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CAT Exam  >  CAT Test  >  Level-wise Tests  >  Test Level 3: Functions - CAT MCQ

Test Level 3: Functions CAT MCQs & solutions - Free


MCQ Practice Test & Solutions: Test Level 3: Functions (10 Questions)

You can prepare effectively for CAT Level-wise Tests for CAT with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test Level 3: Functions". These 10 questions have been designed by the experts with the latest curriculum of CAT 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 10

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Test Level 3: Functions - Question 1
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If f : A → B is defined such that f(x) = then f -1(x) = ?

Detailed Solution: Question 1

f(x) = 
Let f(x) be y.

Test Level 3: Functions - Question 2
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If f(x) = xa + ax and g(x) = xa - ax , then find, when x = 2 and a = 3.

Detailed Solution: Question 2


For x = 2 and a = 3
We have,

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Test Level 3: Functions - Question 3
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When f(x) - f(-x) = 0, the function f(x) is called an even function. Which of the following functions is an even function?

Detailed Solution: Question 3

Going with the options, the second option comes out to be the answer.
f(t) = … (1)
f(-t) = 
Subtracting (2) from (1), we get
f(t) - f(-t) = (1 - et) + t = -t + t = 0

Test Level 3: Functions - Question 4
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A function satisfies f(0, n) = n + 1 and f(m + 1, n) = f(m, f(m, n)). What is the value of f(4, 100)?

Detailed Solution: Question 4

f(4, 100) = f[3, f(3, 100)] …………(1)
f(3, 100) = f[2, f(2, 100)] ……….. (2)
f(2, 100) = f[1, f(1, 100)] …………(3)
f(1, 100) = f[0, f(0, 100)] ……….. (4)
f(0, 100) = 100 + 1 = 101.
f(1, 100) = f(0, 101) = 102
f(2, 100) = f(1, 102) …………. (5)
f(1, 102) = f[0, f(0, 102)] = f(0, 103) = 104
Put in (5) = f(2, 100) = 104
Put this value in (2),
f(3, 100) = f(2, 104) …………… (6)
Now, f(2, 104) = f[1, f(1, 104)] …………… (7)
f(1, 104) = f[0, f(0, 104)] = f(0, 105) = 106
Put in (7)... and so on
f(4, 100) = 116

Test Level 3: Functions - Question 5
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If f(x) satisfies 2f(x) + f(1 – x) = x2 for all x, then f(x) is equal to

Detailed Solution: Question 5

The given equation 2f(x) + f(1 – x) = x2 holds for all x.
2f(x) + f(1 – x) = x2.......(i)
In particular, the equation holds, if we replace x with 1 – x.
Thus, we deduce that 2f(1 – x) + f(x) = (1 – x)2.
2f(1 – x) + f(x) = (1 – x)2..............(ii)
Multiply equation (i) by 2 and then subtract equation (ii) from equation (i).
Thus, we get:
3f(x) = 2x2 – (1 – x)2 = x2 + 2x – 1
f(x) = (x2 + 2x – 1)/3

Test Level 3: Functions - Question 6
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Let f(x) be a polynomial with integer coefficients, for which 3 and 13 are the roots. Which of the following could possibly be the value of f(10)?

Detailed Solution: Question 6

The given information implies that f(x) = (x – 3) (x – 13) g(x), where g(x) is a polynomial with integer coefficients. Hence, f(10) = – 21g(10), so that 21 must be a divisor of f(10). The only choice divisible by 21 is 42, so the correct answer is 42. To check if f(x) exists, consider f(x) = – 2(x – 3) (x – 13). 

Test Level 3: Functions - Question 7
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A function f(x) is called even if f(-x) = f(x) for all x and it is called odd if f(-x) = -f(x) for all x. Which of the following statements is/are true?
(i) The product of an even function and an odd function is even.
(ii) The sum of two even functions is even.
(iii) The product of two odd functions is even.

Detailed Solution: Question 7

Statement (i): Let F(x) be an even function and G(x) be odd.
H(x) = Product of an even function and an odd function
H(x) = F(x) G(x)
Now, H(-x) = F(-x) G(-x) = F(x) (-G(x)) = -F(x) G(x) = -H(x)
Hence, H(x) is an odd function. Statement (i) is not true.
Statement (ii): Let F(x) and G(x) be even functions.
H(x) = F(x) + G(x)
H(-x) = F(-x) + G(-x) = F(x) + G(x) = H(x)
It is an even function; hence, statement (ii) is true.
Statement (iii): Let F(x) and G(x) be odd functions.
H(x) = F(x) G(x)
H(-x) = F(-x) G(-x) = (-F(x))(-G(x)) = F(x) G(x) = H(x)
It is an even function. Hence, statement (iii) is true.
So, statements (ii) and (iii) are true.

Test Level 3: Functions - Question 8
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For all real numbers x, a function f(x) satisfies 2f(x) + f(1 - x) = x2. Find the value of f(5).

Detailed Solution: Question 8

Given: 2f(x) + f(1 - x) = x2 ... (A)
Put x = 5 in (i).
2f(5) + f(-4) = 25
Multiply by 2
4f(5) + 2f(-4) = 50 ... (1)
Put x = -4 in (A).
2f(-4) + f(5) = 16 ... (2)Subtract (2) from (1):
4f(5) - f(5) = 50 - 16
3f(5) = 34
f(5) = 34/3

Test Level 3: Functions - Question 9
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Which of the following is the graph of f(x) = ?

Detailed Solution: Question 9

 f(x) = 1/(x - 2)
When x = 1, f(x) = -1
When x = 2, f(x) = ∞
When x = 3, f(x) = 1, and so on
Therefore, option d is correct.

Test Level 3: Functions - Question 10
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Refer to the following data to answer the question that follows.
F(x) = Modulus of x
G(x) = The largest integer less than or equal to x
H(x) = The smallest integer greater than or equal to x
I(x) = x, a real number
Find the value of H

Detailed Solution: Question 10

 I  = I (-1) = -1

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This test contains 10 MCQs on Test Level 3: Functions with detailed solutions for each question. It is part of the Level-wise Tests for CAT course on EduRev, designed to align with the current CAT syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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