UG Mathematics Booster Test 2 - Continuity Fundamentals -

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UG Mathematics Booster Test 2 - Continuity Fundamentals

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 2 - Continuity Fundamentals

1 / 20

QUESTION 1 OF 20

As x→a, the values of a function f(x) form sequences tending to L.
For the function to be continuous at x=a, what must hold?

Answer Review

1 As x→a, the values of a function f(x) form sequences tending to L.
For the function to be continuous at x=a, what must hold?

A f(a)=0

B f(a)=L

C f(a)=∞

D f(a) is undefined

💡 Short Explanation

Continuity links limit and function value Limit near a must equal actual value Defined equality condition is necessary

✅ Correct Answer Explanation

For continuity at x=a, \({lim⁡}_{x\rightarrow a}f(x)=f(a)\) If nearby function values approach L, then continuity requires f(a)=L. Hence option B is correct. Option A is unrelated. Option C is impossible for finite continuity. Option D violates the requirement that the function must be defined at x=a.

❓ Why Other Options Are Incorrect

Option A → Continuity does not require the function value to be zero unless the limit itself is zero.
Option C → Infinite values do not satisfy ordinary continuity conditions for real-valued functions.
Option D → An undefined function value makes continuity impossible at that point.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Match the statement directly with the formal continuity definition.

Final Logic:

Continuity means function value equals the approaching limit.

🧠 Memory Trick

"Continuous = Limit equals value"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Formal Definition of Continuity

2 For the function
\(f(x)=\frac{1}{x}\)
as x→0^+, the graph approaches +∞.
This implies:

A The function is continuous at x=0

B The graph has a sharp corner at x=0

C The right-hand limit does not exist as a real number

D The graph intersects the y-axis

💡 Short Explanation

Reciprocal grows unbounded near zero Infinite behavior prevents finite limit Hence continuity fails at x=0

✅ Correct Answer Explanation

As x approaches 0 from the right, \(\frac{1}{x}\rightarrow +\infty\) The limit is not a finite real number, so the right-hand limit does not exist in the real-number sense. Hence option C is correct. Option A is false because continuity requires a finite defined value. Options B and D are unrelated graphical interpretations.

❓ Why Other Options Are Incorrect

Option A → The function is undefined at x=0, so continuity is impossible there.
Option B → Infinite discontinuity is not the same as a sharp corner.
Option D → The graph never intersects the y-axis because x=0 is excluded from the domain.

🎯 Strategy

Used: Elimination

Application:

Remove options inconsistent with infinite behavior near x=0.

Final Logic:

Infinite growth means no finite real right-hand limit exists.

🧠 Memory Trick

"1/x near 0 → Infinity"

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

3 Let
\(f(x)=\left\{\begin{pmatrix}x+4, & x<3\\ 2x, & x\geq 3\end{pmatrix}\right.\)
Find the left-hand limit at x=3.

A 9

B 6

C 3

D 7

💡 Short Explanation

LHL uses expression for x<3 Substitute x=3 into x+4 Result equals 7

✅ Correct Answer Explanation

For the left-hand limit at x=3, we use the branch valid for x<3: \(f(x)=x+4\) Thus, \({lim⁡}_{x\rightarrow 3^{-}}f(x)=3+4=7\) Hence option D is correct. The right-side expression 2x is not used for the left-hand limit.

❓ Why Other Options Are Incorrect

Option A → 9 is not obtained from either branch at x=3.
Option B → 6 comes from the right-side branch 2x, not the left-hand limit.
Option C → 3 ignores the additional constant term in x+4.

🎯 Strategy

Used: Substitution

Application:

Substitute x=3 into the expression corresponding to x<3.

Final Logic:

Left-hand limit uses the left-side formula only.

🧠 Memory Trick

"LHL → Left branch"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Left-Hand Limit

4 Let a velocity component be defined as:
\(v(t)=\left\{\begin{pmatrix}t^{2}, & t<2\\ kt, & t\geq 2\end{pmatrix}\right.\)
For continuity at t=2, the right-hand limit must equal:

A 4

B 2

C 0

D ∞

💡 Short Explanation

Continuity requires equal one-sided limits Left-hand limit equals \(2^{2}\) RHS must also equal 4

✅ Correct Answer Explanation

For continuity at t=2, \({lim⁡}_{t\rightarrow 2^{-}}v(t)={lim⁡}_{t\rightarrow 2^{+}}v(t)\) The left-hand limit is: \(2^{2}=4\) Therefore the right-hand limit must also equal 4. Hence option A is correct. The value of k would later be adjusted to satisfy this condition.

❓ Why Other Options Are Incorrect

Option B → Continuity requires equality with the left-hand limit, which is 4.
Option C → Zero does not match the left-side behavior.
Option D → The polynomial expression remains finite near t=2.

🎯 Strategy

Used: Substitution

Application:

Evaluate the left-hand limit and equate it with the right-hand limit.

Final Logic:

Continuity demands both one-sided limits equal 4.

🧠 Memory Trick

"Continuous join → Same value"

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

5 To compute the area under a curve \(y=f(x)\)on [a,b], which condition ensures integrability via continuity?

A f(x) is discontinuous at some point

B f(x) is continuous for all x∈[a,b]

C \(f^{'}(x)\)is undefined at 0

D The graph breaks at some point

💡 Short Explanation

Continuous functions are integrable Unbroken graphs support area calculation NCERT links continuity with integration

✅ Correct Answer Explanation

A function continuous on a closed interval [a,b] is guaranteed to be integrable there. This allows meaningful computation of the area under the curve. Hence option B is correct. Discontinuities or breaks may complicate integration. Derivative behavior is unrelated to basic integrability requirements.

❓ Why Other Options Are Incorrect

Option A → Discontinuities may prevent direct integrability in standard situations.
Option C → Differentiability is not the primary condition for definite integrability.
Option D → Graph breaks indicate discontinuity, opposing smooth area computation.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Connect continuity with the standard NCERT condition for integrability.

Final Logic:

Continuous functions on closed intervals are integrable.

🧠 Memory Trick

"Continuous curve → Integrable area"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Continuity and Integrability

6 Let
\(f(x)=\left\{\begin{pmatrix}x^{2}, & x\neq 1\\ 3, & x=1\end{pmatrix}\right.\)
Here,
\({lim⁡}_{x\rightarrow 1}f(x)=1\neq f(1)\)
This implies:

A The integral is 0

B The function is continuous

C x=1 is a point of discontinuity (removable); integrals may be split if needed

D The area is infinite

💡 Short Explanation

Limit exists but differs from value Continuity condition fails Removable discontinuity occurs at x=1

✅ Correct Answer Explanation

Since \({lim⁡}_{x\rightarrow 1}f(x)=1\) but \(f(1)=3\) the continuity condition fails. The discontinuity is removable because redefining f(1)=1 would restore continuity. Hence option C is correct. Options A and D are unrelated, while option B contradicts the unequal limit and function value.

❓ Why Other Options Are Incorrect

Option A → Discontinuity does not force the integral to become zero.
Option B → Continuity requires limit and function value to be equal.
Option D → A removable discontinuity does not produce infinite area.

🎯 Strategy

Used: Elimination

Application:

Compare the limit with the actual function value.

Final Logic:

Unequal limit and value imply removable discontinuity.

🧠 Memory Trick

"Limit exists, value differs → Removable"

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

7 Assertion (A):
The function
\(f(x)=\left\{\begin{pmatrix}-1, & x<0\\ 1, & x\geq 0\end{pmatrix}\right.\)
is continuous at x=0.
Reason (R):
The LHL at x=0 is −1 and the RHL is 1; since LHL ≠ RHL, the limit does not exist.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

One-sided limits are unequal Overall limit does not exist Therefore continuity fails at 0

✅ Correct Answer Explanation

The left-hand limit at x=0 is −1 while the right-hand limit is 1. Since these are unequal, the overall limit does not exist. Therefore the function cannot be continuous at x=0. Hence Assertion is false, but the Reason is true, making option D correct.

❓ Why Other Options Are Incorrect

Option A → The Reason is mathematically correct.
Option B → The Assertion is false because unequal one-sided limits prevent continuity.
Option C → The Assertion itself is incorrect, so both cannot be true together.

🎯 Strategy

Used: Substitution

Application:

Compute LHL and RHL directly from the piecewise definition.

Final Logic:

Unequal one-sided limits destroy continuity.

🧠 Memory Trick

"LHL ≠ RHL → No continuity"

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

8 Arrange steps to check continuity of f(x) at x=a:
1. Compute \({lim⁡}_{x\rightarrow a}f(x)\)
2. Check if \({lim⁡}_{x\rightarrow a}f(x)=f(a)\)
3. Evaluate f(a)
4. Conclude continuity

A 1, 3, 2, 4

B 4, 3, 1, 2

C 3, 1, 4, 2

D 2, 1, 3, 4

💡 Short Explanation

First find the limit Then evaluate function value Compare and conclude continuity

✅ Correct Answer Explanation

To test continuity: First compute the limit at x=a. Then evaluate f(a). Next verify whether: \({lim⁡}_{x\rightarrow a}f(x)=f(a)\) Finally conclude continuity. Therefore the correct order is 1, 3, 2, 4. This follows the standard continuity checking procedure used in NCERT.

❓ Why Other Options Are Incorrect

Option B → Conclusion cannot come before calculations.
Option C → Continuity cannot be concluded before comparing limit and function value.
Option D → Equality cannot be checked before finding the required quantities.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow the logical sequence used in continuity verification.

Final Logic:

Compute, compare, then conclude.

🧠 Memory Trick

"Limit → Value → Compare → Decide"

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

9 Match each function with the reason for discontinuity at x=0:

List I	List II
1. \(f(x)=1/x\)	a. LHL ≠ RHL (jump discontinuity)
2. \(f(x)=\left\{\begin{pmatrix}1, & x<0\\ 2, & x>0\end{pmatrix}\right.\)	b. Limit exists but ≠f(0) (removable)
3. \(f(x)=\left\{\begin{pmatrix}x, & x\neq 0\\ 5, & x=0\end{pmatrix}\right.\)	c. Limit undefined (infinite discontinuity)
4. \(f(x)=⌊x⌋\)	d. Step/jump behavior (e.g., −1 to 0)

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

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

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

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

💡 Short Explanation

\(1/x\) has infinite discontinuity Unequal one-sided limits create jumps Mismatched value gives removable discontinuity

✅ Correct Answer Explanation

\(1/x\) has infinite discontinuity at 0 → c. The piecewise constant function has unequal LHL and RHL → a. The third function has existing limit 0 but \(f(0)=5\)→ removable discontinuity → b. The greatest integer function shows step behavior → d. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly identifies removable and infinite discontinuities.
Option C → Wrongly classifies the discontinuity types of multiple functions.
Option D → The greatest integer function exhibits step behavior, not ordinary jump matching here.

🎯 Strategy

Used: Option Grouping

Application:

Associate each standard discontinuity type with its defining feature.

Final Logic:

Correct classification uniquely determines the matching.

🧠 Memory Trick

"∞ → Infinite, unequal → Jump, mismatch → Removable"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Types of Discontinuity

10 Which of the following functions have exactly one point of discontinuity?
i. \(f(x)=1/x\)
ii. \(f(x)=\left\{\begin{pmatrix}1, & x\neq 2\\ 5, & x=2\end{pmatrix}\right.\)
iii. \(f(x)=∣x∣\)

A i and ii only

B ii and iii only

C i and iii only

D ii only

💡 Short Explanation

\(1/x\) discontinuous only at 0 Second function discontinuous only at 2 |x| is continuous everywhere

✅ Correct Answer Explanation

\(1/x\) has exactly one discontinuity at x=0. The second function is discontinuous only at x=2 because the limit equals 1 while \(f(2)=5\). The modulus function |x| is continuous for all real numbers. Hence functions i and ii each have exactly one discontinuity. Therefore option A is correct, not D.

❓ Why Other Options Are Incorrect

Option B → |x| is continuous everywhere and has no discontinuity.
Option C → Includes |x| incorrectly while omitting the second discontinuous function.
Option D → Ignores the discontinuity of \(1/x\) at x=0.

🎯 Strategy

Used: Elimination

Application:

Check discontinuity points individually for each function.

Final Logic:

Both i and ii each contain exactly one discontinuity point.

🧠 Memory Trick

"1/x breaks at 0 only"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Points of Discontinuity

11 Identify the INCORRECT statement:

A A continuous graph can be drawn without lifting the pen

B A break in the graph indicates discontinuity

C A discontinuous function requires lifting the pen

D A continuous graph always requires lifting the pen

💡 Short Explanation

Continuous graphs are unbroken Pen lifting indicates breaks Continuity means smooth connection

✅ Correct Answer Explanation

A continuous graph can generally be drawn without lifting the pen, so option D is incorrect. Options A and B correctly describe graphical continuity and discontinuity. Option C is also correct because breaks in discontinuous graphs force interruption while drawing. NCERT informally explains continuity through uninterrupted graph tracing.

❓ Why Other Options Are Incorrect

Option A → Correct because continuity implies no breaks or gaps in the graph.
Option B → Correct since a graphical break represents discontinuity.
Option C → Correct because discontinuous graphs cannot usually be drawn in one stroke.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "always" signals an absolute statement likely to be incorrect.

Final Logic:

Continuous graphs do not require lifting the pen.

🧠 Memory Trick

"Continuous = One-stroke graph"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Informal Idea of Continuity

12 A function has a jump discontinuity at x=a.
If LHL = 5 and RHL = 12, what is the magnitude of the jump?

A 7

B 5

C 12

D 17

💡 Short Explanation

Jump size equals difference of limits Subtract LHL from RHL \(12-5=7\)

✅ Correct Answer Explanation

The magnitude of a jump discontinuity is: \(∣RHL-LHL∣\) Given: \(RHL=12,LHL=5\) So, \(∣12-5∣=7\) Hence option A is correct. The jump represents the vertical gap between left-side and right-side limiting values.

❓ Why Other Options Are Incorrect

Option B → Uses only one-sided limit instead of their difference.
Option C → Incorrectly takes the right-hand limit itself as jump magnitude.
Option D → Adds the limits instead of subtracting them.

🎯 Strategy

Used: Substitution

Application:

Directly apply the jump discontinuity formula.

Final Logic:

Jump magnitude equals absolute difference of one-sided limits.

🧠 Memory Trick

"Jump = Right − Left"

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

13 If f(x) and g(x) are continuous functions, which is guaranteed to be continuous wherever defined?

A \(\frac{f(x)}{g(x)}\)

B \(\sqrt{f(x)}\)

C \(\frac{f(x)}{g(x)}\), provided \(g(x)\neq 0\)

D Only constant functions

💡 Short Explanation

Quotient continuity needs nonzero denominator Continuous functions preserve operations Undefined points must be excluded

✅ Correct Answer Explanation

The quotient of two continuous functions is continuous wherever the denominator is nonzero. Therefore: \(\frac{f(x)}{g(x)}\) is continuous only when \(g(x)\neq 0\). Hence option C is correct. Option A ignores denominator restrictions. Option B is not always defined for negative inputs. Option D is too restrictive.

❓ Why Other Options Are Incorrect

Option A → Division by zero causes discontinuity or undefined behavior.
Option B → Square roots require nonnegative inputs for real-valued continuity.
Option D → Many nonconstant functions are continuous, including polynomials and trigonometric functions.

🎯 Strategy

Used: Elimination

Application:

Remove options missing important domain conditions.

Final Logic:

Quotients remain continuous only where denominator is nonzero.

🧠 Memory Trick

"Quotient continuous if denominator ≠ 0"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Algebra of Continuous Functions

14 Why is the function \(f(x)=∣x^{2}-1∣\)continuous?

A It is undefined at some points

B Subtraction causes discontinuity

C Modulus is continuous only for positive inputs

D It is a composition of continuous functions

💡 Short Explanation

Polynomial functions are continuous Modulus function is continuous Composition preserves continuity

✅ Correct Answer Explanation

The function \(x^{2}-1\) is a polynomial and therefore continuous everywhere. The modulus function is also continuous. The composition: \(∣x^{2}-1∣\) remains continuous for all real x. Hence option D is correct. The function is defined everywhere and modulus works for both positive and negative inputs.

❓ Why Other Options Are Incorrect

Option A → The function is defined for every real number.
Option B → Subtraction of continuous functions preserves continuity.
Option C → Modulus is continuous for all real inputs, not just positive values.

🎯 Strategy

Used: Option Grouping

Application:

Use continuity properties of polynomial and modulus functions together.

Final Logic:

Composition of continuous functions remains continuous.

🧠 Memory Trick

"Continuous inside + continuous outside"

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

15 Let \(f(x)=k\) be constant. If \(f(3)=10\), find k.

A 10

B −2

C 2

D 4

💡 Short Explanation

Constant functions have same value everywhere Given function value fixes constant Thus k equals 10

✅ Correct Answer Explanation

For a constant function, \(f(x)=k\) for all x. Since: \(f(3)=10\) the constant value must be: \(k=10\) Hence option A is correct. The value does not depend on the input variable.

❓ Why Other Options Are Incorrect

Option B → Contradicts the given value \(f(3)=10\).
Option C → Does not satisfy the constant function condition.
Option D → Also inconsistent with the provided function value.

🎯 Strategy

Used: Substitution

Application:

Replace x=3 into the constant function definition.

Final Logic:

Constant function value remains the same for every x.

🧠 Memory Trick

"Constant means same everywhere"

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

16 For \(f(x)=x\), evaluate:
\({lim⁡}_{x\rightarrow a}f(x)\)

A 0

B a

C 1

D Undefined

💡 Short Explanation

Identity function returns input itself As x approaches a, output approaches a Hence limit equals a

✅ Correct Answer Explanation

For the identity function: \(f(x)=x\) the output equals the input. Therefore as x approaches a, \({lim⁡}_{x\rightarrow a}x=a\) Hence option B is correct. The identity function is continuous everywhere and preserves approaching values directly.

❓ Why Other Options Are Incorrect

Option A → The limit equals a generally, not always zero.
Option C → The limit is not fixed at 1 unless \(a=1\).
Option D → The limit clearly exists for all real a.

🎯 Strategy

Used: Substitution

Application:

Observe direct input-output equality in the identity function.

Final Logic:

Approaching x=a gives output approaching a.

🧠 Memory Trick

"Identity keeps input unchanged"

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

17 Which function is continuous for all real x?

A \(1/x\)

B \(\sqrt{x}\)

C \(x^{2}\)

D \(\ln\,x\)

💡 Short Explanation

Polynomials are continuous everywhere Other functions have domain restrictions \(x^{2}\)defined for all real x

✅ Correct Answer Explanation

Polynomial functions are continuous for all real numbers. Hence \(x^{2}\)is continuous everywhere on R. The function \(1/x\) is undefined at 0, \(\sqrt{x}\)is defined only for \(x\geq 0\), and \(\ln\,x\) is defined only for positive x.

❓ Why Other Options Are Incorrect

Option A → Undefined at x=0, causing discontinuity.
Option B → Not defined for negative real numbers.
Option D → Logarithmic functions exist only for positive inputs.

🎯 Strategy

Used: Elimination

Application:

Check domain restrictions of each function.

Final Logic:

Only the polynomial is continuous over all real numbers.

🧠 Memory Trick

"Polynomials are always continuous"

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

18 If a function is defined only at x=5, it is:

A Discontinuous

B Undefined

C Non-real

D Continuous

💡 Short Explanation

Singleton domains have no nearby points Continuity conditions become automatic Vacuous continuity applies

✅ Correct Answer Explanation

If the domain contains only one point, there are no neighboring domain points approaching it. Hence continuity conditions are satisfied trivially. Therefore the function is considered continuous at that point. Thus option D is correct according to the formal definition of continuity on restricted domains.

❓ Why Other Options Are Incorrect

Option A → No discontinuity arises because there are no nearby domain points.
Option B → The function is explicitly defined at x=5.
Option C → Continuity has no relation to being "non-real."

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply the definition of continuity to a singleton domain.

Final Logic:

No nearby points means continuity holds automatically.

🧠 Memory Trick

"One-point domain → Continuous"

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

19 For continuity on [a,b], which is NOT required?

A Continuity outside [a,b]

B Continuity at x=a

C Continuity at x=b

D Continuity for all \(x\in (a,b)\)

💡 Short Explanation

Closed interval continuity concerns interval only Outside behavior is irrelevant Endpoints and interior matter

✅ Correct Answer Explanation

To be continuous on [a,b], a function must be continuous at all interior points and satisfy endpoint continuity conditions at a and b. Behavior outside the interval does not matter. Therefore option A is correct because continuity beyond the interval is unnecessary.

❓ Why Other Options Are Incorrect

Option B → Left endpoint continuity is required on closed intervals.
Option C → Right endpoint continuity is also required.
Option D → Interior continuity is essential throughout the interval.

🎯 Strategy

Used: Odd One Out

Application:

Identify the option unrelated to continuity within the specified interval.

Final Logic:

Outside continuity is irrelevant for interval continuity.

🧠 Memory Trick

"Inside interval matters only"

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

20 For a function on [-2,2], continuity at x=2 requires checking:

A Right-hand limit at 2

B Left-hand limit at 2

C Limit at 0

D Continuity at −2

💡 Short Explanation

Right endpoint has values only to left Endpoint continuity uses one-sided limit Left-hand approach is required

✅ Correct Answer Explanation

At the right endpoint x=2 of the interval [-2,2], values exist only from the left side within the domain. Therefore continuity is checked using: \({lim⁡}_{x\rightarrow 2^{-}}f(x)=f(2)\) Hence option B is correct. Right-hand limits are outside the interval domain.

❓ Why Other Options Are Incorrect

Option A → No domain points exist to the right of 2.
Option C → Continuity at x=0 is unrelated to endpoint continuity at x=2.
Option D → Continuity at −2 is a separate condition.

🎯 Strategy

Used: Elimination

Application:

Remove options involving unavailable domain directions.

Final Logic:

Right endpoint continuity requires the left-hand limit only.

🧠 Memory Trick

"Right endpoint → Left limit"

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

UG Mathematics Booster Test 2 - Continuity Fundamentals

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