# Functions and Graphs Questions for CAT Set-3 PDF

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## Functions and Graphs Questions for CAT Set-3 PDF

Download important CAT Functions and Graphs Set-3 Questions with Solutions PDF based on previously asked questions in CAT exam. Practice Functions and Graphs Set-3 Questions with Solutions for CAT exam.

Question 1: Let $f(x) = ax^2 + bx + c$, where a, b and c are certain constants and $a \neq 0$ ?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?[CAT 2008]

a) -7

b) – 4

c) 2

d) 6

e) cannot be determined

Question 2: 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 3: 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

Question 4: If $f(x)=x^3-4x+p$ , and f(0) and f(1) are of opposite signs, then which of the following is necessarily true[CAT 2004]

a) -1 < p < 2

b) 0 < p < 3

c) -2 < p < 1

d) -3 < p < 0

Question 5: For all non-negative integers x and y, f(x, y) is defined as below:
f(0, y) = y + 1
f(x + 1, 0) = f(x, 1)
f(x+ 1, y+ 1)= f(x, f(x+ 1, y))
Then, what is the value of f(1,2)?

a) Two

b) Four

c) Three

d) Cannot be determined

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 6:

a) a

b) b

c) c

d) d

Question 7:

a) a

b) b

c) c

d) d

Question 8:

a) a

b) b

c) c

d) d

Question 9:

a) a

b) b

c) c

d) d

Question 10: The figure below shows the graph of a function f(x). How many solutions does the equation f(f(x)) = 15 have?

a) 5

b) 6

c) 7

d) 8

e) cannot be determined from the given graph

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 –> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 –> (2)
From (1) and (2), a – b = 0 => a = b
=> c = -12a
The equation is, therefore, $ax^2 + ax – 12a = 0 => x^2 + x – 12 = 0$
=> -4 is a root of the equation.

$f(1)^2$ = f(1) => f(1) = 1

f(2)*(f(1/2) = f(1) => 4x = 1

So, f(1/2) = 1/4

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

f(1) = 1-4+p = p-3
f(0) = p
Since they are of opposite signs, p(p-3) < 0
=> 0 < p < 3

For f(1,2). First consider x=0 and y=1 and use 3rd given equation, we get f(0,f(1,1)) now for f(1,1) take x=0 and y=0 we get f(0,f(1,0)), for f(1,0) which we use 2nd equation we get f(0,1) whose value is 2. So we have f(0,f(1,0))= f(0,2) whose value is 3 then put this in f(0,f(1,1)) we get f(0,3) we get as 4

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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.

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.

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.

F(0) = 1 ; F1(0) = -1

F(1) = 0 ; F1(-1) = 0

F(2) = -1 ; F1(-2) = 1

=> F1(x) = -F(-x).