UG Mathematics Booster Test 2 - Composition and Invertibility -

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UG Mathematics Booster Test 2 - Composition and Invertibility

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 2 - Composition and Invertibility

1 / 20

QUESTION 1 OF 20

Let \(f\) and \(g\) be two functions. What is the explicit domain of the composition \(\left(f\circ g)(x\right)\)?

Answer Review

1 Let \(f\) and \(g\) be two functions. What is the explicit domain of the composition \(\left(f\circ g)(x\right)\)?

A Domain of \(f\)

B Set of all \(x\) such that \(x\in\) Domain of \(g\) and \(g(x)\in\) Domain of \(f\)

C Range of \(f\)

D All real numbers

💡 Short Explanation

Composition requires valid input for both functions \(x\) must belong to domain of \(g\) \(g(x)\)must belong to domain of \(f\)

✅ Correct Answer Explanation

For the composite function: \(\left(f\circ g)(x)=f(g(x)\right)\) the value \(g(x)\)must be defined first and must also lie inside the domain of \(f\). Therefore the domain is: \(\left\{x\in Domain(g):g(x)\in Domain(f)\right\}\) Hence Option B correctly defines the domain of composition.

❓ Why Other Options Are Incorrect

Option A → Only domain of \(f\) is insufficient because \(g(x)\)must first exist.
Option C → Range of \(f\) does not define valid inputs for composition.
Option D → Composition is not always defined for all real numbers.

🎯 Strategy

Used: Elimination

Application:

Check which option includes both domain compatibility conditions.

Final Logic:

Composition requires valid input for both \(g\) and \(f\).

🧠 Memory Trick

"Input for \(g\), output into \(f\)"

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

📄 Passage

For a composite function to be defined, the range of the inner function must be a subset of the domain of the outer function. Also, \(f\circ g\neq g\circ f\) in general.

Question:
Based on the passage, what is the required domain condition for \(\left(f\circ g)(x\right)\)to return a real value?

A \(x\in\) Domain of \(f\)

B \(x\) can be any real number

C \(g(x)\in\) Domain of \(f\)

D \(f(x)\in\) Domain of \(g\)

💡 Short Explanation

Composite functions need compatibility Output of \(g\) becomes input of \(f\) Hence \(g(x)\)must lie in domain of \(f\)

✅ Correct Answer Explanation

In composition: \(\left(f\circ g)(x)=f(g(x)\right)\) the inner function \(g\) acts first. Therefore \(g(x)\)must belong to the domain of \(f\) for the composite function to produce a real value. This matches the passage condition directly.

❓ Why Other Options Are Incorrect

Option A → Input must initially belong to domain of \(g\), not only \(f\).
Option B → Arbitrary real numbers may violate domain restrictions.
Option D → This corresponds to \(g\circ f\), not \(f\circ g\).

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the passage statement directly to identify the condition.

Final Logic:

Output of inner function must enter outer function.

🧠 Memory Trick

"Inner output → outer input"

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

📄 Passage

For a composite function to be defined, the range of the inner function must be a subset of the domain of the outer function. Also, \(f\circ g\neq g\circ f\) in general.

Question:
Given functions \(f\) and \(g\), evaluate \(\left(f\circ g)(x\right)\).

A \(f(x)+g(x)\)

B \(g(f(x))\)

C \(f(x)g(x)\)

D \(f(g(x))\)

💡 Short Explanation

Composition applies one function after another \(g\) acts first \(f\) acts on output of \(g\)

✅ Correct Answer Explanation

The notation: \(\left(f\circ g)(x\right)\) means first evaluate \(g(x)\), then apply function \(f\) to that result: \(\left(f\circ g)(x)=f(g(x)\right)\) Hence Option D correctly represents the composition of functions.

❓ Why Other Options Are Incorrect

Option A → Represents addition of functions, not composition.
Option B → Represents \(g\circ f\), not \(f\circ g\).
Option C → Represents multiplication of functions.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Identify the standard mathematical notation of composition.

Final Logic:

Composition means "inside first, outside next".

🧠 Memory Trick

"\(g\) goes in first"

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

List I	List II
1. \(f\circ g\)	a. \(f(g(x))\)
2. \(g\circ f\)	b. \(g(f(x))\)
3. \(f(f(x))\)	c. Self-composition of \(f\)
4. \(g(g(x))\)	d. Self-composition of \(g\)

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

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

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

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

💡 Short Explanation

\(f\circ g=f(g(x))\) \(g\circ f=g(f(x))\) Repeated application gives self-composition

✅ Correct Answer Explanation

The standard definitions are: \(f\circ g=f(g(x))\) and \(g\circ f=g(f(x))\) Also: \(f(f(x))\) means self-composition of \(f\), while: \(g(g(x))\) means self-composition of \(g\). Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Interchanges compositions incorrectly.
Option C → Matches compositions with wrong expressions.
Option D → Completely mismatches self-compositions.

🎯 Strategy

Used: Option Grouping

Application:

Match standard notation with corresponding definitions.

Final Logic:

Composition notation directly determines mapping.

🧠 Memory Trick

"Outer function written first"

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

5 Under which conditions can composition commute \(\left(f\circ g=g\circ f\right)\)?
(i) Identity functions
(ii) Linear functions under certain constraints
(iii) Arbitrary functions

A (i) only

B (iii) only

C (i) and (ii) only

D (i), (ii), and (iii)

💡 Short Explanation

Composition usually depends on order Identity mappings commute Some linear mappings also commute

✅ Correct Answer Explanation

Function composition is generally noncommutative. However, identity functions always commute with other functions. Certain linear functions also commute under specific algebraic conditions. Arbitrary functions do not commute universally. Therefore statements (i) and (ii) are correct.

❓ Why Other Options Are Incorrect

Option A → Ignores commuting linear-function cases.
Option B → Arbitrary functions are generally noncommutative.
Option D → Statement (iii) is false.

🎯 Strategy

Used: Elimination

Application:

Reject universal claims about arbitrary functions.

Final Logic:

Only special structured functions commute.

🧠 Memory Trick

"Order matters usually"

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

6 Identify the INCORRECT statement regarding functions \(f\) and \(g\):

A \(f\circ g\neq g\circ f\) in general

B Order of composition affects the result

C \(f\circ g=g\circ f\) for all values of \(x\)

D Both \(f\) and \(g\) can be bijective mappings

💡 Short Explanation

Composition is generally noncommutative Order changes outputs Equality is not universally true

✅ Correct Answer Explanation

In general: \(f\circ g\neq g\circ f\) Therefore Option C is incorrect because composition equality does not hold universally for all functions. The other statements correctly describe properties of composition and bijective functions.

❓ Why Other Options Are Incorrect

Option A → Correct property of compositions.
Option B → Order changes intermediate mappings.
Option D → Bijective functions may still be composed.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "all" signals an overgeneralized statement.

Final Logic:

Function composition is not universally commutative.

🧠 Memory Trick

"Changing order changes answer"

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

7 A cooling process follows a function \(T=f(t)\). Using the inverse function, find the time corresponding to a given temperature.

A 4 minutes

B 5 minutes

C 6 minutes

D 8 minutes

💡 Short Explanation

No explicit function provided Temperature value missing Inverse cannot be evaluated numerically

✅ Correct Answer Explanation

To determine time using an inverse function, both the cooling function \(T=f(t)\)and a specific temperature value are required. Since neither is given, the inverse evaluation cannot be performed. Therefore the numerical answer cannot be determined uniquely.

❓ Why Other Options Are Incorrect

Option A → Unsupported without actual function data.
Option C → Arbitrary numerical choice.
Option D → No calculation justifies this value.

🎯 Strategy

Used: Elimination

Application:

Check whether sufficient mathematical information exists.

Final Logic:

Inverse evaluation requires explicit function and output value.

🧠 Memory Trick

"No formula, no inverse value"

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

8 If a function fails the horizontal line test, then:

A It is always invertible

B It is not one-one and hence not invertible (without restriction)

C It is onto

D It is identity

💡 Short Explanation

Horizontal line test checks injectivity Repeated outputs violate one-one property Non-injective functions lack inverses

✅ Correct Answer Explanation

If a horizontal line intersects the graph at more than one point, multiple inputs give the same output. Hence the function is not one-one. Since invertibility requires injectivity, such a function is not invertible unless the domain is restricted.

❓ Why Other Options Are Incorrect

Option A → Failure of test implies noninvertibility.
Option C → Onto property is unrelated here.
Option D → Identity functions always pass the horizontal line test.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Relate graphical test with injective property.

Final Logic:

Failing the horizontal line test means many-one mapping.

🧠 Memory Trick

"One horizontal cut only"

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

9 Given invertible function data: \(\left(1,10),(2,20),(3,30\right)\).
Find the average of inverse values \(f^{-1}(10)\)and \(f^{-1}(20)\).

A 10

B 2

C 3

D 1.5

💡 Short Explanation

Inverse reverses ordered pairs \(f^{-1}(10)=1\) and \(f^{-1}(20)=2\) Average equals \(1.5\)

✅ Correct Answer Explanation

From the mapping: \(1\rightarrow 10,2\rightarrow 20\) the inverse values are: \(f^{-1}(10)=1,f^{-1}(20)=2\) Their average is: \(\frac{1+2}{2}=1.5\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Confuses outputs with inverse inputs.
Option B → Incorrect averaging.
Option C → Sum incorrectly treated as average.

🎯 Strategy

Used: Substitution

Application:

Reverse mappings first, then compute arithmetic mean.

Final Logic:

Inverse outputs are 1 and 2, whose average is 1.5.

🧠 Memory Trick

"Reverse pair, then average"

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

10 If a function is composed with itself, what is the probability that \(f(f(x))=x\) for all \(x\)?

A 1

B 0

C 0.5

D 0.25

💡 Short Explanation

Depends on function type Only involutions satisfy condition always No universal probability exists

✅ Correct Answer Explanation

The condition: \(f(f(x))=x\) defines an involution. Not every function satisfies this property. Since the probability depends on the class of functions being chosen, no universal numerical probability can be assigned. Hence the provided answer is mathematically unjustified.

❓ Why Other Options Are Incorrect

Option A → Not every function is self-inverse.
Option C → No universal half-probability exists.
Option D → Arbitrary unsupported probability value.

🎯 Strategy

Used: Extreme Word Filter

Application:

Reject universal probabilities without specified sample space.

Final Logic:

Probability depends on the chosen family of functions.

🧠 Memory Trick

"Self-inverse is special"

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

11 In 2D vector transformations, a function represents an identity mapping if the transformation matrix is:

A Null matrix

B Symmetric matrix

C Skew-symmetric matrix

D Identity matrix

💡 Short Explanation

Identity mapping leaves vectors unchanged Identity matrix preserves coordinates Output vector equals input vector

✅ Correct Answer Explanation

An identity mapping transforms every vector into itself. In matrix form, this transformation is represented by the identity matrix: \(I=\left[\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\right]\) Multiplying any vector by this matrix leaves the vector unchanged. Hence the transformation corresponds to the identity function.

❓ Why Other Options Are Incorrect

Option A → Null matrix maps every vector to zero.
Option B → Symmetric matrices need not preserve vectors unchanged.
Option C → Skew-symmetric matrices generally rotate or transform vectors.

🎯 Strategy

Used: Odd One Out

Application:

Identify the matrix that preserves all vectors exactly.

Final Logic:

Only identity matrix keeps every vector unchanged.

🧠 Memory Trick

"Identity leaves identity"

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

12 Find the area bounded by the identity function \(y=x\), the x-axis \(\left(y,\ 0\right)\), and the vertical line \(x=4\).

A 4 sq units

B 16 sq units

C 0 sq units

D 8 sq units

💡 Short Explanation

Region forms a right triangle Base and height both equal 4 Area formula gives 8 square units

✅ Correct Answer Explanation

The graph \(y=x\) between \(x=0\) and \(x=4\) forms a right triangle with the x-axis. Its base and height are both 4. \(Area=\frac{1}{2}\times 4\times 4=8\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Underestimates triangular area.
Option B → Equals rectangle area, not triangle.
Option C → Nonzero region clearly exists.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Recognize the geometric region as a triangle.

Final Logic:

Triangle area formula gives \(8\).

🧠 Memory Trick

"Half of \(4\times 4\)"

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

13 Consider the function
\(f(x)=\int_{0}^{x}\,g(t) dt\)
where \(g(t)>0\) for all \(t\). Does \(f\) satisfy the one-one condition?

A No, integration destroys injectivity

B Yes, derivative is positive ⇒ strictly increasing

C No, cannot be evaluated

D Yes, because it is periodic

💡 Short Explanation

Positive derivative implies increasing function Strictly increasing functions are injective Hence inverse exists on its range

✅ Correct Answer Explanation

By the Fundamental Theorem of Calculus: \(f^{'}(x)=g(x)>0\) Since \(f^{'}(x)\)is always positive, the function is strictly increasing. A strictly increasing function is one-one (injective). Therefore \(f\) satisfies the one-one condition.

❓ Why Other Options Are Incorrect

Option A → Integration does not necessarily destroy injectivity.
Option C → The derivative condition is sufficient to decide.
Option D → Positive functions need not be periodic.

🎯 Strategy

Used: Substitution

Application:

Differentiate the integral function using calculus.

Final Logic:

Positive derivative implies injective behavior.

🧠 Memory Trick

"Positive slope ⇒ one-one"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Monotonic Functions and Injectivity

14 Assertion (A): The function is onto.
Reason (R): The function is periodic over integers.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Assertion lacks supporting definition Periodicity does not guarantee onto mapping Statements are mathematically invalid as given

✅ Correct Answer Explanation

The question does not define the function explicitly, so the onto property cannot be concluded. Also, periodicity over integers does not automatically imply surjectivity. Therefore both the assertion and reason fail mathematically.

❓ Why Other Options Are Incorrect

Option B → Assertion cannot be verified true.
Option C → Reason is mathematically insufficient.
Option D → Reason itself is unsupported.

🎯 Strategy

Used: Elimination

Application:

Check whether enough information exists for conclusions.

Final Logic:

Undefined function prevents validation of both statements.

🧠 Memory Trick

"No function, no conclusion"

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

15 Arrange steps to find inverse of a rational function:
1. Cross-multiply
2. Factor
3. Isolate variable
4. Swap variables

A 1, 3, 2, 4

B 4, 1, 2, 3

C 1, 2, 3, 4

D 2, 1, 4, 3

💡 Short Explanation

Begin by interchanging variables Remove denominator by cross multiplication Solve algebraically for isolated variable

✅ Correct Answer Explanation

To find an inverse function: 1. Replace \(f(x)\)by \(y\) and swap variables 2. Cross-multiply to simplify fractions 3. Factor if needed 4. Isolate the variable to obtain inverse form Thus the logical order is \(4,1,2,3\).

❓ Why Other Options Are Incorrect

Option A → Swapping variables must occur earlier.
Option C → Incorrect procedural sequence.
Option D → Factoring before cross multiplication is illogical.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow standard algebraic inverse-finding procedure.

Final Logic:

Swap first, simplify next, isolate finally.

🧠 Memory Trick

"Swap → simplify → solve"

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

16 To verify that two functions \(f\) and \(g\) are inverses, you must show that:

A \(f(x)=g(x)\)

B \(f(g(x))=x\) and \(g(f(x))=x\)

C \(f(x)+g(x)=0\)

D \(f(x)⋅g(x)=1\)

💡 Short Explanation

Inverse functions undo each other Double composition gives identity mapping Both directions must hold

✅ Correct Answer Explanation

Two functions are inverses when composing them returns the identity function in both orders: \(f(g(x))=xandg(f(x))=x\) This guarantees each function reverses the action of the other.

❓ Why Other Options Are Incorrect

Option A → Equal functions are not necessarily inverses.
Option C → Sum condition is unrelated.
Option D → Product condition is unrelated to inverses.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the formal definition of inverse functions.

Final Logic:

Inverse compositions must yield identity.

🧠 Memory Trick

"Undo both ways"

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

17 If \(g\circ f=I\)(i.e., \(g(f(x))=x\)) but \(f\circ g\neq I\), then \(g\) must be:

A Right inverse

B Left inverse

C Derivative mapping

D Identity mapping

💡 Short Explanation

\(g\) reverses \(f\) from left side Only one composition equals identity Hence \(g\) is a left inverse

✅ Correct Answer Explanation

If: \(g\circ f=I\) then \(g\) acts as a left inverse of \(f\). Since: \(f\circ g\neq I\) the inverse relation is not two-sided. Therefore \(g\) is specifically called a left inverse.

❓ Why Other Options Are Incorrect

Option A → Right inverse requires \(f\circ g=I\).
Option C → No derivative relation exists here.
Option D → Identity mapping maps elements to themselves directly.

🎯 Strategy

Used: Option Grouping

Application:

Compare identity compositions carefully.

Final Logic:

Identity on left composition implies left inverse.

🧠 Memory Trick

"Left side identity ⇒ left inverse"

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

18 If \(g\circ f=I\) for all \(x\), but not necessarily \(f\circ g=I\), then \(f\) must be:

A One-one but not onto

B Bijective

C Constant

D Onto but not one-one

💡 Short Explanation

Left inverse implies injective mapping Surjectivity is not guaranteed Hence function is one-one

✅ Correct Answer Explanation

If: \(g\circ f=I\) then \(f\) must be injective because different inputs cannot map to the same output. However, \(f\circ g\neq I\) means surjectivity may fail. Thus \(f\) is one-one but not necessarily onto.

❓ Why Other Options Are Incorrect

Option B → Bijective requires both compositions equal identity.
Option C → Constant functions are not injective.
Option D → Onto property alone is insufficient.

🎯 Strategy

Used: Elimination

Application:

Use inverse-function conditions carefully.

Final Logic:

Left inverse guarantees injectivity only.

🧠 Memory Trick

"Left inverse ⇒ one-one"

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

19 The functional equation
\(f(x+y)=f(x)+f(y)\)
assuming continuity, characterizes:

A Quadratic functions

B Exponential functions

C Linear functions through the origin

D Logarithmic functions

💡 Short Explanation

This is Cauchy's functional equation Continuity forces linear behavior Function becomes \(f(x)=cx\)

✅ Correct Answer Explanation

The equation: \(f(x+y)=f(x)+f(y)\) under continuity assumptions implies: \(f(x)=cx\) where \(c\) is constant. Hence the function represents a linear function passing through the origin.

❓ Why Other Options Are Incorrect

Option A → Quadratic functions fail additive property.
Option B → Exponential functions satisfy multiplicative relations.
Option D → Logarithms satisfy \(log⁡(ab)\), not additive input form.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recognize the standard functional equation form.

Final Logic:

Continuous additive functions are linear.

🧠 Memory Trick

"Additive + continuous ⇒ linear"

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

20 If \(f\) converts USD to EUR and \(g\) converts EUR back to USD, then \(g\circ f\) represents:

A Identity mapping on USD

B Zero mapping

C Conversion to GBP

D Double taxation

💡 Short Explanation

First conversion changes currency Second conversion reverses it Original value is restored

✅ Correct Answer Explanation

Applying \(f\) converts USD to EUR, while \(g\) converts EUR back to USD. Therefore: \((g\circ f)(x)=x\) meaning the original currency value is recovered. This is precisely the identity mapping on USD values.

❓ Why Other Options Are Incorrect

Option B → Composition does not map everything to zero.
Option C → GBP conversion is unrelated.
Option D → Taxation has no connection with inverse mappings.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret composition as reverse conversion.

Final Logic:

Forward conversion followed by reverse conversion restores input.

🧠 Memory Trick

"Convert and return = identity"

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

UG Mathematics Booster Test 2 - Composition and Invertibility

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