UG Mathematics Booster Test 2 - Types of Functions -

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UG Mathematics Booster Test 2 - Types of Functions

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

UG Mathematics Booster Test 2 - Types of Functions

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QUESTION 1 OF 20

Applying the mapping concept mathematically, which statements denote a valid function f from A to B?
I. Every element a in A has exactly one image b in B.
II. One element in A maps to multiple elements in B.
III. The subset of pairs is from A × B.

Answer Review

1 Applying the mapping concept mathematically, which statements denote a valid function f from A to B?
I. Every element a in A has exactly one image b in B.
II. One element in A maps to multiple elements in B.
III. The subset of pairs is from A × B.

A I, II, III

B I and III only

C II and III only

D I only

💡 Short Explanation

Functions assign one output per input Ordered pairs belong to A × B Multiple outputs violate function definition

✅ Correct Answer Explanation

A valid function requires every element of A to have exactly one image in B. Also, all ordered pairs of the function belong to: \(A\times B\) Statement II is false because one input cannot map to multiple outputs. Hence only I and III are correct.

❓ Why Other Options Are Incorrect

Option A → Statement II violates uniqueness of outputs.
Option C → Multiple outputs for one input are not allowed.
Option D → Ignores the Cartesian product requirement.

🎯 Strategy

Used: Elimination

Application:

Reject statements contradicting the definition of a function.

Final Logic:

A function means exactly one image for each input.

🧠 Memory Trick

"One input → one output"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Definition of Function

2 Match the domain-codomain concepts for f: X → Y:

List I	List II
1. X	a. Image of x
2. Y	b. Domain
3. f(X)	c. Range
4. y = f(x)	d. Codomain

A 1-b, 2-c, 3-d, 4-a

B 1-c, 2-d, 3-b, 4-a

C 1-d, 2-b, 3-a, 4-c

D 1-b, 2-d, 3-c, 4-a

💡 Short Explanation

X denotes domain Y denotes codomain f(X) gives range

✅ Correct Answer Explanation

In the notation: \(f:X\rightarrow Y\) X is the domain and Y is the codomain. The set f(X) represents the range. Expression y=f(x) gives the image of x. Thus the correct matching is 1-b, 2-d, 3-c, 4-a.

❓ Why Other Options Are Incorrect

Option A → Range and codomain are interchanged.
Option B → Domain incorrectly matched to range.
Option C → Domain and codomain mismatched.

🎯 Strategy

Used: Option Grouping

Application:

Match standard function notation definitions carefully.

Final Logic:

Domain → X, Codomain → Y, Range → f(X).

🧠 Memory Trick

"X starts, Y receives"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Domain and Codomain

3 The region bounded by the injective definition function f(x) = x³ from x = 0 to x = 2 and the x-axis represents an area. What is this calculated area?

A 8

B 2

C 4

D 16

💡 Short Explanation

Area found using integration Integrate x³ from 0 to 2 Result equals 4 square units

✅ Correct Answer Explanation

The required area is: \(\int_{0}^{2}\,x^{3} dx\) Evaluating: \({\left[\frac{x^{4}}{4}\right]}_{0}^{2}=\frac{16}{4}=4\) Hence the area bounded by the curve and x-axis equals 4 square units. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Double the correct value.
Option B → Too small due to incorrect integration.
Option D → Equals x⁴ value before division by 4.

🎯 Strategy

Used: Substitution

Application:

Use the power-rule integration formula directly.

Final Logic:

Integral of x³ gives x⁴/4.

🧠 Memory Trick

"Power increases, divide next power"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Area Under Curves

4 A condition check is run on all possible functions mapping from set A = {1, 2, 3} to set B = {x, y, z}. What is the probability that a randomly selected function is injective (one-one)?

A 2/9

B 1/3

C 4/9

D 1/9

💡 Short Explanation

Total functions = 3³ Injective functions = 3! Probability = 6/27 = 2/9

✅ Correct Answer Explanation

Total functions from A to B: \(3^{3}=27\) Injective functions require distinct images: \(3!=6\) Therefore probability: \(\frac{6}{27}=\frac{2}{9}\) Hence the required probability is 2/9.

❓ Why Other Options Are Incorrect

Option B → Incorrect simplification.
Option C → Overcounts injective mappings.
Option D → Too few injective functions counted.

🎯 Strategy

Used: Substitution

Application:

Count total and injective mappings separately.

Final Logic:

Injective probability = favorable ÷ total.

🧠 Memory Trick

"3! over 3³"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: One-One Functions

5 Which of the following is INCORRECT regarding a non-injective case?

A The mapping involves distinct elements of domain having the same image.

B The function can be inverted completely without restricting the domain.

C A horizontal line intersects its graph more than once.

D f(x₁) = f(x₂) does not guarantee x₁ = x₂.

💡 Short Explanation

Many-one functions fail injectivity Inverse function requires one-one property Domain restriction is often necessary

✅ Correct Answer Explanation

Non-injective (many-one) functions do not possess proper inverses over the entire domain because multiple inputs give the same output. To define an inverse, the domain must usually be restricted. Therefore Option B is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct description of many-one mapping.
Option C → Horizontal line test fails for non-injective functions.
Option D → Equal outputs may come from distinct inputs.

🎯 Strategy

Used: Elimination

Application:

Check which statement contradicts inverse-function theory.

Final Logic:

Only injective functions have unrestricted inverses.

🧠 Memory Trick

"Many-one blocks inverse"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Many-One Functions

6 A chemical mixture's stability is modeled by a many-one function f(x) = x² - 4. Evaluate the numerical difference between f(-4) and f(4) examples to verify it maps identically. What is the value of f(4) - f(-4)?

A 8

B 16

C 0

D 32

💡 Short Explanation

Square removes sign difference f(4)=f(-4) Their difference equals zero

✅ Correct Answer Explanation

Given: \(f(x)=x^{2}-4\) Then: \(f(4)=16-4=12\) and \(f(-4)=16-4=12\) Therefore: \(f(4)-f(-4)=0\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect arithmetic subtraction.
Option B → Uses squared value only.
Option D → Ignores cancellation.

🎯 Strategy

Used: Substitution

Application:

Evaluate function at ±4 directly.

Final Logic:

Equal outputs imply zero difference.

🧠 Memory Trick

"Squares erase signs"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Many-One Functions

7 Arrange the following surjective definition examples in increasing order of the size of their codomains, given they are all onto functions:
1. f: X → {1}
2. f: X → {a, b, c}
3. f: X → {True, False}

A 1, 3, 2

B 2, 3, 1

C 3, 1, 2

D 1, 2, 3

💡 Short Explanation

Codomain sizes are 1,2,3 Arrange smallest to largest Onto functions cover all codomain elements

✅ Correct Answer Explanation

Codomain sizes are: \(∣1∣=1,∣True,False∣=2,∣a,b,c∣=3\) Increasing order is therefore: 1 → 3 → 2. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Largest codomain placed first.
Option C → Two-element codomain incorrectly placed before one-element.
Option D → Three-element codomain misplaced.

🎯 Strategy

Used: Odd One Out

Application:

Compare codomain cardinalities directly.

Final Logic:

Arrange according to codomain size.

🧠 Memory Trick

"1 then 2 then 3"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Onto Functions

8 A transformation maps a set of position vectors into a vector space V. If the mapping is an onto function where Range equals codomain, it implies the mapped vectors:

A Are all zero vectors.

B Cannot be added together.

C Must form a subset smaller than V.

D Span the entire vector space V.

💡 Short Explanation

Onto means full codomain coverage Every vector in V has a pre-image Hence image spans whole vector space

✅ Correct Answer Explanation

An onto transformation satisfies: \(Range=Codomain\) Thus every vector in V is obtained from some input vector. Therefore the mapped vectors span the entire vector space V. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Onto mappings need not produce only zero vectors.
Option B → Vector addition property is unrelated.
Option C → Onto mappings cover all of V, not smaller subsets.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret "onto" in vector-space language.

Final Logic:

Onto transformations cover entire codomain.

🧠 Memory Trick

"Onto means full coverage"

📚 Class XII – NCERT Mathematics – Chapter: Vector Algebra – Topic: Onto Mapping

9 Assertion (A): The function f(x) = x² from R to R is an onto function.
Reason (R): The not onto case is defined when every element in the codomain has a pre-image.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

x² misses negative outputs Hence function is not onto Reason incorrectly defines onto instead of into

✅ Correct Answer Explanation

The function: \(f(x)=x^{2}\) never produces negative real numbers, so it is not onto R. The Reason is also false because "every codomain element has a pre-image" defines onto functions, not not-onto functions. Therefore both statements are false.

❓ Why Other Options Are Incorrect

Option B → Assertion is false because negatives are missing.
Option C → Reason incorrectly defines onto condition.
Option D → Reason itself is false.

🎯 Strategy

Used: Elimination

Application:

Test onto definition against modulus-like behavior.

Final Logic:

Missing negative outputs make x² not onto.

🧠 Memory Trick

"Squares never go negative"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Into Functions

10 A function maps a dataset of 5 inputs to a codomain. The missing elements in the codomain tracking over a series are 10, 20, 30, 40. What is the 2-period moving average of the first two missing elements?

A 20

B 10

C 15

D 25

💡 Short Explanation

Use first two elements only Average of 10 and 20 Result equals 15

✅ Correct Answer Explanation

The 2-period moving average of the first two missing elements is: \(\frac{10+20}{2}=15\) Hence the moving average equals 15. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Equals second term, not average.
Option B → Equals first term only.
Option D → Incorrect addition/division.

🎯 Strategy

Used: Substitution

Application:

Apply moving-average formula directly.

Final Logic:

Average of first two terms equals 15.

🧠 Memory Trick

"Average sits in middle"

📚 Class XI – NCERT Mathematics – Chapter: Statistics – Topic: Moving Average

11 When verifying if a region-based function graph represents a one-one and onto function (bijective), which graphical tests must it pass on its entire domain and codomain?

A Vertical line test only.

B Both vertical and horizontal line tests.

C Only the horizontal line test.

D It must not pass through the origin.

💡 Short Explanation

Vertical line test checks function validity Horizontal line test checks injectivity Bijective functions satisfy both conditions

✅ Correct Answer Explanation

A bijective function must be both a valid function and one-one. The vertical line test confirms each x has one y-value, while the horizontal line test ensures different x-values produce distinct y-values. Together they verify bijection graphically. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Vertical line test alone checks only function validity.
Option C → Horizontal line test alone cannot confirm function definition.
Option D → Passing through origin is irrelevant to bijection.

🎯 Strategy

Used: Elimination

Application:

Separate conditions for "function" and "one-one" graphically.

Final Logic:

Bijective = valid function + injective graph.

🧠 Memory Trick

"Vertical for function, horizontal for one-one"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Bijective Functions

12 For the bijective definition function f(x) = 3x² on the interval, find the definite integral value to measure its cumulative application.

A 6

B 4

C 12

D 8

💡 Short Explanation

Integrate 3x² over given interval Use power rule of integration Result evaluates to 8

✅ Correct Answer Explanation

The integral is: \(\int_{0}^{2}\,3x^{2} dx\) Evaluating: \(={\left[x^{3}\right]}_{0}^{2}=8-0=8\) Hence the definite integral equals 8. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect evaluation of upper limit.
Option B → Result from wrong power handling.
Option C → Overestimation of area.

🎯 Strategy

Used: Substitution

Application:

Apply standard integration formula directly.

Final Logic:

Integral of 3x² becomes x³.

🧠 Memory Trick

"3x² integrates neatly to x³"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Definite Integral

13 The identity function applied to an element x returns:

A The element x itself.

B The inverse of x.

C Zero.

D A constant c.

💡 Short Explanation

Identity function preserves input Output equals input exactly No transformation occurs

✅ Correct Answer Explanation

The identity function is defined as: \(I(x)=x\) It maps every element to itself without modification. Hence the output of x under the identity function is x itself. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Inverse value means reciprocal, not identity.
Option C → Identity does not always output zero.
Option D → Constant output defines constant function.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall standard definition of identity mapping.

Final Logic:

Identity leaves elements unchanged.

🧠 Memory Trick

"Identity = unchanged"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Identity Function

14 The constant function defined by f(x) = c maps all inputs to a single output. It is classified as:

A One-one and onto.

B One-one and into.

C Many-one and onto.

D Many-one and into (if codomain has >1 element).

💡 Short Explanation

All inputs give same output Hence function is many-one Codomain larger than range makes it into

✅ Correct Answer Explanation

A constant function maps every input to the same fixed value c. Thus multiple inputs share one output, making it many-one. If the codomain has more than one element, not every codomain element gets mapped, so the function is into. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Constant functions are never one-one.
Option B → One-one condition fails immediately.
Option C → Constant functions are generally not onto.

🎯 Strategy

Used: Elimination

Application:

Check injective and onto conditions separately.

Final Logic:

Single repeated output implies many-one into.

🧠 Memory Trick

"One constant for all"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Constant Function

15 The polynomial function f(x) = x³ - x maps R to R. Evaluating its behavior, it is:

A One-one but not onto.

B Into but not one-one.

C Many-one and onto.

D Bijective.

💡 Short Explanation

Polynomial covers all real outputs Different inputs may give same output Hence onto but not injective

✅ Correct Answer Explanation

The function: \(f(x)=x^{3}-x\) is onto because cubic polynomials attain all real values. However: \(f(-1)=0,f(0)=0,f(1)=0\) so distinct inputs share the same output. Therefore it is many-one and onto. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Function is not injective.
Option B → Cubic polynomials are onto R.
Option D → Bijective requires injectivity too.

🎯 Strategy

Used: Substitution

Application:

Test equal outputs at different x-values.

Final Logic:

Same output from multiple inputs breaks one-one property.

🧠 Memory Trick

"−1, 0, 1 all give 0"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Polynomial Functions

16 Evaluating the rational function f(x) = (x+1)/(x-2), its domain must exclude:

A x = 0

B x = 2

C x = 1

D x = -1

💡 Short Explanation

Denominator cannot be zero x−2 = 0 gives undefined value Hence x = 2 excluded

✅ Correct Answer Explanation

For rational functions, denominator must not vanish: \(x-2\neq 0\) Solving gives: \(x\neq 2\) Therefore x = 2 must be excluded from the domain. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Denominator remains nonzero at x=0.
Option C → Function defined at x=1.
Option D → Function defined at x=-1.

🎯 Strategy

Used: Substitution

Application:

Set denominator equal to zero.

Final Logic:

Undefined denominator restricts domain.

🧠 Memory Trick

"Denominator zero forbidden"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Rational Functions

17 Applying the modulus function f(x) = |x-3| to evaluate x = -2 yields:

A -5

B 1

C -1

D 5

💡 Short Explanation

Substitute x = −2 Compute modulus value Absolute value becomes positive

✅ Correct Answer Explanation

Given: \(f(x)=∣x-3∣\) Substituting x = −2: \(∣-2-3∣=∣-5∣=5\) Hence the function value equals 5. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Modulus cannot be negative.
Option B → Arithmetic error in subtraction.
Option C → Absolute value never negative.

🎯 Strategy

Used: Substitution

Application:

Directly evaluate expression inside modulus.

Final Logic:

Absolute value converts −5 to 5.

🧠 Memory Trick

"Modulus removes sign"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Modulus Function

18 The signum function yields what output for x = 100?

A 1

B 100

C 0

D -1

💡 Short Explanation

Signum depends on sign only Positive numbers map to 1 100 is positive

✅ Correct Answer Explanation

The signum function is: \(sgn⁡(x)=\left\{\begin{pmatrix}1, & x>0\\ 0, & x=0\\ -1, & x<0\end{pmatrix}\right.\) Since 100 is positive: \(sgn⁡(100)=1\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Signum gives sign, not value itself.
Option C → Output 0 occurs only at x=0.
Option D → Negative outputs occur only for x<0.

🎯 Strategy

Used: Elimination

Application:

Check sign of the given number.

Final Logic:

Positive input gives output 1.

🧠 Memory Trick

"Positive → +1"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Signum Function

📄 Passage

The results mentioned are also true for an arbitrary finite set X, i.e., a one-one function f: X → X is necessarily onto and an onto map f: X → X is necessarily one-one, for every finite set X. In contrast to this, for an infinite set, this may not be true. In fact, this is a characteristic difference between a finite and an infinite set.

Based on the finite set behavior described in the passage, if a function f: {1, 2, 3} → {1, 2, 3} is known to be one-one, it must also be:

A An into function

B A many-one function

C A constant function

D An onto function

💡 Short Explanation

Finite sets connect injective and onto properties One-one automatically implies onto Domain and codomain have equal size

✅ Correct Answer Explanation

For finite sets with equal cardinality, an injective mapping automatically covers the entire codomain. Thus a one-one function: \(f:X\rightarrow X\) must also be onto. Therefore the function becomes bijective. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Into contradicts finite injective property.
Option B → Many-one opposes one-one assumption.
Option C → Constant functions are not injective.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the exact finite-set theorem from passage.

Final Logic:

Finite equal-sized sets force injective ⇒ onto.

🧠 Memory Trick

"Finite one-one means onto"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Finite Set Functions

📄 Passage

According to the passage, infinite set behavior differs because:

A Infinite sets cannot form functions.

B Functions on infinite sets are always bijective.

C A one-one function on an infinite set is not necessarily onto.

D The domain and codomain cannot be defined.

💡 Short Explanation

Infinite sets behave differently Injective need not imply onto Example: f(x)=x+1 on N

✅ Correct Answer Explanation

For infinite sets, one-one functions may fail to cover the entire codomain. Example: \(f(n)=n+1\) on natural numbers is injective but misses 1 in the codomain. Hence injective does not always imply onto for infinite sets. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Infinite sets can define functions normally.
Option B → Infinite-set functions need not be bijective.
Option D → Domains and codomains are well-defined for infinite sets.

🎯 Strategy

Used: Elimination

Application:

Compare finite-set property with infinite-set exception.

Final Logic:

Infinite sets break injective ⇒ onto equivalence.

🧠 Memory Trick

"Infinite sets break the rule"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Infinite Set Functions

UG Mathematics Booster Test 2 - Types of Functions

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