UG Mathematics Booster Test 3 - Continuity Analysis & Functions -

🚀

UG Mathematics Booster Test 3 - Continuity Analysis & Functions

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 3 - Continuity Analysis & Functions

1 / 20

QUESTION 1 OF 20

Arrange the logical steps to prove continuity of a polynomial function \(p(x)\)at \(x=c\):
1. Find \(p(c)\)
2. Evaluate \({lim⁡}_{x\rightarrow c}p(x)\)
3. Show \({lim⁡}_{x\rightarrow c}p(x)=p(c)\)
4. State that \(p(x)\)is defined for all real numbers

Answer Review

1 Arrange the logical steps to prove continuity of a polynomial function \(p(x)\)at \(x=c\):
1. Find \(p(c)\)
2. Evaluate \({lim⁡}_{x\rightarrow c}p(x)\)
3. Show \({lim⁡}_{x\rightarrow c}p(x)=p(c)\)
4. State that \(p(x)\)is defined for all real numbers

A 4, 1, 2, 3

B 1, 2, 3, 4

C 4, 2, 1, 3

D 2, 3, 1, 4

💡 Short Explanation

Polynomial exists for all real numbers Compute value and limit separately Equality proves continuity

✅ Correct Answer Explanation

To prove continuity, first note that polynomial functions are defined for all real numbers. Then compute \(p(c)\), evaluate: \({lim⁡}_{x\rightarrow c}p(x)\) and finally show: \({lim⁡}_{x\rightarrow c}p(x)=p(c)\) Thus the correct logical order is 4,1,2,3. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Definition over all real numbers should be stated first.
Option C → Function value should be identified before comparing equality.
Option D → Equality comparison requires both quantities beforehand.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow the formal continuity proof structure.

Final Logic:

Defined value, limit, and equality establish continuity.

🧠 Memory Trick

"Defined → Value → Limit → Equal"

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

📄 Passage

Consider the continuity of composite functions.
Theorem: Suppose f and g are real-valued functions such that \(\left(f\circ g)(x\right)\)is defined at \(x=c\). If g is continuous at \(x=c\) and f is continuous at \(g(c)\), then the composite function \(\left(f\circ g)(x\right)\)is continuous at \(x=c\).
This explains why functions like
\(sin⁡(x^{2}), cos⁡(x^{2}), e^{x^{2}}\)
remain continuous without separately checking limits at each point.

A rational function
\(f(x)=\frac{p(x)}{q(x)}\)
can be analyzed as a quotient of two polynomials. It is continuous everywhere except where:

A Numerator is zero

B \(p(x)\)is positive

C Denominator is zero \(\left(q(x)=0\right)\)

D x is an integer

💡 Short Explanation

Rational functions involve division Division by zero is undefined Continuity fails only there

✅ Correct Answer Explanation

A rational function is continuous wherever it is defined. Discontinuity occurs when: \(q(x)=0\) because division by zero is undefined. The numerator becoming zero does not create discontinuity. Hence option C is correct according to the continuity property of rational functions.

❓ Why Other Options Are Incorrect

Option A → Zero numerator only makes the function value zero.
Option B → Positivity of numerator does not affect continuity.
Option D → Integers are not automatic discontinuity points.

🎯 Strategy

Used: Elimination

Application:

Remove conditions unrelated to undefined expressions.

Final Logic:

Denominator zero alone breaks rational continuity.

🧠 Memory Trick

"Rational breaks at denominator 0"

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

📄 Passage

Using the theorem from the passage for
\(f(x)=sin⁡(x^{2})\)
since both \(\sin\,x\) and \(x^{2}\)are continuous:

A \(f(x)\)is discontinuous at x=0

B \(f(x)\)is continuous only for x>0

C Limit does not exist at x=0

D \(f(x)\)is continuous for all real numbers

💡 Short Explanation

Inner and outer functions continuous Composition preserves continuity Valid for all real x

✅ Correct Answer Explanation

The function: \(f(x)=sin⁡(x^{2})\) is a composition of: \(g(x)=x^{2}\) and: \(f(t)=sin⁡t\) Since both are continuous for all real numbers, their composition is also continuous everywhere. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → No discontinuity occurs at x=0.
Option B → Continuity is not restricted to positive x-values.
Option C → The limit exists at every real point.

🎯 Strategy

Used: Option Grouping

Application:

Use continuity theorem for composite functions.

Final Logic:

Continuous inside and outside gives global continuity.

🧠 Memory Trick

"Continuous composed stays continuous"

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

4 Match functions with their domains of continuity:

List I	List II
1. \(\sin\,x\)	a. All real numbers
2. \(\cos\,x\)	b. All real numbers except \(x=\frac{(2n+1)\pi }{2}\)
3. \(\tan\,x\)	c. All real numbers
4. \(\frac{1}{x}\)	d. Undefined only at x=0

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

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

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

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

💡 Short Explanation

Sine and cosine continuous everywhere Tangent undefined at odd \(\pi /2\) Reciprocal undefined at zero

✅ Correct Answer Explanation

\(\sin\,x\) and \(\cos\,x\) are continuous for all real numbers. The tangent function is undefined at: \(x=\frac{(2n+1)\pi }{2}\) and: \(\frac{1}{x}\) is undefined at x=0. Thus the correct matching is option B.

❓ Why Other Options Are Incorrect

Option A → Incorrectly swaps tangent and sine continuity domains.
Option C → Cosine continuity is incorrectly matched.
Option D → Reciprocal function is not undefined at tangent discontinuity points.

🎯 Strategy

Used: Option Grouping

Application:

Recall standard continuity domains of elementary functions.

Final Logic:

Match undefined points with corresponding functions.

🧠 Memory Trick

"tan breaks at \(\pi /2\), 1/x breaks at 0"

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

5 For
\(f(x)=\frac{1}{x}\)
which statements are true?
I. It is continuous for all non-zero real numbers
II. x=0 is not in its domain
III. It is continuous for all real numbers

A I only

B I and II

C II and III

D I, II, and III

💡 Short Explanation

Reciprocal undefined at zero Continuous elsewhere on domain Not continuous on all real numbers

✅ Correct Answer Explanation

The reciprocal function: \(\frac{1}{x}\) is undefined at x=0, so 0 is excluded from the domain. It remains continuous for all nonzero real numbers. Statement III is false because continuity cannot hold at a point outside the domain. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Statement II is also true.
Option C → Statement III is false since x=0 is excluded.
Option D → The function is not continuous on all real numbers.

🎯 Strategy

Used: Elimination

Application:

Check domain restrictions carefully.

Final Logic:

Continuity exists only where the function is defined.

🧠 Memory Trick

"1/x continuous except 0"

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

6 Choose the incorrect statement:

A \(\infty\) is treated as a real number

B Limit does not exist as a real number

C Function values exceed any given bound

D Graph approaches axis but never meets it

💡 Short Explanation

Infinity is not a real number Infinite limits mean unbounded growth Real-valued limit does not exist

✅ Correct Answer Explanation

Infinity represents unbounded behavior and is not a real number. Therefore statements like: \({lim⁡}_{x\rightarrow 0}\frac{1}{x}=\infty\) mean the function grows without bound, not that the limit equals a real value. Hence option A is incorrect. The remaining statements correctly describe infinite behavior.

❓ Why Other Options Are Incorrect

Option B → Infinite limits do not exist as finite real numbers.
Option C → Infinite growth exceeds every finite bound.
Option D → Reciprocal graphs approach asymptotes without touching them.

🎯 Strategy

Used: Extreme Word Filter

Application:

Identify misuse of "real number" with infinity.

Final Logic:

Infinity symbolizes unboundedness, not a real value.

🧠 Memory Trick

"∞ ≠ Real number"

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

7 If
\(f(x)=\frac{1}{x}\)
for input \(x=0.01\), the value of the function is:

A 0.01

B 1

C 100

D 10

💡 Short Explanation

Reciprocal means divide 1 by x \(1/0.01=100\) Small positive inputs give large outputs

✅ Correct Answer Explanation

Substituting: \(x=0.01\) gives: \(f(x)=\frac{1}{0.01}=100\) Hence option C is correct. This demonstrates how reciprocal values increase rapidly near zero from the positive side.

❓ Why Other Options Are Incorrect

Option A → Gives the input instead of reciprocal.
Option B → \(\frac{1}{0.01}\neq 1\).
Option D → \(\frac{1}{0.01}=100\), not 10.

🎯 Strategy

Used: Substitution

Application:

Directly evaluate the reciprocal expression.

Final Logic:

Small decimal reciprocals become large numbers.

🧠 Memory Trick

"0.01 flips to 100"

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

8 In the third quadrant, the graph of
\(f(x)=\frac{1}{x}\)
approaches:

A Positive infinity

B Origin

C Positive values

D Negative infinity

💡 Short Explanation

Third quadrant has negative x and y Reciprocal stays negative there Values decrease without bound near zero

✅ Correct Answer Explanation

In the third quadrant: \(x<0,y=\frac{1}{x}<0\) As x approaches 0 from the negative side: \(\frac{1}{x}\rightarrow -\infty\) Hence the graph approaches negative infinity. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → Positive infinity occurs in the first quadrant near zero.
Option B → The graph approaches axes but not the origin directly.
Option C → Third quadrant values are negative.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use quadrant sign conventions.

Final Logic:

Negative x gives negative reciprocal values.

🧠 Memory Trick

"Third quadrant → Negative reciprocal"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Reciprocal Graph Behavior

9 Let
\(f(x)=\left\{\begin{pmatrix}x+2, & x\leq 1\\ x-2, & x>1\end{pmatrix}\right.\)
Is the function continuous at x=1?

A Yes

B Cannot be determined

C No, LHL = 3, RHL = -1

D Limit equals 0

💡 Short Explanation

Compute left and right limits separately One-sided limits differ Hence discontinuity occurs at x=1

✅ Correct Answer Explanation

For x≤1: \({lim⁡}_{x\rightarrow 1^{-}}f(x)=1+2=3\) For x>1: \({lim⁡}_{x\rightarrow 1^{+}}f(x)=1-2=-1\) Since: \(3\neq -1\) the limit does not exist. Therefore the function is discontinuous at x=1. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Equal one-sided limits are required for continuity.
Option B → Continuity can be directly verified from definitions.
Option D → Neither one-sided limit equals zero.

🎯 Strategy

Used: Substitution

Application:

Evaluate LHL and RHL using separate branches.

Final Logic:

Unequal one-sided limits imply discontinuity.

🧠 Memory Trick

"LHL ≠ RHL → Break"

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

10 Let a function have discontinuity only at a finite number of points in the real line.
The probability of randomly selecting a discontinuity point from all real numbers is:

A 1

B 1/2

C Depends on function

D 0

💡 Short Explanation

Finite points have zero length Real line contains infinitely many points Probability of exact point is zero

✅ Correct Answer Explanation

A finite set of discontinuity points has measure zero on the real line. Therefore the probability of randomly selecting exactly one discontinuity point is: \(0\) Hence option D is correct. This is a standard geometric probability interpretation.

❓ Why Other Options Are Incorrect

Option A → Discontinuity points form only a tiny finite subset.
Option B → No finite subset occupies half the real line.
Option C → Finite discontinuity sets always have zero probability.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Compare finite points with infinitely large continuous interval.

Final Logic:

Single or finite points contribute zero probability.

🧠 Memory Trick

"Finite points → Zero probability"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Geometric Probability

11 A vector magnitude is defined by
\(f(x)=∣x∣\)
Over its domain, this function is:

A Continuous everywhere

B Discontinuous at x=0

C Differentiable everywhere

D Discontinuous at negative real numbers

💡 Short Explanation

Modulus function has no breaks LHL and RHL match at x=0 Continuous across all real numbers

✅ Correct Answer Explanation

The modulus function: \(f(x)=∣x∣\) is continuous for every real number because: \({lim⁡}_{x\rightarrow a}∣x∣=∣a∣=f(a)\) for all \(a\in R\). At x=0, both one-sided limits equal 0. Hence option A is correct. The function is continuous though not differentiable at x=0.

❓ Why Other Options Are Incorrect

Option B → Continuity holds at x=0 since both one-sided limits are equal.
Option C → The modulus function is not differentiable at x=0.
Option D → Negative inputs do not create discontinuity.

🎯 Strategy

Used: Elimination

Application:

Remove discontinuity claims using continuity definition.

Final Logic:

Equal limits and defined values ensure continuity everywhere.

🧠 Memory Trick

"|x| joins smoothly"

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

12 Let
\(f(x)=∣x∣\)
While evaluating the left-hand limit at x=0, we use:

A \(f(x)=x\)

B Direct substitution without limit

C Undefined expression

D \(f(x)=-x\)

💡 Short Explanation

Left side means x<0 Modulus becomes negative of x Evaluate using \(-x\) expression

✅ Correct Answer Explanation

For negative values: \(∣x∣=-x\) Hence while evaluating: \({lim⁡}_{x\rightarrow 0^{-}}∣x∣\) we use: \(∣x∣=-x\) This gives: \({lim⁡}_{x\rightarrow 0^{-}}(-x)=0\) Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → \(∣x∣=x\) only for nonnegative x.
Option B → One-sided limits require proper branch selection.
Option C → The modulus function is always defined.

🎯 Strategy

Used: Substitution

Application:

Replace modulus using the negative-side definition.

Final Logic:

Left-hand limit uses x<0 behavior.

🧠 Memory Trick

"Left of 0 → −x"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: One-Sided Limits of Modulus Function

13 While evaluating a definite integral involving
\(f(x)=[x]\)
the interval is split at integer values because:

A It is continuous at integers

B It has a jump discontinuity at every integer

C It evaluates to a constant

D It becomes a polynomial

💡 Short Explanation

Greatest integer changes abruptly Jumps occur at every integer Integrals split at discontinuities

✅ Correct Answer Explanation

The greatest integer function: \(\left[x\right]\) has jump discontinuities at all integers because: \({lim⁡}_{x\rightarrow n^{-}}[x]=n-1,{lim⁡}_{x\rightarrow n^{+}}[x]=n\) Thus the interval must be divided at integer points while integrating. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → The function is discontinuous, not continuous, at integers.
Option C → Greatest integer values vary across intervals.
Option D → The function is not polynomial in nature.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Relate integration splitting with jump discontinuities.

Final Logic:

Piecewise jumps require interval subdivision.

🧠 Memory Trick

"GIF jumps at integers"

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

14 Assertion (A): The function \(f(x)=[x]\)is discontinuous at all integers.
Reason (R): The function \(f(x)=[x]\)is continuous at all integers.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

GIF jumps at integers One-sided limits differ Hence continuity fails there

✅ Correct Answer Explanation

For every integer n: \({lim⁡}_{x\rightarrow n^{-}}[x]=n-1\) and \({lim⁡}_{x\rightarrow n^{+}}[x]=n\) Since these limits are unequal, the greatest integer function is discontinuous at integers. Thus Assertion is true. The Reason incorrectly states continuity, so it is false. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Assertion is mathematically correct.
Option C → Reason is false and cannot explain Assertion.
Option D → Greatest integer function is indeed discontinuous at integers.

🎯 Strategy

Used: Elimination

Application:

Compare one-sided limits at integer points.

Final Logic:

Unequal one-sided limits imply discontinuity.

🧠 Memory Trick

"Integer jump → Discontinuous"

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

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

A 4

B 5

C 9

D 3

💡 Short Explanation

Compute inner function first \(g(2)=3\) Then square the result

✅ Correct Answer Explanation

By composition: \(\left(f\circ g)(2)=f(g(2)\right)\) Since: \(g(2)=2+1=3\) then: \(f(3)=3^{2}=9\) Therefore: \((f\circ g)(2)=9\) Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Squares 2 directly without composition.
Option B → Gives only the intermediate value incorrectly.
Option D → Represents only \(g(2)\), not final composition.

🎯 Strategy

Used: Substitution

Application:

Apply the inner function before the outer function.

Final Logic:

Composition works sequentially.

🧠 Memory Trick

"Inside first, outside next"

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

16 To prove
\(f(x)=sin⁡(x^{2})\)
is continuous, we decompose it as:

A \(f(x)=sin⁡(t), t=x^{2}\)

B \(f(x)=xsin⁡x\)

C \(f(x)=sin⁡x^{2}\)directly

D Cannot be expressed as a composition

💡 Short Explanation

Composite function theorem applies Inner function is \(x^{2}\) Outer function is sine

✅ Correct Answer Explanation

Write: \(t=x^{2}\) and: \(f(x)=sin⁡t\) Since both \(x^{2}\)and \(\sin\,t\) are continuous, their composition: \(sin⁡(x^{2})\) is continuous. Hence option A correctly expresses the function as a composition.

❓ Why Other Options Are Incorrect

Option B → Represents multiplication, not composition.
Option C → Does not explicitly show the composition structure.
Option D → The function clearly is a composite function.

🎯 Strategy

Used: Option Grouping

Application:

Separate inner and outer functions clearly.

Final Logic:

Continuous inner and outer functions imply continuous composite.

🧠 Memory Trick

"Square inside sine"

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

17 If the domain of a function consists of a single point \(\left\{a\right\}\), then the function is:

A Continuous at a

B Discontinuous

C Undefined in limit sense

D Infinite

💡 Short Explanation

No nearby domain points exist Continuity condition holds trivially Singleton domains are continuous

✅ Correct Answer Explanation

If the domain contains only one point \(\left\{a\right\}\), there are no nearby domain points approaching a. Thus continuity conditions are automatically satisfied. Hence every function on a singleton domain is continuous at that point. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → No discontinuity arises in a singleton domain.
Option C → Continuity is defined vacuously in this case.
Option D → Continuity has no relation to infinity here.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the topological idea of isolated points.

Final Logic:

No neighboring domain points means automatic continuity.

🧠 Memory Trick

"One-point domain → Always continuous"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Continuity on Restricted Domains

18 A function defined on \(\left[a,\ b\right]\)is continuous if:

A Limits exist only at interior points

B Only right-hand limit at a exists

C Continuous in \(\left(a,\ b\right)\), RHL at a equals \(f(a)\), and LHL at b equals \(f(b)\)

D Derivative exists everywhere

💡 Short Explanation

Closed intervals include endpoints Endpoint continuity uses one-sided limits Interior continuity also required

✅ Correct Answer Explanation

For continuity on \(\left[a,\ b\right]\): • Function must be continuous on \(\left(a,\ b\right)\) • Right-hand limit at a must equal \(f(a)\) • Left-hand limit at b must equal \(f(b)\) Hence option C correctly gives the rigorous definition of continuity on a closed interval.

❓ Why Other Options Are Incorrect

Option A → Endpoint conditions are also necessary.
Option B → Left-end continuity alone is insufficient.
Option D → Differentiability is stronger than continuity and not required.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Match the formal interval continuity definition.

Final Logic:

Closed intervals require endpoint continuity too.

🧠 Memory Trick

"Closed interval → Check endpoints"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Continuity on Closed Intervals

19 A discontinuity at \(x=a\) implies:

A Graph is a straight line

B Graph has constant slope

C Graph forms a parabola

D Graph cannot be drawn without lifting the pen

💡 Short Explanation

Discontinuity creates graph interruption Continuous curves stay connected Broken graph requires pen lifting

✅ Correct Answer Explanation

The graphical interpretation of continuity says a continuous graph can be drawn without lifting the pen. Therefore a discontinuity implies a break or interruption in the graph. Hence option D is correct. Straight lines and parabolas may still be continuous.

❓ Why Other Options Are Incorrect

Option A → Straight lines can be continuous.
Option B → Constant slope does not define discontinuity.
Option C → Parabolas are continuous polynomial graphs.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Connect graphical interpretation with continuity.

Final Logic:

Broken graph means discontinuity.

🧠 Memory Trick

"Lift pen = Break"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Graphical Meaning of Continuity

20 Let
\(f(x)=\left\{\begin{pmatrix}x^{2}, & x\leq 1\\ x^{2}, & x>1\end{pmatrix}\right.\)
Checking continuity at x=1:

A Limit does not exist

B Function is continuous

C LHL ≠ RHL

D Requires derivative

💡 Short Explanation

Both branches are identical One-sided limits are equal Polynomial remains continuous

✅ Correct Answer Explanation

Both parts define the same function: \(f(x)=x^{2}\) At x=1: \({lim⁡}_{x\rightarrow 1^{-}}x^{2}=1\) and \({lim⁡}_{x\rightarrow 1^{+}}x^{2}=1\) Also: \(f(1)=1\) Thus the function is continuous at x=1. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → The limit exists and equals 1.
Option C → Left-hand and right-hand limits are equal.
Option D → Continuity does not require differentiability testing.

🎯 Strategy

Used: Substitution

Application:

Evaluate one-sided limits and function value at x=1.

Final Logic:

Equal limits matching function value imply continuity.

🧠 Memory Trick

"Same branches → Continuous"

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

UG Mathematics Booster Test 3 - Continuity Analysis & Functions

Test Complete!

Answer Review