**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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