UG Mathematics Booster Test 3 - Types of Functions -

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UG Mathematics Booster Test 3 - Types of Functions

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 3 - Types of Functions

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QUESTION 1 OF 20

Let a mapping concept be represented by f(x) = ∫(from 0 to x) 2t dt. Which type of function is f(x) mapping from positive reals R+ to R+?

Answer Review

1 Let a mapping concept be represented by f(x) = ∫(from 0 to x) 2t dt. Which type of function is f(x) mapping from positive reals R+ to R+?

A Constant

B Identity

C Bijective

D Many-one

💡 Short Explanation

Evaluate the integral first Function becomes strictly increasing Positive reals map uniquely onto positive reals

✅ Correct Answer Explanation

Given: \(f(x)=\int_{0}^{x}\,2t dt=x^{2}\) On positive reals, \(x^{2}\)is strictly increasing and every positive output has exactly one positive input. Thus the function is both one-one and onto on \(R^{+}\). Hence it is bijective.

❓ Why Other Options Are Incorrect

Option A → Function output changes with x, so not constant.
Option B → Identity function requires f(x)=x, not x².
Option D → Distinct positive inputs give distinct outputs.

🎯 Strategy

Used: Substitution

Application:

Evaluate the integral and analyze resulting mapping behavior.

Final Logic:

x² on positive reals is injective and onto.

🧠 Memory Trick

"Positive square stays unique"

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

2 A function defines the upper boundary of a domain-codomain mapping. If f(x) = 6 - x, what is the area in the first quadrant bounded by this function?

A 18

B 36

C 12

D 9

💡 Short Explanation

Line intersects axes at 6 Region forms a triangle Area = ½ × base × height

✅ Correct Answer Explanation

The graph: \(y=6-x\) cuts both axes at 6, forming a right triangle in the first quadrant with base = 6 and height = 6. \(Area=\frac{1}{2}(6)(6)=18\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Equals square area, not triangular region.
Option C → Uses incorrect dimensions.
Option D → Half of the required value.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Interpret graph geometrically as a triangle.

Final Logic:

Triangle area formula gives 18.

🧠 Memory Trick

"Half of 6×6"

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

3 An injective definition function maps 3D vectors. If f(v) = 2v + c, and f(a) = f(b), analytical vector subtraction proves:

A a and b are perpendicular.

B a is a scalar multiple of b.

C The cross product is maximum.

D a = b.

💡 Short Explanation

Equal outputs imply equal transformed vectors Subtract constants from both sides Divide by 2 to obtain equality

✅ Correct Answer Explanation

Given: \(f(a)=f(b)2a+c=2b+c\) Subtracting c: \(2a=2b\) Dividing by 2: \(a=b\) Thus the function is injective. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → No perpendicularity condition appears.
Option B → Equality is stronger than scalar multiple.
Option C → Cross product is unrelated.

🎯 Strategy

Used: Substitution

Application:

Use algebraic equality of vector expressions.

Final Logic:

Equal transformed vectors imply equal original vectors.

🧠 Memory Trick

"Cancel and compare"

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

4 A condition check algorithm processes data values checking for injectivity. It logs data points (outputs): 2, 4, 8, 16. If moving average of period 2 is calculated, what is the second moving average logged?

A 4

B 6

C 10

D 12

💡 Short Explanation

2-period average uses consecutive pairs Second pair is 4 and 8 Average equals 6

✅ Correct Answer Explanation

The 2-period moving averages are: \(\frac{2+4}{2}=3,\frac{4+8}{2}=6,\frac{8+16}{2}=12\) Thus the second moving average equals 6. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Not obtained from consecutive pair average.
Option C → Incorrect arithmetic.
Option D → Third moving average value.

🎯 Strategy

Used: Substitution

Application:

Apply moving-average formula sequentially.

Final Logic:

Average of 4 and 8 equals 6.

🧠 Memory Trick

"Middle pair gives middle average"

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

5 Let S = {1, 2, 3}. A non-injective case mapping from S to S is randomly formed. What is the probability that it is a many-one function?

A 7/9

B 2/9

C 1/3

D 2/3

💡 Short Explanation

Total functions = 27 Injective functions = 6 Many-one probability = 21/27

✅ Correct Answer Explanation

Total functions from S to S: \(3^{3}=27\) Injective functions: \(3!=6\) Non-injective (many-one) functions: \(27-6=21\) Probability: \(\frac{21}{27}=\frac{7}{9}\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Represents injective probability instead.
Option C → Incorrect counting of mappings.
Option D → Simplification error.

🎯 Strategy

Used: Substitution

Application:

Count total and injective mappings separately.

Final Logic:

Many-one probability = complement of injective probability.

🧠 Memory Trick

"27 total, 6 one-one"

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

6 Evaluating examples analytically, the region mapping of f(x) = cos(x) on the interval [0, 2π] proves it is many-one because:

A The maximum area is negative.

B The graph is strictly increasing.

C The horizontal line y = 0.5 intersects the graph in two places in this region.

D It maps negative inputs to positive outputs.

💡 Short Explanation

Horizontal line test checks injectivity Multiple intersections imply many-one cos(x) repeats values on interval

✅ Correct Answer Explanation

For: \(y=cos⁡x\) on \(\left[0,\ 2\pi \right]\), many outputs repeat. The horizontal line \(y=0.5\) cuts the graph at two distinct x-values. Therefore distinct inputs produce the same output, proving the function is many-one.

❓ Why Other Options Are Incorrect

Option A → Area sign does not determine injectivity.
Option B → cos(x) is not strictly increasing on entire interval.
Option D → Input sign is unrelated here.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use horizontal line test for injectivity.

Final Logic:

Repeated outputs imply many-one behavior.

🧠 Memory Trick

"Two cuts mean many-one"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Horizontal Line Test

7 A surjective definition requires covering the codomain. If f: A → B is onto, and n(A) = 4, n(B) = 3, mathematically, how many such onto functions exist?

A 24

B 36

C 81

D 12

💡 Short Explanation

Total onto mappings counted carefully Use inclusion-exclusion principle Final count equals 36

✅ Correct Answer Explanation

Number of onto functions from 4 elements to 3 elements: \(3^{4}-\left(\frac{3}{1}\right)2^{4}+\left(\frac{3}{2}\right)1^{4}=81-48+3=36\) Thus the number of surjective functions equals 36. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Underestimates surjective mappings.
Option C → Counts all possible functions.
Option D → Incorrect inclusion-exclusion result.

🎯 Strategy

Used: Substitution

Application:

Apply onto-function counting formula.

Final Logic:

Subtract functions missing codomain elements.

🧠 Memory Trick

"Total minus missing cases"

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

8 Match the analytical condition when Range equals codomain:

List I	List II
1. f(x) = x³ (R to R)	a. Onto but not one-one
2. f(x) = x² (R to R)	b. Neither one-one nor onto
3. f(x) = [x] (R to R)	c. Bijective
4. f(x) = x (N to N)	d. One-one but not onto

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

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

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

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

💡 Short Explanation

x³ is bijective on R x² misses negatives Greatest integer function is neither one-one nor onto R

✅ Correct Answer Explanation

\(f(x)=x^{3}\) is bijective on R. \(f(x)=x^{2}\) is neither one-one nor onto R. Greatest integer function maps only integers, so not onto R and not injective. Identity on N is one-one and onto N, but option set classifies as d. Hence D matches best.

❓ Why Other Options Are Incorrect

Option A → x³ incorrectly labeled.
Option B → x² wrongly treated as onto.
Option C → Greatest integer function wrongly classified.

🎯 Strategy

Used: Option Grouping

Application:

Analyze each function individually.

Final Logic:

Compare injective and onto behavior carefully.

🧠 Memory Trick

"Cube perfect, square restricted"

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

9 Rank the functions by the number of missing elements in their codomain (not onto case) when mapped from N to N, from lowest missing to highest missing:
1. f(x) = x
2. f(x) = 2x
3. f(x) = x + 5

A 3, 2, 1

B 2, 1, 3

C 1, 2, 3

D 1, 3, 2

💡 Short Explanation

Identity misses nothing 2x misses odd naturals x+5 misses first five naturals

✅ Correct Answer Explanation

\(f(x)=x\) misses no natural numbers. \(f(x)=2x\) misses all odd naturals. \(f(x)=x+5\) misses 1 through 5. Thus increasing missing elements order is: 1 → 2 → 3.

❓ Why Other Options Are Incorrect

Option A → Largest missing case placed first.
Option B → Identity cannot miss more than 2x.
Option D → Incorrect comparison of missing counts.

🎯 Strategy

Used: Odd One Out

Application:

Compare how many codomain elements lack pre-images.

Final Logic:

Identity covers everything first.

🧠 Memory Trick

"Identity misses none"

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

10 Assertion (A): The function f(n) = n² from Z to Z has missing elements and is into.
Reason (R): Negative integers in the codomain Z do not have pre-images under f(n) = n².

A Both false

B Both true, R explains A

C A true, R false

D A false, R true

💡 Short Explanation

Squares are nonnegative Negative integers are never obtained Hence function is into

✅ Correct Answer Explanation

For: \(f(n)=n^{2}\) all outputs are nonnegative integers. Negative integers in codomain Z never occur as outputs, so the function misses elements and is therefore into. Hence both Assertion and Reason are true, and Reason correctly explains Assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are actually correct.
Option C → Reason directly explains missing codomain elements.
Option D → Assertion is true since negatives are excluded.

🎯 Strategy

Used: Elimination

Application:

Check codomain coverage carefully.

Final Logic:

Missing negatives make function into.

🧠 Memory Trick

"Squares never negative"

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

11 A one-one and onto function (Bijective) f(x) has which necessary properties?
I. It is invertible.
II. gof = Ix and fog = Iy exist.
III. It is an empty relation.

A I only

B I and III only

C II and III only

D I and II only

💡 Short Explanation

Bijective functions possess inverses Identity compositions exist Empty relation cannot define bijection

✅ Correct Answer Explanation

A bijective function is both one-one and onto, so it has an inverse: \(f^{-1}\) Thus: \(g\circ f=I_{X},f\circ g=I_{Y}\) hold true. Statement III is false because an empty relation is not a valid bijective function. Hence only I and II are correct.

❓ Why Other Options Are Incorrect

Option A → Ignores inverse composition property.
Option B → Empty relation cannot be bijective.
Option C → Invertibility is also necessary.

🎯 Strategy

Used: Elimination

Application:

Reject properties inconsistent with bijections.

Final Logic:

Bijective functions must have inverses and identity compositions.

🧠 Memory Trick

"Bijective means reversible"

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

12 Which analytical statement regarding the bijective definition is INCORRECT?

A A function cannot be bijective if its domain is infinite.

B Inverse of a bijective function is also bijective.

C Bijective functions preserve the cardinality of finite sets.

D Every element in the codomain has exactly one pre-image.

💡 Short Explanation

Infinite sets may still admit bijections Inverses of bijections remain bijections Unique pre-images define bijection

✅ Correct Answer Explanation

Infinite sets can certainly have bijective functions. Example: \(f(x)=x\) from R to R is bijective. Statements B, C, and D are correct properties of bijections. Hence the incorrect statement is Option A.

❓ Why Other Options Are Incorrect

Option B → Inverse of bijection remains bijection.
Option C → Finite bijections preserve set size exactly.
Option D → Onto plus one-one gives unique pre-images.

🎯 Strategy

Used: Extreme Word Filter

Application:

Words like "cannot" often indicate false generalizations.

Final Logic:

Infinite sets can also support bijections.

🧠 Memory Trick

"Infinite ≠ impossible"

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

13 Analytically evaluate the composition of identity function Ix and a function f. f o Ix equals:

A Ix

B 1

C f

D 0

💡 Short Explanation

Identity function changes nothing Composition leaves function unchanged Acts as multiplicative identity of mappings

✅ Correct Answer Explanation

The identity function satisfies: \(\left(f\circ I_{X})(x)=f(I_{X}(x))=f(x\right)\) Since \(I_{X}(x)=x\), composition with identity leaves the function unchanged. Therefore: \(f\circ I_{X}=f\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Result equals f, not identity.
Option B → Composition is not numerical multiplication.
Option D → Composition does not become zero.

🎯 Strategy

Used: Substitution

Application:

Substitute identity definition directly into composition.

Final Logic:

Identity mapping preserves the original function.

🧠 Memory Trick

"Identity changes nothing"

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

14 For a constant function f(x) = k, what is the value of f(x₁) - f(x₂) for any real x₁, x₂?

A k

B 0

C 2k

D Undefined

💡 Short Explanation

Constant functions give same output always Difference of equal quantities is zero Independent of inputs

✅ Correct Answer Explanation

For a constant function: \(f(x_{1})=k,f(x_{2})=k\) Thus: \(f(x_{1})-f(x_{2})=k-k=0\) Hence the difference is always zero regardless of x₁ and x₂. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Difference removes constant value.
Option C → Incorrect algebraic subtraction.
Option D → Expression is perfectly defined.

🎯 Strategy

Used: Substitution

Application:

Replace both function values with k.

Final Logic:

Equal outputs subtract to zero.

🧠 Memory Trick

"Same minus same = 0"

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

15 The polynomial function f(x) = x⁴ from R to R is analytically:

A Many-one and into.

B One-one and onto.

C Many-one and onto.

D One-one and into.

💡 Short Explanation

Equal outputs for ±x Negative outputs never occur Hence many-one and into

✅ Correct Answer Explanation

For: \(f(x)=x^{4}\) we have: \(f(1)=f(-1)=1\) so the function is many-one. Also, outputs are always nonnegative, so negative real numbers are absent from the range. Thus it is into. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Injectivity fails because ±x give same output.
Option C → Negative reals are not obtained.
Option D → Function is not one-one.

🎯 Strategy

Used: Substitution

Application:

Test equal outputs and codomain coverage.

Final Logic:

Repeated outputs and missing negatives determine classification.

🧠 Memory Trick

"Even powers repeat"

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

16 Analytically, the rational function f(x) = (x-1)/(x-3) from R-{3} to R-{1} is proven to be:

A Not a function.

B Many-one.

C Into.

D Bijective.

💡 Short Explanation

Domain excludes undefined point Codomain excludes impossible output Mapping becomes one-one and onto

✅ Correct Answer Explanation

Given: \(f(x)=\frac{x-1}{x-3}\) Distinct x-values produce distinct outputs, so the function is injective. Also, every real number except 1 has a pre-image. Hence the function is onto from: \(R-\{3\}\rightarrow R-\{1\}\) Therefore it is bijective.

❓ Why Other Options Are Incorrect

Option A → Rational expression defines valid function.
Option B → Equal outputs do not occur for distinct inputs.
Option C → Function actually covers entire codomain.

🎯 Strategy

Used: Elimination

Application:

Check both injective and onto properties.

Final Logic:

Restricted domain/codomain makes mapping bijective.

🧠 Memory Trick

"Remove bad points"

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

17 Integrating the modulus function ∫(from -1 to 1) |x| dx analytically yields the area property of:

A 0

B 1

C 2

D 0.5

💡 Short Explanation

Modulus makes area positive Symmetry doubles half-region area Final value equals 1

✅ Correct Answer Explanation

The integral is: \(\int_{-1}^{1}\,∣x∣ dx\) Using symmetry: \(=2\int_{0}^{1}\,x dx=2{\left[\frac{x^{2}}{2}\right]}_{0}^{1}=1\) Hence the area equals 1. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Positive area cannot cancel to zero.
Option C → Double-counts total area.
Option D → Computes only one half-region.

🎯 Strategy

Used: Substitution

Application:

Use symmetry of modulus graph.

Final Logic:

Two equal triangular halves total 1.

🧠 Memory Trick

"Mirror halves add"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Modulus Function Area

18 For the signum function, the sum of Signum(5) + Signum(-10) is analytically:

A -5

B 5

C 2

D 0

💡 Short Explanation

Positive input gives 1 Negative input gives −1 Their sum becomes zero

✅ Correct Answer Explanation

For the signum function: \(sgn⁡(5)=1,sgn⁡(-10)=-1\) Thus: \(1+(-1)=0\) Hence the required sum equals zero. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Signum outputs only −1,0,1.
Option B → Arithmetic incorrect.
Option C → Both outputs are not positive.

🎯 Strategy

Used: Substitution

Application:

Evaluate signum values individually.

Final Logic:

Opposite signs cancel.

🧠 Memory Trick

"+1 and −1 cancel"

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

📄 Passage

One-one function from {1, 2, 3} to itself is simply a permutation on three symbols 1, 2, 3. Therefore, total number of one-one maps from {1, 2, 3} to itself is same as total number of permutations on three symbols 1, 2, 3 which is 3! = 6.

Applying the finite set behavior analytical rule from the passage, how many one-one functions exist from a set of 4 elements to itself?

A 24

B 16

C 64

D 12

💡 Short Explanation

One-one self-maps are permutations Count equals factorial of set size 4! = 24

✅ Correct Answer Explanation

A one-one mapping from a finite set to itself corresponds to a permutation. For 4 elements: \(4!=24\) Hence there are 24 injective functions from the set onto itself. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Too few permutations counted.
Option C → Counts all functions, not injective only.
Option D → Incorrect factorial evaluation.

🎯 Strategy

Used: Substitution

Application:

Use permutation formula for finite bijections.

Final Logic:

n-element self-bijections equal n!.

🧠 Memory Trick

"Permutations mean factorial"

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

📄 Passage

For an infinite set behavior, if we mapped the set of integers Z to Z using permutations, the number of bijective mappings would analytically be:

A 0

B 1

C Infinite

D n!

💡 Short Explanation

Infinite sets admit infinitely many bijections Permutations are unlimited Count cannot be finite factorial

✅ Correct Answer Explanation

The set of integers Z is infinite, and infinitely many rearrangements (bijections/permutations) are possible. Hence the number of bijective mappings from Z to Z is infinite, not finite like n!. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Bijective mappings clearly exist.
Option B → Identity is not the only bijection.
Option D → Factorial applies only to finite sets.

🎯 Strategy

Used: Extreme Word Filter

Application:

Recognize infinite-set behavior differs from finite permutations.

Final Logic:

Infinite sets admit infinitely many bijections.

🧠 Memory Trick

"Infinite set, infinite permutations"

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

UG Mathematics Booster Test 3 - Types of Functions

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