UG Mathematics Booster Test 2 - Algebra of Continuous Functions -

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CUET UG Mathematics Booster Test 2 - Algebra of Continuous Functions

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 2 - Algebra of Continuous Functions

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QUESTION 1 OF 20

The graphs of \(f(x)\)and \(g(x)\)are continuous at \(x=c\).
By the sum rule, the function \(\left(f+g)(x\right)\)at \(x=c\) is:

Answer Review

1 The graphs of \(f(x)\)and \(g(x)\)are continuous at \(x=c\).
By the sum rule, the function \(\left(f+g)(x\right)\)at \(x=c\) is:

A Discontinuous due to a sharp corner

B Continuous because both f and g are continuous at c

C A vertical line with multiple values

D Undefined

💡 Short Explanation

Sum rule preserves continuity Both functions are continuous at c Their sum remains continuous

✅ Correct Answer Explanation

If: \(f(x)\) and: \(g(x)\) are continuous at \(x=c\), then: \({lim⁡}_{x\rightarrow c}[f(x)+g(x)]=f(c)+g(c)\) Thus: \(\left(f+g)(x\right)\) is continuous at c. Therefore option B correctly applies the sum rule for continuity.

❓ Why Other Options Are Incorrect

Option A → Sharp corners do not automatically arise from adding continuous functions.
Option C → Continuous functions still satisfy vertical line test.
Option D → Sum of defined continuous functions remains defined.

🎯 Strategy

Used: Option Grouping

Application:

Apply continuity sum theorem directly.

Final Logic:

Continuous + Continuous = Continuous.

🧠 Memory Trick

"Sum stays smooth"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Sum Rule of Continuity

2 Let \(f(x)\)and \(g(x)\)be continuous functions such that
\(\int (f(x)+g(x))dx=\int f(x)dx+\int g(x)dx\)
If
\(\int f(x)dx=5,\int g(x)dx=-3\)
find the value of the total integral.

A 8

B -15

C 2

D -8

💡 Short Explanation

Use linearity of integration Add individual integrals directly \(5+(-3)=2\)

✅ Correct Answer Explanation

By the linearity property: \(\int (f+g) dx=\int f dx+\int g dx\) Substitute the given values: \(5+(-3)=2\) Hence the total integral equals: \(2\) Therefore option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect addition sign handling.
Option B → Multiplication is not involved.
Option D → Arithmetic error in combining values.

🎯 Strategy

Used: Substitution

Application:

Replace given integral values directly.

Final Logic:

Integral addition follows ordinary addition.

🧠 Memory Trick

"Integrals add directly"

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

3 Let \(f(x)\)and \(g(x)\)be continuous functions. For
\(h(x)=f(x)-g(x)\)
which statements are true?
I. \(h(x)\)is continuous everywhere
II. \(h(x)\)is continuous only for \(x>0\)
III. Continuity follows from the difference rule

A I only

B II only

C I and II

D I and III

💡 Short Explanation

Difference rule preserves continuity No restriction to positive x only Difference remains continuous everywhere

✅ Correct Answer Explanation

If: \(f(x)\) and: \(g(x)\) are continuous, then: \(h(x)=f(x)-g(x)\) is also continuous everywhere in their domain. Thus statements I and III are true. Statement II is false because continuity is not restricted to positive values only. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Statement III is also true.
Option B → Continuity is not limited to positive x-values.
Option C → Statement II is false.

🎯 Strategy

Used: Option Grouping

Application:

Evaluate each statement using difference continuity theorem.

Final Logic:

Continuous − Continuous remains continuous.

🧠 Memory Trick

"Difference keeps continuity"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Difference Rule

4 Evaluate limits at \(x=c\):

List 1	List 2
1. \(lim⁡(f-g)\)	a. -1
2. \(lim⁡(g-f)\)	b. 0
3. \(lim⁡(f-f)\)	c. 3
4. \(lim⁡(g-g)\)	d. 5

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

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

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

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

💡 Short Explanation

Difference limits follow subtraction rules Same-function subtraction gives zero Matching follows algebraic properties

✅ Correct Answer Explanation

Using limit laws: \(lim⁡(f-f)=0\) and: \(lim⁡(g-g)=0\) The remaining expressions correspond to the assigned numerical values in the list. Therefore the correct matching is option D according to subtraction properties of limits.

❓ Why Other Options Are Incorrect

Option A → Incorrectly assigns nonzero value to identical subtraction.
Option B → Matching violates subtraction identities.
Option C → Swaps required pairings incorrectly.

🎯 Strategy

Used: Option Grouping

Application:

Match zero-producing expressions first.

Final Logic:

Same-function differences always equal zero.

🧠 Memory Trick

"Same minus same = 0"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Difference Law of Limits

5 Assertion (A): The function \(f(x)g(x)\)is continuous for all real x.
Reason (R): The product of two continuous functions is continuous at every point in their domain.

A Both false

B Both true, R explains A

C A true, R false

D A false, R true

💡 Short Explanation

Product rule preserves continuity Continuous factors give continuous product Reason directly explains assertion

✅ Correct Answer Explanation

If: \(f(x)\) and: \(g(x)\) are continuous functions, then: \(f(x)g(x)\) is continuous throughout their domain. This follows directly from the product rule of continuity. Therefore both Assertion and Reason are true, and Reason correctly explains Assertion. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Both mathematical statements are valid.
Option C → Reason is also true.
Option D → Assertion correctly follows from product continuity theorem.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Check whether theorem directly explains statement.

Final Logic:

Product continuity theorem proves assertion immediately.

🧠 Memory Trick

"Continuous × Continuous = Continuous"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Product Rule

6 If \(f(x)\)and \(g(x)\)are continuous functions and
\(h(x)=f(x)g(x)\)
then
\({lim⁡}_{x\rightarrow c}h(x)=?\)

A Undefined

B \(f(c)g(c)\)

C 0

D 1

💡 Short Explanation

Product limit equals product of limits Continuity allows direct substitution Evaluate functions at c

✅ Correct Answer Explanation

Since: \(h(x)=f(x)g(x)\) and both functions are continuous at c, \({lim⁡}_{x\rightarrow c}h(x)=\left({lim⁡}_{x\rightarrow c}f(x)\right)\left({lim⁡}_{x\rightarrow c}g(x)\right)\) Thus: \(=f(c)g(c)\) Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Product limit exists under continuity.
Option C → Product need not equal zero generally.
Option D → Product need not equal one.

🎯 Strategy

Used: Substitution

Application:

Replace limits by function values using continuity.

Final Logic:

Product of limits equals limit of product.

🧠 Memory Trick

"Multiply the limits"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Product Law of Limits

7 Let \(f(x)\)and \(g(x)\)be continuous. The function
\(\frac{f(x)}{g(x)}\)
is continuous everywhere except where:

A \(f(x)=0\)

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

C \(g(x)=0\)

D x is constant

💡 Short Explanation

Quotient requires nonzero denominator Division by zero undefined Continuity fails only there

✅ Correct Answer Explanation

The quotient: \(\frac{f(x)}{g(x)}\) remains continuous wherever: \(g(x)\neq 0\) because division by zero is undefined. Thus discontinuity occurs only where the denominator equals zero. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Zero numerator is allowed.
Option B → Equal numerator and denominator do not affect continuity.
Option D → Constant x-values are irrelevant here.

🎯 Strategy

Used: Elimination

Application:

Identify condition creating undefined quotient.

Final Logic:

Denominator zero destroys continuity.

🧠 Memory Trick

"Bottom zero → Break"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Quotient Rule

8 Which statement is incorrect?

A Quotient is continuous if f,g are continuous and \(g(c)\neq 0\)

B If \(g(c)=0\), function is undefined at c

C Function remains continuous even if \(g(c)=0\)

D Denominator must be non-zero

💡 Short Explanation

Zero denominator makes quotient undefined Undefined points cannot stay continuous Continuity needs nonzero denominator

✅ Correct Answer Explanation

For: \(\frac{f(x)}{g(x)}\) continuity requires: \(g(c)\neq 0\) If: \(g(c)=0\) the quotient becomes undefined, so continuity fails. Hence option C is incorrect. The remaining statements correctly describe quotient continuity conditions.

❓ Why Other Options Are Incorrect

Option A → Correct quotient continuity theorem.
Option B → Division by zero is undefined.
Option D → Nonzero denominator is necessary.

🎯 Strategy

Used: Extreme Word Filter

Application:

Detect impossible continuity claim despite zero denominator.

Final Logic:

Undefined quotient cannot be continuous.

🧠 Memory Trick

"No zero in denominator"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Quotient Continuity

9 If \(f(x)\)is continuous and \(\lambda =0\), then \(\lambda f(x)\)is:

A Discontinuous

B Continuous only for positive x

C Undefined

D Constant zero function, hence continuous

💡 Short Explanation

Multiplying by zero gives zero function Constant functions are continuous Continuity preserved everywhere

✅ Correct Answer Explanation

If: \(\lambda =0\) then: \(\lambda f(x)=0\) for every x. This becomes the constant zero function, which is continuous over all real numbers. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Zero function never becomes discontinuous.
Option B → Continuity is not restricted to positive x.
Option C → Multiplication by zero is perfectly defined.

🎯 Strategy

Used: Substitution

Application:

Replace λ with zero directly.

Final Logic:

Zero times any function gives continuous constant zero.

🧠 Memory Trick

"0×f = Continuous zero"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Scalar Multiplication Rule

10 If a continuous function is scaled by a constant factor, the new function is:

A Discontinuous

B Continuous only if positive

C Step-wise continuous

D Continuous everywhere

💡 Short Explanation

Constant scaling preserves continuity Sign of constant irrelevant Continuity remains unchanged

✅ Correct Answer Explanation

If: \(f(x)\) is continuous and: \(\lambda\) is a constant, then: \(\lambda f(x)\) is also continuous. Multiplication by a constant does not introduce breaks or discontinuities. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Constant scaling cannot destroy continuity.
Option B → Negative constants also preserve continuity.
Option C → Step behavior is unrelated to scalar multiplication.

🎯 Strategy

Used: Option Grouping

Application:

Apply scalar multiplication continuity theorem.

Final Logic:

Continuous functions remain continuous after scaling.

🧠 Memory Trick

"Scale keeps smoothness"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Scalar Multiplication of Functions

11 Arrange steps to prove \(1/x\) is continuous for \(x\neq 0\):
1. Note that \(x\neq 0\)
2. Assume x is non-zero
3. State limit exists
4. Conclude continuity

A 2, 1, 3, 4

B 1, 3, 4, 2

C 4, 3, 2, 1

D 1, 2, 3, 4

💡 Short Explanation

Start with nonzero assumption Reciprocal defined only there Existing limit proves continuity

✅ Correct Answer Explanation

To prove continuity of: \(f(x)=\frac{1}{x}\) first assume x is nonzero. Then note: \(x\neq 0\) so the function is defined. Next establish that the limit exists, and finally conclude continuity. Therefore the correct logical sequence is 2,1,3,4. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assumption should precede later deductions.
Option C → Conclusion cannot appear before proof.
Option D → Nonzero assumption should logically come first.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange statements in mathematical proof order.

Final Logic:

Definition, limit existence, then continuity conclusion.

🧠 Memory Trick

"Assume → Define → Limit → Conclude"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Reciprocal Function Continuity

12 If
\(f(x)=\frac{1}{x}\)
as \(x\rightarrow 0^{+}\), the function:

A Tends to \(+\infty\), not a real limit

B Becomes 0

C Remains constant

D Becomes 1

💡 Short Explanation

Reciprocal grows unbounded near zero Positive side gives positive infinity Infinity is not a real limit

✅ Correct Answer Explanation

As: \(x\rightarrow 0^{+}\) the denominator becomes a very small positive number. Therefore: \(\frac{1}{x}\rightarrow +\infty\) This means values grow beyond every finite bound. Since \(+\infty\) is not a real number, the limit does not exist as a real limit. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Reciprocal increases, not decreases to zero.
Option C → Function values change rapidly near zero.
Option D → No fixed constant value is approached.

🎯 Strategy

Used: Extreme Word Filter

Application:

Identify unbounded behavior near denominator zero.

Final Logic:

Positive tiny denominators create huge positive values.

🧠 Memory Trick

"0⁺ below fraction → +∞"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Infinite Limits

13 Let
\(f(x)=x^{2}+1\)
Find \(f(2)\):

A 0

B 4

C 5

D -1

💡 Short Explanation

Substitute x=2 directly Square first, then add 1 \(4+1=5\)

✅ Correct Answer Explanation

Substitute: \(x=2\) into: \(f(x)=x^{2}+1\) Then: \(f(2)=2^{2}+1=4+1=5\) Hence option C is correct. Polynomial functions allow direct evaluation at every real number.

❓ Why Other Options Are Incorrect

Option A → Incorrect arithmetic evaluation.
Option B → Gives only \(2^{2}\), missing +1.
Option D → Polynomial value cannot become negative here.

🎯 Strategy

Used: Substitution

Application:

Directly replace x with 2.

Final Logic:

Evaluate square before addition.

🧠 Memory Trick

"Square then add one"

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

14 A polynomial function is continuous over:

A Positive integers

B Integers only

C Rational numbers

D All real numbers

💡 Short Explanation

Polynomials have no breaks Defined for every real number Continuous on entire real line

✅ Correct Answer Explanation

Polynomial functions are formed using powers, constants, addition, subtraction, and multiplication. These operations preserve continuity. Therefore polynomial functions remain continuous for every: \(x\in R\) Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Continuity is not restricted to positive integers.
Option B → Real numbers beyond integers are included.
Option C → Irrational numbers are also included in continuity domain.

🎯 Strategy

Used: Elimination

Application:

Remove unnecessarily restricted domains.

Final Logic:

Polynomial continuity holds everywhere on \(R\).

🧠 Memory Trick

"Polynomial = Always smooth"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Polynomial Continuity

📄 Passage

A rational function
\(f(x)=\frac{p(x)}{q(x)}\)
is continuous wherever \(q(x)\neq 0\).

If \(q(x)=0\) at some point, then the function:

A Is continuous there

B Is not rational

C Is discontinuous there

D Becomes polynomial

💡 Short Explanation

Denominator zero makes function undefined Undefined points break continuity Rational continuity fails there

✅ Correct Answer Explanation

A rational function: \(\frac{p(x)}{q(x)}\) is continuous only where: \(q(x)\neq 0\) If: \(q(x)=0\) the function becomes undefined and continuity fails. Therefore the function is discontinuous at that point. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Undefined points cannot be continuous.
Option B → Function still remains rational in form.
Option D → Zero denominator does not convert it into a polynomial.

🎯 Strategy

Used: Elimination

Application:

Identify the condition creating undefined values.

Final Logic:

Denominator zero implies discontinuity.

🧠 Memory Trick

"Zero denominator → Break"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Rational Functions

📄 Passage

A rational function
\(f(x)=\frac{p(x)}{q(x)}\)
is continuous wherever \(q(x)\neq 0\).

Roots of denominator represent:

A Discontinuity points

B Maximum points

C Zero of numerator

D Continuous points

💡 Short Explanation

Denominator roots make division undefined Undefined points create discontinuities Rational graphs break there

✅ Correct Answer Explanation

If: \(q(x)=0\) then: \(\frac{p(x)}{q(x)}\) becomes undefined. Such points are excluded from the domain and become discontinuity points of the rational function. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Roots do not automatically indicate maxima.
Option C → Denominator roots differ from numerator roots.
Option D → Continuity fails at denominator zeros.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Relate denominator roots to continuity restrictions.

Final Logic:

Undefined denominator creates discontinuity.

🧠 Memory Trick

"Denominator roots = Break points"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Rational Function Discontinuity

17 Which limits are used to prove continuity of \(\sin\,x\)?
I. \({lim⁡}_{x\rightarrow 0}sin⁡x=0\)
II. \({lim⁡}_{x\rightarrow 0}\frac{\sin\,x}{x}=1\)
III. \({lim⁡}_{x\rightarrow 0}cos⁡x=1\)

A I only

B I, II, and III

C II and III

D III only

💡 Short Explanation

Fundamental trigonometric limits used Sine and cosine behavior near zero important All three statements support continuity proofs

✅ Correct Answer Explanation

The continuity proofs of trigonometric functions use standard limits: \({lim⁡}_{x\rightarrow 0}sin⁡x=0{lim⁡}_{x\rightarrow 0}\frac{\sin\,x}{x}=1\) and: \({lim⁡}_{x\rightarrow 0}cos⁡x=1\) These establish continuity of sine and related trigonometric functions. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores other important trigonometric limits.
Option C → First sine limit is also essential.
Option D → Cosine limit alone is insufficient.

🎯 Strategy

Used: Option Grouping

Application:

Group all standard trigonometric limits together.

Final Logic:

All three limits are foundational continuity tools.

🧠 Memory Trick

"sin, sin/x, cos → Core trig limits"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Trigonometric Limits

18 For \(\tan\,x\), incorrect statement:

A \(tan⁡x=\frac{\sin\,x}{\cos\,x}\)

B It is quotient of continuous functions

C It is continuous for all real numbers

D Undefined at odd multiples of \(\pi /2\)

💡 Short Explanation

Tangent denominator becomes zero sometimes Undefined at odd \(\pi /2\) multiples Hence not continuous everywhere

✅ Correct Answer Explanation

Since: \(tan⁡x=\frac{\sin\,x}{\cos\,x}\) it becomes undefined where: \(cos⁡x=0\) i.e. \(x=\frac{(2n+1)\pi }{2}\) Thus tangent is not continuous for all real numbers. Hence option C is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A → Correct trigonometric identity.
Option B → Quotient of continuous functions where denominator is nonzero.
Option D → Correct discontinuity locations for tangent.

🎯 Strategy

Used: Extreme Word Filter

Application:

"All real numbers" signals overgeneralization.

Final Logic:

Undefined points prevent global continuity.

🧠 Memory Trick

"tan breaks at odd \(\pi /2\)"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Tangent Function

19 If g is continuous at c, then for \(f(g(x))\)to be continuous at c, we need:

A f continuous at c

B g continuous at \(g(c)\)

C \(f=g\)

D f continuous at \(g(c)\)

💡 Short Explanation

Composite continuity uses image point Outer function checked at \(g(c)\) Inner continuity already given

✅ Correct Answer Explanation

The composition theorem states: If: \(g(x)\) is continuous at c and: \(f(x)\) is continuous at: \(g(c)\) then: \(f(g(x))\) is continuous at c. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → Outer continuity must be checked at \(g(c)\), not c generally.
Option B → g is already continuous at c.
Option C → Equality between functions is unnecessary.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply composite continuity theorem carefully.

Final Logic:

Outer continuity required at the transformed point.

🧠 Memory Trick

"Outer checks inner output"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Composite Functions

20 Let
\(f(x)=x+1,g(x)=x^{2}\)
Find
\(\left(f\circ g)(2\right)\)

A 5

B 1

C 3

D 0

💡 Short Explanation

Compute inner function first \(g(2)=4\) Then apply outer function

✅ Correct Answer Explanation

By composition: \(\left(f\circ g)(2)=f(g(2)\right)\) Now: \(g(2)=2^{2}=4\) Then: \(f(4)=4+1=5\) Therefore: \((f\circ g)(2)=5\) Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect substitution into functions.
Option C → Gives incomplete evaluation.
Option D → No valid composition produces zero.

🎯 Strategy

Used: Substitution

Application:

Evaluate inner function before outer function.

Final Logic:

Composition works step-by-step.

🧠 Memory Trick

"Inside first, outside next"

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

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