UG Mathematics Booster Test 2 - Advanced Concepts and Exercises -

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UG Mathematics Booster Test 2 - Advanced Concepts and Exercises

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 2 - Advanced Concepts and Exercises

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QUESTION 1 OF 20

Which of the following is INCORRECT about the empty relation \(ϕ\)on a non-empty set \(A\)?

Answer Review

1 Which of the following is INCORRECT about the empty relation \(ϕ\)on a non-empty set \(A\)?

A It is symmetric

B It is reflexive

C It is transitive

D It is a subset of \(A\times A\)

💡 Short Explanation

Empty relation has no ordered pairs Reflexive relation needs all \(\left(a,\ a\right)\)pairs Empty relation cannot satisfy reflexivity

✅ Correct Answer Explanation

For a relation to be reflexive, every element \(a\in A\) must satisfy: \((a,a)\in R\) The empty relation contains no ordered pairs. Hence it cannot be reflexive on a non-empty set. However, it is symmetric and transitive vacuously and is always a subset of \(A\times A\).

❓ Why Other Options Are Incorrect

Option A → Empty relation is symmetric because no violating pair exists.
Option C → It is transitive vacuously since no chains exist.
Option D → Every relation on \(A\) must be a subset of \(A\times A\).

🎯 Strategy

Used: Elimination

Application:

Check definition of reflexivity directly.

Final Logic:

Missing all self-pairs means not reflexive.

🧠 Memory Trick

"Reflexive needs mirrors"

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

2 Let \(L\) be the set of lines in a plane. \(R=\{(L_{1},L_{2}):L_{1} is perpendicular to L_{2}\}\). Which is logically true?

A \(R\) is symmetric but not reflexive.

B \(R\) is reflexive and symmetric.

C \(R\) is transitive.

D \(R\) is an equivalence relation.

💡 Short Explanation

Perpendicularity reverses naturally No line is perpendicular to itself Hence relation is symmetric but not reflexive

✅ Correct Answer Explanation

If: \(L_{1}⊥L_{2}\) then automatically: \(L_{2}⊥L_{1}\) so the relation is symmetric. However, no line is perpendicular to itself, so reflexivity fails. Therefore it cannot be an equivalence relation.

❓ Why Other Options Are Incorrect

Option B → Reflexivity fails because \(L⊥̸L\).
Option C → Perpendicularity is not transitive.
Option D → Equivalence relations must be reflexive and transitive too.

🎯 Strategy

Used: Elimination

Application:

Test symmetry and reflexivity separately.

Final Logic:

Perpendicularity reverses but does not self-apply.

🧠 Memory Trick

"Perpendicular flips, never reflects"

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

3 Data is recorded on consecutive days. Let \(R\) be a relation where day \(xRy\) if the moving average of day \(x\leq\) moving average of day \(y\). Is \(R\) transitive?

A No, because averages fluctuate.

B No, it is symmetric only.

C Yes, but only for positive data.

D Yes, because \(a\leq b\) and \(b\leq c\Rightarrow a\leq c\).

💡 Short Explanation

Transitivity follows inequality property Less-than-or-equal relation chains naturally Therefore relation is transitive

✅ Correct Answer Explanation

The relation depends on: \(a\leq b\) Since real-number inequalities satisfy: \(a\leq b, b\leq c\Rightarrow a\leq c\) the relation is transitive regardless of data type or fluctuations.

❓ Why Other Options Are Incorrect

Option A → Fluctuations do not affect transitivity logic.
Option B → Relation is not symmetric in general.
Option C → Positivity is unnecessary for transitivity.

🎯 Strategy

Used: Substitution

Application:

Replace moving averages with real numbers.

Final Logic:

Inequality relations are inherently transitive.

🧠 Memory Trick

"≤ always chains"

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

4 Which relations provide a mathematical counterexample to transitivity?
I. \(R=\{(x,y):y=x+1\}\)
II. \(R=\{(x,y):x\leq y\}\)
III. \(R=\{(x,y):x=y\}\)

A II only

B III only

C I only

D I and II

💡 Short Explanation

Transitivity requires chaining \(y=x+1\) fails after repeated application Equality and ≤ remain transitive

✅ Correct Answer Explanation

For relation I: \(xRy, yRz\) means: \(y=x+1, z=y+1=x+2\) but \(z\neq x+1\). Hence transitivity fails. Relations II and III satisfy transitivity naturally.

❓ Why Other Options Are Incorrect

Option A → \(\leq\) is transitive.
Option B → Equality relation is transitive.
Option D → Relation II should not be included.

🎯 Strategy

Used: Counterexample Method

Application:

Test chained ordered pairs.

Final Logic:

"+1 relation" breaks after second step.

🧠 Memory Trick

"One-step jump breaks chaining"

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

5 \(R\) on \(\left\{1,\ 2,\ 3,\ 4,\ 5\right\}\)is given by \(R=\{(a,b):∣a-b∣ is even\}\). Is \(R\) an equivalence relation?

A No, it is not reflexive

B Yes, it is an equivalence relation

C No, it is not symmetric

D No, it is not transitive

💡 Short Explanation

Difference of identical numbers is zero Evenness property is symmetric Sum of even differences remains even

✅ Correct Answer Explanation

The relation satisfies: Reflexive: \(∣a-a∣=0\), which is even Symmetric: \(∣a-b∣=∣b-a∣\) Transitive: even + even = even Therefore the relation is reflexive, symmetric, and transitive, making it an equivalence relation.

❓ Why Other Options Are Incorrect

Option A → Reflexivity clearly holds.
Option C → Absolute value guarantees symmetry.
Option D → Even differences preserve transitivity.

🎯 Strategy

Used: Option Grouping

Application:

Check all three equivalence properties.

Final Logic:

All required properties are satisfied.

🧠 Memory Trick

"Even differences stay together"

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

6 For the relation \(R=\{(a,b):∣a-b∣ is even\}\)on \(\left\{1,\ 2,\ 3,\ 4,\ 5\right\}\), what is the specific equivalence class containing 1?

A \(\left\{1,\ 3,\ 5\right\}\)

B \(\left\{2,\ 4\right\}\)

C \(\left\{1,\ 2,\ 3\right\}\)

D \(\left\{1\right\}\)

💡 Short Explanation

Difference with 1 must be even Odd numbers differ by even values Hence class contains all odd elements

✅ Correct Answer Explanation

Numbers related to 1 satisfy: \(∣a-1∣ is even\) This occurs for: \(a=1,3,5\) since: \(∣1-1∣=0, ∣3-1∣=2, ∣5-1∣=4\) all even.

❓ Why Other Options Are Incorrect

Option B → Even numbers differ from 1 by odd values.
Option C → Contains 2, which violates condition.
Option D → Ignores other valid odd members.

🎯 Strategy

Used: Substitution

Application:

Test each set element with relation condition.

Final Logic:

Odd numbers belong to same equivalence class.

🧠 Memory Trick

"Odd stays with odd"

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

7 Let \(f(x)=\int_{0}^{x}\,t dt=x^{2}/2\) defined from \(R\rightarrow R\). Is \(f\) an injective function?

A Yes, because the integral is unique

B Yes, because it's a continuous polynomial

C No, because \(f(0)=0\)

D No, because \(f(2)=f(-2)=2\)

💡 Short Explanation

Injective functions give distinct outputs Squaring removes sign information Positive and negative inputs can match

✅ Correct Answer Explanation

Since: \(f(x)=\frac{x^{2}}{2}\) we get: \(f(2)=\frac{4}{2}=2,f(-2)=\frac{4}{2}=2\) Different inputs produce the same output, so the function is not injective.

❓ Why Other Options Are Incorrect

Option A → Uniqueness of integration is irrelevant.
Option B → Continuity does not imply injectivity.
Option C → Single point gives no injectivity failure.

🎯 Strategy

Used: Counterexample Method

Application:

Find unequal inputs with equal outputs.

Final Logic:

Equal squares break injectivity.

🧠 Memory Trick

"Square hides sign"

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

8 Let \(f:R\rightarrow R\) be defined by \(f(x)=[x]\)(Greatest Integer Function). Is \(f\) surjective?

A Yes, because it maps to all integers.

B Yes, because its domain is \(R\).

C No, because fractional values in the co-domain have no pre-image.

D No, because \(f(1.5)=2\).

💡 Short Explanation

GIF outputs only integers Codomain contains all real numbers Fractional reals lack pre-images

✅ Correct Answer Explanation

The greatest integer function maps every real number to an integer. Hence values like: \(1.5, 2.7\) in the codomain \(R\) are never achieved. Therefore the function is not surjective onto \(R\).

❓ Why Other Options Are Incorrect

Option A → Surjective onto integers, not all reals.
Option B → Large domain does not guarantee surjectivity.
Option D → \([1.5]=1\), not 2.

🎯 Strategy

Used: Elimination

Application:

Find real numbers without pre-images.

Final Logic:

Missing fractional outputs imply not onto.

🧠 Memory Trick

"GIF gives integers only"

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

9 Arrange the logical steps to verify \(f(x)=x/(x-1)\)from \(R-\{1\}\rightarrow R-\{1\}\)is bijective:
I. Show \(f(x_{1})=f(x_{2})\Rightarrow x_{1}=x_{2}\)
II. State \(f\) is bijective
III. Assume an arbitrary \(y\in R-\{1\}\)
IV. Find \(x=y/(y-1)\)and show \(f(x)=y\)

A I, II, III, IV

B I, III, IV, II

C III, IV, I, II

D IV, I, III, II

💡 Short Explanation

First prove injectivity Then prove surjectivity Finally conclude bijection

✅ Correct Answer Explanation

Logical proof order: 1. Prove injectivity using equal outputs 2. Assume arbitrary codomain value 3. Construct pre-image \(x=y/(y-1)\) 4. Conclude function is bijective Thus correct order: \(I, III, IV, II\)

❓ Why Other Options Are Incorrect

Option A → Conclusion appears too early.
Option C → Injectivity should be proved first.
Option D → Surjectivity step cannot precede assumption.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange proof in mathematical sequence.

Final Logic:

Proof precedes final conclusion.

🧠 Memory Trick

"Injective → onto → bijective"

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

10 Consider \(f:N\rightarrow N\) defined by
\(f(n)=\frac{n+1}{2} if n is odd, and \frac{n}{2} if n is even\)
What is its nature?

A Onto but not one-one

B One-one and onto

C Neither one-one nor onto

D One-one but not onto

💡 Short Explanation

Different inputs may share outputs Every natural number still appears Hence onto but not injective

✅ Correct Answer Explanation

Observe: \(f(1)=1,f(2)=1\) so function is not one-one. But every natural number \(k\) has preimages: \(2k-1, 2k\) Hence every natural number is achieved, making the function onto.

❓ Why Other Options Are Incorrect

Option B → Injectivity fails immediately.
Option C → Surjectivity actually holds.
Option D → Function is onto as every natural number appears.

🎯 Strategy

Used: Counterexample Method

Application:

Check injectivity and surjectivity separately.

Final Logic:

Repeated outputs destroy one-one property.

🧠 Memory Trick

"Odd-even pair gives same value"

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

11 If set \(A\) has 3 elements and set \(B\) has 4 elements, how many unique mathematical functions exist from \(A\rightarrow B\)?

A 12

B 7

C 81

D 64

💡 Short Explanation

Each element of \(A\) has 4 choices Total mappings multiply independently Hence total functions \(=4^{3}\)

✅ Correct Answer Explanation

If \(∣A∣=3\) and \(∣B∣=4\), then each element of \(A\) can map to any of 4 elements in \(B\). Thus total functions are: \(4\times 4\times 4=4^{3}=64\) Hence the correct number of possible functions is 64.

❓ Why Other Options Are Incorrect

Option A → 12 comes from multiplication without full combinations.
Option B → 7 has no combinatorial relevance.
Option C → 81 equals \(3^{4}\), not \(4^{3}\).

🎯 Strategy

Used: Substitution

Application:

Use formula \(n(B)^{n(A)}\).

Final Logic:

4 choices for each of 3 inputs gives \(4^{3}\).

🧠 Memory Trick

"Functions = outputs raised to inputs"

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

12 How many purely bijective functions exist from a set of 4 elements to itself?

A 4

B 16

C 24

D 256

💡 Short Explanation

Bijective functions are permutations Number of permutations of 4 elements is \(4!\) \(4!=24\)

✅ Correct Answer Explanation

A bijection from a set to itself rearranges elements uniquely. Therefore counting bijections equals counting permutations: \(4!=4\times 3\times 2\times 1=24\) Hence there are 24 bijective functions.

❓ Why Other Options Are Incorrect

Option A → Only counts single-step mappings.
Option B → Not permutation count.
Option D → Equals \(4^{4}\), total functions, not bijections.

🎯 Strategy

Used: Option Grouping

Application:

Recognize bijections as permutations.

Final Logic:

Bijective maps on 4 elements equal \(4!\).

🧠 Memory Trick

"Bijective means permutation"

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

13 Let \(A=\{1,2\}\). How many reflexive relations are possible on set \(A\)?

A 2

B 4

C 8

D 16

💡 Short Explanation

Reflexive relations require all diagonal pairs \(\left(1,\ 1\right)\)and \(\left(2,\ 2\right)\)must exist Remaining pairs vary freely

✅ Correct Answer Explanation

For set \(A=\{1,2\}\), \(A\times A=\{(1,1),(1,2),(2,1),(2,2)\}\) Reflexivity forces inclusion of: \(\left(1,1),(2,2\right)\) The remaining two pairs may either appear or not. Thus: \(2^{2}=4\) reflexive relations exist.

❓ Why Other Options Are Incorrect

Option A → Under-counts possibilities.
Option C → Assumes three free choices.
Option D → Counts all relations on \(A\).

🎯 Strategy

Used: Substitution

Application:

Fix required reflexive pairs, vary remaining pairs.

Final Logic:

Two unrestricted pairs give \(2^{2}\).

🧠 Memory Trick

"Fix diagonal, vary others"

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

14 Assertion (A): The total number of equivalence relations on \(\left\{1,\ 2,\ 3\right\}\)containing \(\left(1,\ 2\right)\)and \(\left(2,\ 1\right)\)is 1.
Reason (R): We cannot add \(\left(2,\ 3\right)\)without adding \(\left(3,\ 2\right)\), \(\left(1,\ 3\right)\), and \(\left(3,\ 1\right)\).

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

More than one equivalence relation exists Symmetry and transitivity force extra pairs Hence reason is true but assertion fails

✅ Correct Answer Explanation

The assertion is false because multiple equivalence relations can contain: \(\left(1,2),(2,1\right)\) For example: Partition \(\left\{\{1,2\},\{3\}\right\}\) Partition \(\left\{\{1,2,3\}\right\}\) Both generate valid equivalence relations. The reason is true since adding \(\left(2,\ 3\right)\)forces further symmetric and transitive pairs.

❓ Why Other Options Are Incorrect

Option A → Reason is actually true.
Option B → Assertion is false.
Option C → Assertion itself is incorrect.

🎯 Strategy

Used: Elimination

Application:

Test assertion and reason separately.

Final Logic:

Multiple valid equivalence relations exist.

🧠 Memory Trick

"Equivalence comes from partitions"

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

15 If \(f(x)=∣x∣\)and \(g(x)=sin⁡(x)\), what is the composite evaluation of \(f(g(x))\)?

A \(sin⁡(∣x∣)\)

B \(-sin⁡(x)\)

C \(∣x∣sin⁡(x)\)

D \(∣sin⁡(x)∣\)

💡 Short Explanation

Composition means outer after inner First apply sine function Then modulus acts on result

✅ Correct Answer Explanation

Since: \(f(x)=∣x∣,g(x)=sin⁡x\) then: \(f(g(x))=f(sin⁡x)=∣sin⁡x∣\) The modulus function takes the absolute value of the sine output.

❓ Why Other Options Are Incorrect

Option A → Represents \(sin⁡(∣x∣)\), different composition.
Option B → Negation is unrelated.
Option C → Product of functions, not composition.

🎯 Strategy

Used: Substitution

Application:

Replace input of outer function with inner output.

Final Logic:

Outer modulus acts on sine result.

🧠 Memory Trick

"Outer waits for inner"

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

16 Match the composite functions for \(f(x)=2x, g(x)=x+3\):

List I	List II
1. \(f(g(x))\)	a. \(4x\)
2. \(g(f(x))\)	b. \(2x+6\)
3. \(f(f(x))\)	c. \(2x+3\)

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

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

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

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

💡 Short Explanation

Substitute carefully in compositions Evaluate inner function first Self-composition doubles repeatedly

✅ Correct Answer Explanation

Compute each: \(f(g(x))=f(x+3)=2(x+3)=2x+6g(f(x))=g(2x)=2x+3f(f(x))=f(2x)=4x\) Hence matching is: \(1-b, 2-c, 3-a\)

❓ Why Other Options Are Incorrect

Option A → Swaps second and third mappings.
Option B → Incorrect first composition.
Option D → Incorrect arrangement of outputs.

🎯 Strategy

Used: Substitution

Application:

Apply inner function first in every composition.

Final Logic:

Direct substitution gives exact matching.

🧠 Memory Trick

"Inside first, outside next"

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

17 Let \(f:R\rightarrow R\) be \(f(x)=10x+7\). If its inverse \(f^{-1}(y)=ay+b\), find the numerical value of \(a+b\). (Case/Numerical)

A \(1/10\)

B \(-6/10\)

C \(7/10\)

D \(10\)

💡 Short Explanation

Solve linear equation for inverse Express x in terms of y Add coefficients \(a\) and \(b\)

✅ Correct Answer Explanation

Given: \(y=10x+7\) solve for \(x\): \(x=\frac{y-7}{10}=\frac{1}{10}y-\frac{7}{10}\) Thus: \(a=\frac{1}{10},b=-\frac{7}{10}\) So: \(a+b=-\frac{6}{10}\)

❓ Why Other Options Are Incorrect

Option A → Only coefficient \(a\).
Option C → Incorrect sign handling.
Option D → Original slope, not inverse sum.

🎯 Strategy

Used: Substitution

Application:

Interchange variables and isolate \(x\).

Final Logic:

Inverse gives coefficients directly.

🧠 Memory Trick

"Swap and solve"

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

18 If \(f\) is a strictly invertible function, then mathematically \({\left(f^{-1}\right)}^{-1}\)simplifies to: (MCQ)

A \(f\)

B \(f^{-2}\)

C \(I\)

D \(1/f\)

💡 Short Explanation

Inverse reverses mapping Applying inverse twice restores original Hence double inverse equals original function

✅ Correct Answer Explanation

The inverse function reverses the action of \(f\). Taking inverse again restores the original mapping: \(\left(f^{-1})^{-1},\ f\right.\) This is analogous to reversing a reversal.

❓ Why Other Options Are Incorrect

Option B → Exponent notation is invalid here.
Option C → Identity occurs in composition, not inversion.
Option D → Reciprocal differs from inverse function.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use fundamental inverse-function property.

Final Logic:

Inverse of inverse restores original function.

🧠 Memory Trick

"Undo the undo"

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

📄 Passage

The composition of \(f:A\rightarrow B\) and \(g:B\rightarrow C\) is \(gof:A\rightarrow C\) defined as \(gof(x)=g(f(x))\). A function \(f:X\rightarrow Y\) is invertible if there exists \(g:Y\rightarrow X\) such that \(gof=I_{x}\)and \(fog=I_{y}\).

If \(f:\{2,3,4\}\rightarrow \{3,4,5\}\)has \(f(2)=3, f(3)=4, f(4)=5\) and \(g:\{3,4,5\}\rightarrow \{7,11\}\)has \(g(3)=7, g(4)=g(5)=11\). What is the value of \(gof(3)\)?

A 3

B 4

C 7

D 11

💡 Short Explanation

First apply \(f\) Then apply \(g\) Composition follows inner-to-outer order

✅ Correct Answer Explanation

Compute stepwise: \(f(3)=4\) Now apply \(g\): \(g(4)=11\) Therefore: \((gof)(3)=g(f(3))=g(4)=11\)

❓ Why Other Options Are Incorrect

Option A → Original input value only.
Option B → Intermediate function value.
Option C → Would occur if \(f(3)=3\).

🎯 Strategy

Used: Substitution

Application:

Evaluate inner function before outer function.

Final Logic:

Composition means "output becomes next input."

🧠 Memory Trick

"Inside first, outside second"

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

📄 Passage

If a function \(h:X\rightarrow Y\) is invertible, it must unconditionally be: (MCQ)

A Only one-one

B Only onto

C Both one-one and onto

D Neither one-one nor onto

💡 Short Explanation

Invertible functions require uniqueness Every output must be covered Hence bijectivity is necessary

✅ Correct Answer Explanation

An invertible function must satisfy: One-one → inverse assigns unique inputs Onto → every codomain element has pre-image Therefore invertibility requires bijection, meaning both injective and surjective.

❓ Why Other Options Are Incorrect

Option A → One-one alone is insufficient.
Option B → Onto alone cannot guarantee inverse.
Option D → Opposite of invertibility conditions.

🎯 Strategy

Used: Option Grouping

Application:

Combine injective and surjective conditions.

Final Logic:

Inverse exists only for bijections.

🧠 Memory Trick

"Invertible = Bijective"

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

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