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.
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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
Answers & Solutions:
1) Answer (B)
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.
2) 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
3) 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
4) Answer (B)
f(1) = 1-4+p = p-3
f(0) = p
Since they are of opposite signs, p(p-3) < 0
=> 0 < p < 3
5) Answer (B)
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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6) 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.
7) 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.
8) 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.
9) Answer (C)
F(0) = 1 ; F1(0) = -1
F(1) = 0 ; F1(-1) = 0
F(2) = -1 ; F1(-2) = 1
=> F1(x) = -F(-x).
10) Answer (C)
It has been given that f(f(x)) = 15.
From the graph, we can see that the value of f(4) = 15 and f(12) = 15
Therefore, f(x) can be 4 or 12.
When f(x) = 4, x can take 4 values
When f(x) = 12, x can take 3 values.
Therefore, there are 4+3 = 7 solutions in total.
Therefore, option C is the right answer.
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