UG Mathematics Booster Test 3 - Composition and Invertibility -

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UG Mathematics Booster Test 3 - Composition and Invertibility

⏱️ 30 Minutes

❓ 20 Questions

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UG Mathematics Booster Test 3 - Composition and Invertibility

1 / 20

QUESTION 1 OF 20

Let \(f\) and \(g\) be two mappings. If the composite function \(f\circ g\) is one-one, what can be concluded?

Answer Review

1 Let \(f\) and \(g\) be two mappings. If the composite function \(f\circ g\) is one-one, what can be concluded?

A \(f\) must be one-one

B Both \(f\) and \(g\) must be onto

C \(g\) must be one-one

D \(f\) must be onto

💡 Short Explanation

Injectivity of composition depends on inner mapping Distinct inputs must remain distinct through \(g\) Hence \(g\) must be injective

✅ Correct Answer Explanation

If \(f\circ g\) is one-one, then different inputs cannot produce the same intermediate value through \(g\). Suppose \(g(x_{1})=g(x_{2})\). Then: \(f(g(x_{1}))=f(g(x_{2}))\) which contradicts injectivity unless \(x_{1}=x_{2}\). Thus \(g\) must necessarily be one-one.

❓ Why Other Options Are Incorrect

Option A → \(f\) need not be injective on its entire domain.
Option B → Onto property is unrelated here.
Option D → Surjectivity of \(f\) is unnecessary for injectivity of composition.

🎯 Strategy

Used: Elimination

Application:

Check which property directly preserves distinct inputs.

Final Logic:

Inner function must avoid collapsing distinct elements.

🧠 Memory Trick

"Injective composition protects inputs"

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

2 Identify the correct statements regarding the domain of \(f\circ g\), given functions \(f\) and \(g\):
(i) \(x\in\) Domain of \(g\)
(ii) \(g(x)\in\) Domain of \(f\)
(iii) \(x\in\) Range of \(f\)

A (i) only

B (ii) only

C (i) and (iii) only

D (i) and (ii) only

💡 Short Explanation

Input must belong to domain of \(g\) Output of \(g\) must enter domain of \(f\) Range of \(f\) is irrelevant

✅ Correct Answer Explanation

For composition: \(\left(f\circ g)(x)=f(g(x)\right)\) to exist, \(x\) must first belong to the domain of \(g\). Then the value \(g(x)\)must lie inside the domain of \(f\). Thus both conditions (i) and (ii) are required.

❓ Why Other Options Are Incorrect

Option A → Ignores compatibility with domain of \(f\).
Option B → Input must first belong to domain of \(g\).
Option C → Range of \(f\) has no role in defining composition domain.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow the order of evaluation in composition.

Final Logic:

Input enters \(g\) first, then \(f\).

🧠 Memory Trick

"First \(g\), then \(f\)"

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

3 Let \(f\) and \(g\) be functions. Which statement is incorrect?

A \(f\circ g\) is defined for all \(x\)

B \(g\circ f\) is defined for all \(x\)

C Domain of \(f\circ g\) can be all real numbers

D Domain may be restricted depending on \(g(x)\)

💡 Short Explanation

Composition may fail for some inputs Domain depends on compatibility Not always defined universally

✅ Correct Answer Explanation

The composition \(f\circ g\) exists only when: \(g(x)\in Domain of f\) for the chosen input \(x\). Therefore it is incorrect to claim composition is always defined for all values of \(x\).

❓ Why Other Options Are Incorrect

Option B → Can hold for suitable functions.
Option C → Some compositions indeed have full real domains.
Option D → Correctly describes domain restrictions.

🎯 Strategy

Used: Extreme Word Filter

Application:

The phrase "for all \(x\)" signals overgeneralization.

Final Logic:

Composition depends on domain compatibility.

🧠 Memory Trick

"Composition needs permission"

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

4 Question (Match the Following):

List I	List II
1. Minimum of \(f\circ g\)	a. 0
2. Minimum of \(g\circ f\)	b. -1
3. Value of \(f(g(0))\)	c. 0
4. Value of \(g(f(0))\)	d. -1

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

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

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

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

💡 Short Explanation

Matching follows composite evaluation Minimum values differ by order Composition is order-sensitive

✅ Correct Answer Explanation

The correct arrangement aligns each composition and evaluated value with its corresponding minimum or numerical result. Since \(f\circ g\neq g\circ f\) generally, their outputs and minimum values differ accordingly.

❓ Why Other Options Are Incorrect

Option A → Interchanges evaluated values incorrectly.
Option C → Swaps minima and outputs inconsistently.
Option D → Composition order is mismatched.

🎯 Strategy

Used: Option Grouping

Application:

Track each composition carefully and compare outputs.

Final Logic:

Composition order determines mapping outcomes.

🧠 Memory Trick

"Order changes answers"

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

5 (Case/Numerical):
Let \(f(x)=ax+b\) and \(g(x)=cx+d\). For \(f\circ g=g\circ f\), what condition must hold?

A \(a=c\)

B \(b=d\)

C \(ad=bc\)

D \(a=c\) and \(b=d\)

💡 Short Explanation

Equal compositions require identical coefficients Compare expanded forms Coefficients must match termwise

✅ Correct Answer Explanation

Compute: \(f(g(x))=a(cx+d)+b=acx+ad+b\) and \(g(f(x))=c(ax+b)+d=acx+bc+d\) For equality for all \(x\): \(ad+b=bc+d\) The strongest matching condition among options is \(a=c\) and \(b=d\).

❓ Why Other Options Are Incorrect

Option A → Equal slopes alone are insufficient.
Option B → Equal constants alone do not ensure commutativity.
Option C → Only partial coefficient condition.

🎯 Strategy

Used: Substitution

Application:

Expand both compositions algebraically.

Final Logic:

Equal functions require coefficient equality.

🧠 Memory Trick

"Expand and compare"

📚 Class XI – NCERT Mathematics – Chapter: Composite Functions – Topic: Non-Commutativity

6 When do regions defined by \(f(g(x))\)and \(g(f(x))\)coincide?

A Both functions are zero

B Functions are inverses

C Function is even

D Functions commute

💡 Short Explanation

Same composition outputs imply commutativity Graphs overlap only when compositions equal Order independence is required

✅ Correct Answer Explanation

The regions defined by \(f(g(x))\)and \(g(f(x))\)coincide precisely when: \(f\circ g=g\circ f\) This property is called commutativity of composition.

❓ Why Other Options Are Incorrect

Option A → Zero functions are only special cases.
Option B → Inverse functions need not produce identical compositions.
Option C → Evenness is unrelated.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret equality of regions as equality of compositions.

Final Logic:

Equal composite outputs imply commuting functions.

🧠 Memory Trick

"Same graph = commute"

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

7 A sequence undergoes a moving average transformation. Is this transformation globally invertible to uniquely recover the original sequence?

A Yes, each mean is unique

B No, reconstruction requires an initial boundary value

C Yes, using discrete differentiation

D No, moving averages are not functions

💡 Short Explanation

Moving averages lose information Boundary data is needed Exact recovery is not automatic

✅ Correct Answer Explanation

A moving average combines neighboring values and removes some original information. Different sequences may produce the same averages unless an additional initial condition or boundary value is known. Hence global invertibility fails.

❓ Why Other Options Are Incorrect

Option A → Same averages can arise from different sequences.
Option C → Differentiation alone cannot fully recover data.
Option D → Moving averages are valid functions.

🎯 Strategy

Used: Elimination

Application:

Check whether unique reconstruction is guaranteed.

Final Logic:

Loss of information prevents perfect inversion.

🧠 Memory Trick

"Averages blur data"

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

8 Let \(A\) be a finite set. A function is chosen randomly from all possible functions \(A\rightarrow A\). What is the probability that it is invertible?

A 0

B 1

C \(n!/n^{n}\)(for \(∣A∣=n\))

D \(1/n\)

💡 Short Explanation

Invertible functions are permutations Total permutations equal \(n!\) Total functions equal \(n^{n}\)

✅ Correct Answer Explanation

For a finite set with \(n\) elements: • Total functions \(=n^{n}\) • Invertible functions \(=n!\) Hence probability: \(\frac{n!}{n^{n}}\)

❓ Why Other Options Are Incorrect

Option A → Invertible functions certainly exist.
Option B → Not every function is invertible.
Option D → Formula is incomplete and incorrect.

🎯 Strategy

Used: Substitution

Application:

Use counting principle for functions and permutations.

Final Logic:

Probability equals favorable outcomes over total outcomes.

🧠 Memory Trick

"Permutations are bijections"

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

9 A 2D linear transformation is given by
\(T(x,y)=(ax+by, cx+dy)\)
with non-zero determinant. What is the inverse transformation?

A \(T^{-1}=\frac{1}{ad-bc}\left[\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}\right]\)

B \(T^{-1}=\left[\begin{pmatrix}a & b\\ c & d\end{pmatrix}\right]\)

C \(T^{-1}=\frac{1}{a+d}\times\) identity

D No inverse exists

💡 Short Explanation

Matrix inverse formula applies Determinant must be non-zero Adjoint divided by determinant gives inverse

✅ Correct Answer Explanation

For matrix: \(A=\left[\begin{pmatrix}a & b\\ c & d\end{pmatrix}\right]\) if determinant \(ad-bc\neq 0\), then inverse exists and equals: \(A^{-1}=\frac{1}{ad-bc}\left[\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}\right]\)

❓ Why Other Options Are Incorrect

Option B → Original matrix is not inverse generally.
Option C → Incorrect inverse formula.
Option D → Non-zero determinant guarantees invertibility.

🎯 Strategy

Used: Formula Recall

Application:

Apply standard inverse matrix formula.

Final Logic:

Non-zero determinant ensures inverse transformation exists.

🧠 Memory Trick

"Swap-diagonal, change signs"

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Inverse of Matrix

10 What is the relationship between the area under a function \(f\) and the area under its inverse \(f^{-1}\)over corresponding intervals?

A Area under \(f\) is always greater

B Area under \(f^{-1}\)is zero

C Both areas are equal due to symmetry about \(y=x\)

D Total area is zero

💡 Short Explanation

Function and inverse reflect about \(y=x\) Corresponding regions are symmetric Areas remain equal

✅ Correct Answer Explanation

Graphs of a function and its inverse are mirror images about the line: \(y=x\) Hence corresponding enclosed regions have equal area due to geometric symmetry.

❓ Why Other Options Are Incorrect

Option A → No general inequality exists.
Option B → Inverse area is not necessarily zero.
Option D → Symmetry does not imply cancellation.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use geometric interpretation of inverse functions.

Final Logic:

Reflection symmetry preserves area.

🧠 Memory Trick

"Inverse mirrors across \(y=x\)"

📚 Class XI – NCERT Mathematics – Chapter: Inverse Functions – Topic: Graph of Inverse Functions

📄 Passage

If fis invertible, then f∘f^(-1)=Iand f^(-1)∘f=I. Invertible functions must be bijective (one-one and onto). The identity function behaves like the multiplicative identity under composition.

Using the property of an odd invertible function \(f(-x)=-f(x)\), evaluate:
\(\int_{-a}^{a}\,f(x) dx\)

A 0

B 1

C 2

D 4

💡 Short Explanation

Odd functions show symmetry about origin Positive and negative areas cancel Integral over symmetric interval becomes zero

✅ Correct Answer Explanation

For an odd function: \(f(-x)=-f(x)\) The graph is symmetric about the origin. Over the interval \(\left[-a,a\right]\), the area above the x-axis equals the area below the x-axis in magnitude. Hence both cancel and: \(\int_{-a}^{a}\,f(x) dx=0\) Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → No unit-area condition is given.
Option C → Odd symmetry causes cancellation, not doubling.
Option D → No constant positive accumulation occurs.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recognize the standard odd-function integral property.

Final Logic:

Odd functions integrate to zero over symmetric limits.

🧠 Memory Trick

"Odd + symmetric limits = 0"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Properties of Definite Integrals

📄 Passage

If fis invertible, then f∘f^(-1)=Iand f^(-1)∘f=I. Invertible functions must be bijective (one-one and onto). The identity function behaves like the multiplicative identity under composition.

What structure is formed by all invertible functions from a set to itself under composition?

A A group

B A ring

C A field

D A vector space

💡 Short Explanation

Composition acts as operation Identity function exists Every invertible function has inverse

✅ Correct Answer Explanation

Invertible functions under composition satisfy all group properties: • Closure under composition • Associativity • Identity mapping exists • Every element has an inverse Thus invertible functions from a set to itself form a group under composition.

❓ Why Other Options Are Incorrect

Option B → Rings require two operations.
Option C → Fields need multiplication and addition inverses.
Option D → Vector spaces require scalar multiplication structure.

🎯 Strategy

Used: Option Grouping

Application:

Match algebraic properties with standard structures.

Final Logic:

Identity and inverses under composition imply group structure.

🧠 Memory Trick

"Composition + inverse = group"

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

📄 Passage

If fis invertible, then f∘f^(-1)=Iand f^(-1)∘f=I. Invertible functions must be bijective (one-one and onto). The identity function behaves like the multiplicative identity under composition.

Which function over \(R\) fails the one-one condition?

A \(f(x)=x^{3}\)

B \(f(x)=e^{x}\)

C \(f(x)=x\)

D Greatest Integer Function

💡 Short Explanation

Many inputs share same output Function is not injective Horizontal line intersects repeatedly

✅ Correct Answer Explanation

The Greatest Integer Function maps all real numbers within an interval like \([1,2)\)to the same integer 1. Hence different inputs give identical outputs, violating injectivity. Functions \(x^{3}\), \(e^{x}\), and \(x\) are strictly monotonic and therefore one-one.

❓ Why Other Options Are Incorrect

Option A → \(x^{3}\)is strictly increasing on \(R\).
Option B → Exponential functions are injective.
Option C → Identity function is perfectly one-one.

🎯 Strategy

Used: Elimination

Application:

Check whether distinct inputs can share outputs.

Final Logic:

Only GIF maps intervals to single values.

🧠 Memory Trick

"Step functions fail one-one"

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

14 Assertion (A): A polynomial function is onto.
Reason (R): Its derivative is non-negative.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Not all polynomial functions are onto Derivative need not be non-negative Both statements fail generally

✅ Correct Answer Explanation

The assertion is false because not every polynomial is onto over \(R\rightarrow R\). Example: \(f(x)=x^{2}\) does not produce negative outputs. The reason is also false because polynomial derivatives may be positive, negative, or changing sign. Hence both statements are false.

❓ Why Other Options Are Incorrect

Option B → Assertion fails for even-degree polynomials.
Option C → Derivative condition is incorrect generally.
Option D → Reason is mathematically false.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "all" in polynomial behavior makes statement suspicious.

Final Logic:

Counterexample \(x^{2}\)disproves the assertion.

🧠 Memory Trick

"\(x^{2}\) misses negatives"

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

15 Arrange steps to find inverse of an exponential function:
1. Substitute \(y=a^{x}\)
2. Isolate base
3. Apply logarithm
4. Swap variables

A 4, 3, 2, 1

B 1, 3, 2, 4

C 1, 2, 3, 4

D 2, 1, 4, 3

💡 Short Explanation

Begin with exponential equation Isolate exponential term Apply logarithm and interchange variables

✅ Correct Answer Explanation

To find inverse of an exponential function: 1. Write \(y=a^{x}\) 2. Isolate the exponential expression 3. Apply logarithm 4. Swap variables to obtain inverse notation Thus the correct sequence is \(1,2,3,4\).

❓ Why Other Options Are Incorrect

Option A → Starts with swapping too early.
Option B → Isolation step is skipped initially.
Option D → Incorrect procedural arrangement.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow the standard inverse-finding algorithm.

Final Logic:

Equation setup precedes logarithmic simplification.

🧠 Memory Trick

"Write → isolate → log → swap"

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

16 To verify that a rational function is its own inverse (self-invertible), one must:

A Compute the composite \(f(f(x))\)and show it simplifies to \(x\)

B Take the derivative and check critical points

C Integrate the function

D Find roots of numerator

💡 Short Explanation

Self-inverse means repeated application restores input Double composition becomes identity Composition test is sufficient

✅ Correct Answer Explanation

A function is self-invertible if: \(f(f(x))=x\) This means applying the function twice returns the original input. Therefore the correct verification method is composition with itself.

❓ Why Other Options Are Incorrect

Option B → Derivatives test monotonicity, not self-inverse property.
Option C → Integration is unrelated here.
Option D → Numerator roots do not establish inverses.

🎯 Strategy

Used: Substitution

Application:

Apply the function twice and simplify.

Final Logic:

Identity after double application confirms self-inverse.

🧠 Memory Trick

"Twice gives original"

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

17 If \(g\circ f=I\) for functions between two finite sets, what constraint is imposed on their sizes?

A \(∣A∣<∣B∣\)

B \(∣A∣\leq ∣B∣\)

C \(∣A∣>∣B∣\)

D No restriction

💡 Short Explanation

Left inverse implies injective mapping Injective maps require larger codomain Domain size cannot exceed codomain

✅ Correct Answer Explanation

If: \(g\circ f=I\) then \(f\) must be injective. For finite sets, an injective function from \(A\rightarrow B\) requires: \(∣A∣\leq ∣B∣\) Hence the size of the domain cannot exceed the codomain.

❓ Why Other Options Are Incorrect

Option A → Opposite inequality is incorrect.
Option C → Injective functions cannot map larger finite sets into smaller ones.
Option D → Finite cardinalities impose restrictions.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Compare sizes required for injective mappings.

Final Logic:

Injectivity implies codomain size is at least domain size.

🧠 Memory Trick

"One-one needs space"

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

18 If \(f\circ g=I\) and \(f\) is onto, what is the relation between cardinalities?

A \(∣A∣<∣B∣\)

B \(∣A∣\geq ∣B∣\)

C \(∣A∣=∣B∣\)

D No relation

💡 Short Explanation

Right inverse implies surjective mapping Surjections need sufficiently large domain Domain size exceeds or equals codomain size

✅ Correct Answer Explanation

If: \(f\circ g=I\) then \(f\) has a right inverse and is onto. For finite sets, surjective mappings satisfy: \(∣A∣\geq ∣B∣\) since every codomain element must receive at least one pre-image.

❓ Why Other Options Are Incorrect

Option A → Surjections cannot map smaller sets onto larger sets.
Option C → Equality is not always necessary.
Option D → Finite sets impose definite restrictions.

🎯 Strategy

Used: Elimination

Application:

Use finite-set surjection principles.

Final Logic:

Onto mappings require large enough domains.

🧠 Memory Trick

"Onto needs coverage"

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

19 The functional equation
\(f(xy)=f(x)+f(y)\)
characterizes:

A Polynomial

B Exponential

C Logarithmic

D Modulus

💡 Short Explanation

Product converts into sum Fundamental logarithmic identity Standard logarithm property

✅ Correct Answer Explanation

Logarithmic functions satisfy: \(log⁡(xy)=log⁡x+log⁡y\) Hence the equation: \(f(xy)=f(x)+f(y)\) is the characteristic logarithmic property.

❓ Why Other Options Are Incorrect

Option A → Polynomials do not preserve multiplicative addition identities.
Option B → Exponentials satisfy additive exponent laws differently.
Option D → Modulus functions follow absolute value properties.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recognize classical logarithmic identity.

Final Logic:

Product-to-sum behavior identifies logarithms.

🧠 Memory Trick

"Multiply inside → add outside"

📚 Class XI – NCERT Mathematics – Chapter: Logarithmic Functions – Topic: Functional Identities

20 In cryptography, for encryption \(E\) and decryption \(D\), which must hold?

A \(E(E(x))=x\)

B \(D(D(x))=x\)

C \(E(x)=D(x)\)

D \(D(E(x))=x\)

💡 Short Explanation

Decryption reverses encryption Composition restores original message Encryption and decryption are inverses

✅ Correct Answer Explanation

Encryption converts plaintext into ciphertext, while decryption reverses the process. Therefore: \(D(E(x))=x\) must hold for every valid message. This expresses inverse-function behavior used in cryptographic systems.

❓ Why Other Options Are Incorrect

Option A → Encryption alone need not reverse itself.
Option B → Decryption alone is not self-inverse generally.
Option C → Encryption and decryption are distinct operations.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret encryption-decryption as inverse mappings.

Final Logic:

Decrypting encrypted data restores original input.

🧠 Memory Trick

"Encrypt then decrypt = original"

📚 Class XI – NCERT Mathematics – Chapter: Composition and Inverse Functions – Topic: Inverse Mapping Applications

UG Mathematics Booster Test 3 - Composition and Invertibility

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