:
For x = r,
f(r) = 2r – r = r
So f(f(r)) = f(r) = r
For x < r,
f(x) = r
f(f(x)) = f(r) = r
For x > r,
Let x = kr, where k is some number greater than 1.
f(x) = f(kr) = 2kr – r = (2k – 1)r
Since k > 0, 2k – 1 > k
So for any x > r, f(f(x)) ≠ f(f(r))
Hence, for f(f(x)) = f(x) to be satisfied, the necessary condition is, x ≤ r.
Q.No: 51
Test Name : CAT Actual Paper 2023 Slot 3

Suppose f(x, y) is a real-valued function such that f(3x + 2y, 2x – 5y) = 19x, for all real numbers x and y. The value of x for which f(x, 2x) = 27, is

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Solution:
From Question 87, fog(x) = gof (x) = x, you will realise that if you were to form a chain of these functions for even number of times, you would still end up getting x.
E.g. fogofog(x) = fog(x) = x. Since both the brackets have the functions repeated for even number of times, each of their value will be x and their product will be x2.


Solution:
From question 87, gof(x) = fog(x) = x.
fo(fog)o(gof)(x) = fo(fog)(x) = f(x) = 2x + 3.


Solution:
If a = –2 and b = –3, then our expression will be me(–2 + mo(le(–2, –3)), mo(–2 + me(mo(–2), mo(–3)))
= me(–2 + mo(–3), mo(–2 + me(2, 3)))
= me(–2 + 3, mo(–2 + 3))
= me(1, mo(1)) = me(1, 1) = 1.


Solution:
Please note that the fastest way to solve these sums is the method of simulation, i.e., select any arbitrary values in the range given and verify whether the option holds good. E.g. a = 2, b = 3.
In this case, option (a) LHS = mo(le(2, 3)) = mo(2) = 2.
RHS = (me(mo(2), mo(3)) = (me(2, 3)) = 3.
Hence, LHS < RHS.
(b) LHS = mo(le(2, 3)) = mo(2) = 2. RHS = (me(mo(2), mo(3)) = (me(2, 3)) = 3.
Hence, LHS < RHS.
(c) LHS = mo(le(2, 3)) = mo(2) = 2. RHS = (le(mo(2), mo(3)) = le(2, 3) = 2.
Hence, LHS = RHS.
(d) LHS = mo(le(2, 3)) = mo(2) = 2. RHS = (le(mo(2), mo(3)) = le(2, 3) = 2.
Hence, LHS = RHS.


Solution:


Solution:


Solution:


Solution:


Solution:
The least number of edges will be when one point is
connected to each of the other 11 points, giving a total
of 11 lines. One can move from any point to any other
point via the common point.
The maximum edges will be when a line exists between
any two points. Two points can be selected from 12
points in 12C2 i.e. 66 lines.


Solution:


Solution:


Solution:


Solution:


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Solution:
For x = r,
f(r) = 2r – r = r
So f(f(r)) = f(r) = r
For x < r,
f(x) = r
f(f(x)) = f(r) = r
For x > r,
Let x = kr, where k is some number greater than 1.
f(x) = f(kr) = 2kr – r = (2k – 1)r
Since k > 0, 2k – 1 > k
So for any x > r, f(f(x)) ≠ f(f(r))
Hence, for f(f(x)) = f(x) to be satisfied, the necessary condition is, x ≤ r.


Solution:


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