Functions and Graphs Questions for IIFT PDF
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Question 1: A function $f (x)$ satisfies $f(1) = 3600$, and $f (1) + f(2) + … + f(n) =n^2f(n)$, for all positive integers $n > 1$. What is the value of $f (9)$ ?
a) 80
b) 240
c) 200
d) 100
e) 120
Instructions
Directions for the next 3 questions:
Given below are three graphs made up of straight-line segments shown as thick lines. In each case choose the answer as:
a) if f(x)=3f(-x)
b) if f(x)= -f(-x)
c) if f(x) = f(-x)
d) if 3f(x) = 6f(-x), for x >= 0
Question 2: 
a) if f(x)=3f(-x)
b) if f(x)= -f(-x)
c) if f(x) = f(-x)
d) if 3f(x) = 6f(-x), for x >= 0
Question 3:
a) if f(x)=3f(-x)
b) if f(x)= -f(-x)
c) if f(x) = f(-x)
d) if 3f(x) = 6f(-x), for x >= 0
Instructions
DIRECTIONS for the following questions: These questions are based on the situation given below: In each of the questions a pair of graphs F(x) and F1(x) is given. These are composed of straight-line segments, shown as solid lines, in the domain $x\epsilon (-2, 2)$.
choose the answer as
a. If F1(x) = – F(x)
b. if F1(x) = F(- x)
c. if F1(x) = – F(- x)
d. if none of the above is true
Question 4: 
a) a
b) b
c) c
d) d
Question 5: 
a) a
b) b
c) c
d) d
Question 6: 
a) a
b) b
c) c
d) d
Question 7: 
a) a
b) b
c) c
d) d
Question 8: Let $f(x)\neq0$ for any ‘x’ be a function satisfying $f(x)f(y) = f(xy)$ for all real x, y. If $f(2) = 4$, then what is the value of $f(\frac{1}{2})$?
a) 0
b) 1/4
c) 1/2
d) 1
e) cannot be determined
Question 9: The function f(x) = |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at
a) x = 2.3
b) x = 2.5
c) x = 2.7
d) None of the above
Instructions
Directions for the following two questions:
Answer the questions on the basis of the information given below.
$f_1(x) = x$ if $0 \leq x \leq 1$
$f_1(x) = 1$ if x >= 1
$f_1(x) = 0$ otherwise
$f_2(x) = f_1(-x)$ for all x
$f_3(x) = -f_2(x)$ for all x
$f_4(x) = f_3(-x)$ for all x
Question 10: How many of the following products are necessarily zero for every x:$f_1(x)f_2(x), f_2(x)f_3(x), f_2(x)f_4(x)$
a) 0
b) 1
c) 2
d) 3
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Answers & Solutions:
1) Answer (A)
According to given conditions we get f(2)=f(3)/3 , then f(3)=f(1)/6, then f(4)=f(1)/10 , then f(5)=f(1)/15 .
We can see the pattern here that the denominator goes on increasing from 3,3+3,6+4,10+5,15+6,.. so for the f(9) the denominator will be same as 15+6+7+8+9=45 .
So f(9)=3600/45 = 80
2) Answer (D)
In this graph, f(-1) = 1 and f(1) = 2 and since the two lines pass origin, their values increase linearly.
3f(x) = 6f(-x)
3) Answer (B)
In this graph, we can see that the graph on LHS of y axis is symmetrical to the graph on RHS of y axis about origin.
Hence f(x) = -f(-x)
4) Answer (D)
The correct relation between the two is: F(x) = | F1(x) |
So, all the three options a), b) and c) can be ruled out. Option d) is the correct answer.
5) Answer (B)
The value of F(x) for x < 0 is the same as the value of F1(x) for x > 0.
So, F1(x) = F(-x)
Option b) is the correct answer.
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6) Answer (B)
The value of F(x) for x > 0 is the same as the value of F1(x) for x < 0.
So, F1(x) = F(-x)
Option b) is the correct answer.
7) Answer (C)
F(0) = 1 ; F1(0) = -1
F(1) = 0 ; F1(-1) = 0
F(2) = -1 ; F1(-2) = 1
=> F1(x) = -F(-x).
8) Answer (B)
$f(1)^2$ = f(1) => f(1) = 1
f(2)*(f(1/2) = f(1) => 4x = 1
So, f(1/2) = 1/4
9) Answer (B)
f(x) = |x – 2| + |2.5 – x| + |3.6 – x|
For x belonging to (-infinity to 2), f(x) = 2-x + 2.5-x + 3.6-x = 8.1-3x
This attains the minimum value at x=2. Value = 2.1
For x belonging to (2 to 2.5), f(x) = x-2 + 2.5-x + 3.6-x = 4.1-x
Attains the minimum value at x = 2.5. Value = 1.6
For x belonging to (2.5 to 3.6), f(x) = x-2 + x-2.5 + 3.6-x = x-0.9
Attains the minimum at x=2.5, value = 1.6
For x > 3.6, f(x) = x-2+x-2.5+x-3.6 = 3x – 8.1
Attains the minimum at x= 3.6, value = 2.7
So, min value of the function is 1.6 at x=2.5
10) Answer (C)
Checking for different values of x . Suppose x= -0.5 we get
$f_1(x)f_2(x) = 0*0.5 = 0$
$f_2(x)f_4(x) = 0.5*0 = 0$ .
But $f_2(x)f_3(x)$ is not equal to zero.
Hence two functions are necessarily equal to zero and two products given above are equal to zero.
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