UG Mathematics Booster Test 3 - Continuity Fundamentals -

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UG Mathematics Booster Test 3 - Continuity Fundamentals

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 3 - Continuity Fundamentals

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QUESTION 1 OF 20

If a function is continuous on a closed interval [a,b] (i.e., its graph can be drawn without lifting the pen), then the function is necessarily:

Answer Review

1 If a function is continuous on a closed interval [a,b] (i.e., its graph can be drawn without lifting the pen), then the function is necessarily:

A Un-integrable

B A straight line

C Integrable on [a,b]

D Symmetrical

💡 Short Explanation

Continuous functions are integrable Closed intervals ensure finite behavior NCERT links continuity with integration

✅ Correct Answer Explanation

A standard theorem states that every function continuous on a closed interval [a,b] is integrable there. Hence option C is correct. Continuity does not imply the graph is a straight line or symmetric. Option A is opposite to the theorem because continuity guarantees integrability.

❓ Why Other Options Are Incorrect

Option A → Continuous functions on closed intervals are always integrable.
Option B → Continuous functions may be curved, polynomial, trigonometric, or exponential.
Option D → Symmetry is unrelated to continuity or integrability.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Match the statement with the standard continuity-integrability theorem.

Final Logic:

Continuous on closed interval implies integrable.

🧠 Memory Trick

"Continuous ⇒ Integrable"

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

2 Arrange the following discontinuities at x=a in order of increasing severity:
1. Sharp corner (continuous but not differentiable)
2. Vertical asymptote (infinite discontinuity)
3. Removable hole
4. Finite jump

A 1, 2, 3, 4

B 2, 4, 3, 1

C 3, 1, 4, 2

D 1, 3, 4, 2

💡 Short Explanation

Removable holes are mildest Infinite discontinuities are most severe Jump discontinuity exceeds sharp corner

✅ Correct Answer Explanation

A removable hole is the mildest defect because continuity can be restored by redefining one value. A sharp corner still preserves continuity. A finite jump destroys limit equality. A vertical asymptote creates unbounded behavior and is most severe. Therefore the correct increasing order is 3, 1, 4, 2.

❓ Why Other Options Are Incorrect

Option A → Places infinite discontinuity too early and removable discontinuity too late.
Option B → Starts with the most severe discontinuity incorrectly.
Option D → Sharp corners are less severe than removable discontinuities in continuity context.

🎯 Strategy

Used: Option Grouping

Application:

Rank discontinuities based on continuity and boundedness behavior.

Final Logic:

Severity increases from removable to infinite discontinuity.

🧠 Memory Trick

"Hole → Corner → Jump → Infinity"

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

3 If a function satisfies
\({lim⁡}_{x\rightarrow 2^{-}}f(x)=4\)
and a sequence approaches 2 from the left, what is the probability that the sequence converges to 4?

A 1

B 0

C 0.5

D Depends on sequence

💡 Short Explanation

Left-hand limit fixes nearby behavior All left-approaching sequences converge similarly Probability becomes certain

✅ Correct Answer Explanation

If \({lim⁡}_{x\rightarrow 2^{-}}f(x)=4\) then every sequence approaching 2 from the left produces function values approaching 4. Hence convergence to 4 is guaranteed. Therefore the probability is 1. This follows directly from the sequential interpretation of limits.

❓ Why Other Options Are Incorrect

Option B → Contradicts the definition of left-hand limit.
Option C → Limit behavior is deterministic, not partial probability.
Option D → The limit condition guarantees convergence for all left-approaching sequences.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply the sequential definition of limits.

Final Logic:

Left-hand limit guarantees convergence to the same value.

🧠 Memory Trick

"Limit fixed → Sequence fixed"

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

4 Let
\(\vec{r}(t)=⟨cos⁡t,sin⁡t⟩\)
Find the magnitude of the vector.

A 1

B \(\sqrt{2}\)

C 2

D 0

💡 Short Explanation

Use vector magnitude formula Apply trigonometric identity \({cos⁡}^{2}t+{sin⁡}^{2}t=1\)

✅ Correct Answer Explanation

The magnitude is: \(∣\vec{r}(t)∣=\sqrt{{cos⁡}^{2}t+{sin⁡}^{2}t}\) Using the identity: \({cos⁡}^{2}t+{sin⁡}^{2}t=1\) we get: \(∣\vec{r}(t)∣=1\) Hence option A is correct. The vector lies on the unit circle for all t.

❓ Why Other Options Are Incorrect

Option B → Would arise only if the sum inside the root were 2.
Option C → Magnitude never becomes 2 for unit circle coordinates.
Option D → Zero magnitude would require both coordinates to vanish simultaneously.

🎯 Strategy

Used: Substitution

Application:

Apply the standard vector magnitude formula directly.

Final Logic:

Trigonometric identity reduces magnitude to 1.

🧠 Memory Trick

"sin² + cos² = 1"

📚 Class XI – NCERT Mathematics – Chapter: Introduction to Three Dimensional Geometry – Topic: Magnitude of Vector

5 If a function has a removable discontinuity at x=a, what happens to the definite integral
\(\int_{p}^{q}\,f(x) dx\)
?

A Becomes infinite

B Cannot be determined

C Remains unaffected by the single point discontinuity

D Equals zero

💡 Short Explanation

Single points contribute zero area Removable discontinuities do not alter integral Integration depends on interval behavior

✅ Correct Answer Explanation

Changing or removing a function value at a single point does not affect the definite integral because a single point contributes zero area. Therefore removable discontinuities leave the integral unchanged. Hence option C is correct. Infinite area or undefined integration does not arise from isolated removable discontinuities.

❓ Why Other Options Are Incorrect

Option A → Removable discontinuities do not create infinite area.
Option B → The integral is still well-defined despite a single removable hole.
Option D → A single discontinuity does not force the entire integral to become zero.

🎯 Strategy

Used: Elimination

Application:

Recognize that isolated points have zero contribution to area.

Final Logic:

Single-point discontinuities do not change definite integrals.

🧠 Memory Trick

"One point → Zero area effect"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integrability with Removable Discontinuity

6 A model evaluates
\({lim⁡}_{x\rightarrow 2}(x^{2})\)
For continuity at x=2, what must be the value of f(2)?

A 0

B 1

C Undefined

D 4

💡 Short Explanation

Continuity requires limit equals value \(2^{2}=4\) Function value must match limit

✅ Correct Answer Explanation

For continuity: \(f(2)={lim⁡}_{x\rightarrow 2}x^{2}\) Since: \({lim⁡}_{x\rightarrow 2}x^{2}=4\) the function value must also be 4. Hence option D is correct. Any other value would violate the continuity condition.

❓ Why Other Options Are Incorrect

Option A → Does not equal the computed limit.
Option B → Incorrect evaluation of the quadratic limit.
Option C → Continuity requires the function to be defined at x=2.

🎯 Strategy

Used: Substitution

Application:

Substitute x=2 into the polynomial expression.

Final Logic:

Function value must equal the limit 4.

🧠 Memory Trick

"Continuous ⇒ Value = Limit"

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

7 Assertion (A):
The greatest integer function \(f(x)=⌊x⌋\)has equal LHL and RHL at every integer x=n.
Reason (R):
At an integer n,
\({lim⁡}_{x\rightarrow n^{-}}f(x)=n-1,{lim⁡}_{x\rightarrow n^{+}}f(x)=n\)

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Greatest integer function jumps at integers One-sided limits differ Hence continuity fails at integers

✅ Correct Answer Explanation

For the greatest integer function: \({lim⁡}_{x\rightarrow n^{-}}⌊x⌋=n-1\) and \({lim⁡}_{x\rightarrow n^{+}}⌊x⌋=n\) Since LHL and RHL are unequal, the function has jump discontinuities at integers. Thus the Assertion is false while the Reason is true. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → The Reason statement is mathematically correct.
Option B → Assertion is false because one-sided limits are unequal.
Option C → The Reason actually disproves the Assertion rather than explaining it.

🎯 Strategy

Used: Substitution

Application:

Evaluate one-sided limits near integer points.

Final Logic:

Unequal one-sided limits imply discontinuity.

🧠 Memory Trick

"GIF jumps at integers"

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

8 Let
\(f(x)=\left\{\begin{pmatrix}x^{2}, & x\neq 2\\ k, & x=2\end{pmatrix}\right.\)
To make f(x) continuous at x=2, what must k equal?
i. \({lim⁡}_{x\rightarrow 2}x^{2}\)
ii. 4
iii. \(f(2)\)

A i only

B i, ii, and iii

C ii and iii only

D i and ii only

💡 Short Explanation

Continuity requires common equality Limit equals 4 Function value must also equal 4

✅ Correct Answer Explanation

For continuity: \(f(2)={lim⁡}_{x\rightarrow 2}x^{2}\) Since: \({lim⁡}_{x\rightarrow 2}x^{2}=4\) we must have: \(k=4=f(2)\) Thus statements i, ii, and iii all represent the same required value. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores the explicit numerical value 4 and function value condition.
Option C → Omits the equivalent limit expression.
Option D → Excludes the equality with \(f(2)\), essential for continuity.

🎯 Strategy

Used: Option Grouping

Application:

Identify equivalent continuity statements.

Final Logic:

Limit, numerical value, and function value must coincide.

🧠 Memory Trick

"Limit = Value = 4"

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

9 The function
\(f(x)=\frac{x^{2}-4}{x-2}\)
is discontinuous at x=2.
To remove the discontinuity, define \(f(2)=k\). Find k.

A 8

B 0

C 4

D 16

💡 Short Explanation

Factor numerator expression Simplify function to x+2 Substitute x=2 to remove discontinuity

✅ Correct Answer Explanation

\(\frac{x^{2}-4}{x-2}=\frac{\left(x-2)(x+2\right)}{x-2}=x+2,x\neq 2\) Thus, \({lim⁡}_{x\rightarrow 2}f(x)=2+2=4\) To remove the discontinuity, define: \(f(2)=4\) Hence option C is correct. The provided answer A is incorrect.

❓ Why Other Options Are Incorrect

Option A → Incorrect arithmetic; the limit evaluates to 4, not 8.
Option B → Does not match the limiting value near x=2.
Option D → No valid simplification produces 16.

🎯 Strategy

Used: Substitution

Application:

Factor and simplify before evaluating the limit.

Final Logic:

Continuity requires assigning the limit value at x=2.

🧠 Memory Trick

"Cancel first, then substitute"

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

10 Match each function with its discontinuity set:

List I	List II
1. \(f(x)=1/x\)	a. x=0 only
2. \(f(x)=x^{2}\)	b. No discontinuity
3. \(f(x)=⌊x⌋\)	c. \(x=\pi /2+n\pi\)
4. \(f(x)=tan⁡x\)	d. All integers

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

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

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

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

💡 Short Explanation

Reciprocal undefined at zero Polynomials are continuous Greatest integer jumps at integers

✅ Correct Answer Explanation

\(1/x\) is discontinuous only at x=0 → a. \(x^{2}\) is continuous everywhere → b. The greatest integer function jumps at all integers → d. \(\tan\,x\) is discontinuous at: \(x=\pi /2+n\pi\) → c. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly assigns discontinuity sets for GIF and tan x.
Option B → Misclassifies polynomial continuity and reciprocal discontinuity.
Option C → Completely mismatches standard discontinuity locations.

🎯 Strategy

Used: Option Grouping

Application:

Match each standard function with its known discontinuity behavior.

Final Logic:

Known domain restrictions determine discontinuity sets.

🧠 Memory Trick

"1/x → 0, GIF → Integers, tan → π/2+nπ"

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

11 Identify the INCORRECT statement linking graph drawing to calculus:

A If a curve can be drawn without lifting the pen, the function is everywhere differentiable

B Drawing without lifting the pen is a heuristic for continuity (limit equals function value)

C The graph of \(f(x)=x^{2}\)can be drawn without lifting the pen, showing continuity

D Sharp points do not require lifting the pen, hence continuity is preserved

💡 Short Explanation

Continuity does not guarantee differentiability Sharp corners may exist in continuous graphs Drawing rule indicates continuity only

✅ Correct Answer Explanation

A graph drawable without lifting the pen suggests continuity, not differentiability. Functions like \(f(x)=∣x∣\)are continuous but not differentiable at sharp points. Therefore option A is incorrect. Options B, C, and D correctly connect graphical continuity with calculus concepts.

❓ Why Other Options Are Incorrect

Option B → Correct because uninterrupted drawing is a standard intuitive idea of continuity.
Option C → The parabola \(x^{2}\)is continuous and can indeed be drawn continuously.
Option D → Sharp points may destroy differentiability but continuity still remains intact.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "everywhere differentiable" makes the statement too strong.

Final Logic:

Continuity does not always imply differentiability.

🧠 Memory Trick

"Continuous ≠ Differentiable"

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

12 Consider the function
\(f(x)=\left\{\begin{pmatrix}x, & x<1\\ x+2, & x\geq 1\end{pmatrix}\right.\)
What occurs at x=1?

A Removable hole

B Vertical jump gap of 2 units

C Vertical asymptote

D Sharp corner

💡 Short Explanation

Left and right values differ Difference equals 2 units Jump discontinuity occurs

✅ Correct Answer Explanation

For \(x<1\), \({lim⁡}_{x\rightarrow 1^{-}}f(x)=1\) For \(x\geq 1\), \({lim⁡}_{x\rightarrow 1^{+}}f(x)=3\) The difference is: \(3-1=2\) Hence a vertical jump discontinuity of 2 units occurs at x=1. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Limits from both sides are unequal, so the discontinuity is not removable.
Option C → No unbounded behavior occurs near x=1.
Option D → Sharp corners preserve continuity, unlike this jump discontinuity.

🎯 Strategy

Used: Substitution

Application:

Evaluate LHL and RHL separately at x=1.

Final Logic:

Unequal one-sided limits create a jump discontinuity.

🧠 Memory Trick

"Unequal limits → Jump"

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

13 The proof that polynomials are continuous relies on which property of limits?
\(lim⁡(x^{n})=(lim⁡x)^{n} for all n\in N\)

A Denominators never become zero

B Polynomials behave like exponentials

C Polynomials are bounded

D Limits preserve sums and products:

💡 Short Explanation

Polynomial terms use powers and sums Limit laws preserve algebraic operations Hence continuity follows directly

✅ Correct Answer Explanation

Polynomials are built from sums, products, and powers of x. Limit laws ensure: \(lim⁡(x^{n})=(lim⁡x)^{n}\) and preserve addition and multiplication. Therefore polynomial limits equal polynomial values, proving continuity. Hence option D is correct. The other options are unrelated or incomplete.

❓ Why Other Options Are Incorrect

Option A → Zero denominators concern rational functions, not general polynomials.
Option B → Polynomial continuity does not depend on exponential behavior.
Option C → Boundedness is unnecessary because many polynomials are unbounded.

🎯 Strategy

Used: Option Grouping

Application:

Identify which limit property directly supports polynomial continuity proofs.

Final Logic:

Algebraic limit laws guarantee polynomial continuity.

🧠 Memory Trick

"Limits preserve algebra"

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

14 If f(x) and g(x) are continuous, why is \(∣f(g(x))∣\)continuous?

A Composition of continuous functions remains continuous

B Sine overrides modulus

C Domain becomes positive only

D Modulus is differentiable everywhere

💡 Short Explanation

Composition preserves continuity Modulus function is continuous Continuous inside gives continuous outside

✅ Correct Answer Explanation

If \(g(x)\)is continuous and \(f(x)\)is continuous, then the composite function \(f(g(x))\)is continuous. Since modulus is also continuous, \(∣f(g(x))∣\)remains continuous. Hence option A is correct. Differentiability and positivity conditions are unnecessary here.

❓ Why Other Options Are Incorrect

Option B → Sine is unrelated to the given function.
Option C → Modulus accepts both positive and negative inputs.
Option D → Modulus is not differentiable at 0, though it is continuous there.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply the continuity rule for composition of functions.

Final Logic:

Continuous compositions remain continuous.

🧠 Memory Trick

"Continuous inside → Continuous outside"

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

15 Let \(f(x)=4\)(constant) and \(g(x)\)be any function. Evaluate:
\({lim⁡}_{x\rightarrow a}f(g(x))\)

A 0

B 4

C Does not exist

D Infinite

💡 Short Explanation

Constant functions never change Composition still remains constant Limit equals the same constant

✅ Correct Answer Explanation

Since \(f(x)=4\) for every input, \(f(g(x))=4\) regardless of the function \(g(x)\). Therefore: \({lim⁡}_{x\rightarrow a}f(g(x))=4\) Hence option B is correct. Constant functions preserve the same value under all compositions.

❓ Why Other Options Are Incorrect

Option A → The constant value is 4, not 0.
Option C → Constant functions always possess finite limits.
Option D → No unbounded behavior occurs for a constant function.

🎯 Strategy

Used: Substitution

Application:

Replace the composite expression with its constant value.

Final Logic:

Constant functions keep the same limit everywhere.

🧠 Memory Trick

"Constant stays constant"

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

16 The identity function \(f(x)=x\) is a straight line. Its inverse function is:

A Discontinuous everywhere

B A horizontal constant line

C Exactly the same continuous identity function

D A parabola

💡 Short Explanation

Identity maps each number to itself Inverse reverses same mapping Hence inverse remains identical

✅ Correct Answer Explanation

For the identity function: \(f(x)=x\) every input equals its output. Reversing this mapping produces the same function: \(f^{-1}(x)=x\) Thus the inverse is again the continuous identity function. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Identity functions are continuous everywhere.
Option B → The inverse is not a constant function.
Option D → A parabola is not the inverse of the line \(y=x\).

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret inverse functions as reversal of input-output mapping.

Final Logic:

Identity reversed remains identity.

🧠 Memory Trick

"Identity inverse = Identity"

📚 Class XII – NCERT Mathematics – Chapter: Inverse Trigonometric Functions – Topic: Inverse Function Basics

17 The function \(f(x)=e^{x}\)is continuous for all real x. Its inverse function \(f^{-1}(x)=ln⁡x\) is:

A Continuous for all real numbers

B Discontinuous everywhere

C Continuous only for \(x\leq 0\)

D Continuous at all points in its domain \(\left(0,\ \infty \right)\)

💡 Short Explanation

\(\ln\,x\) defined only for positive x Continuous throughout its domain Inverse preserves continuity locally

✅ Correct Answer Explanation

The logarithmic function: \(\ln\,x\) is defined only for: \(x>0\) and is continuous on this interval. Therefore option D is correct. It is not continuous for all real numbers because it is undefined for nonpositive values.

❓ Why Other Options Are Incorrect

Option A → \(\ln\,x\) is undefined for \(x\leq 0\).
Option B → Logarithmic functions are continuous wherever defined.
Option C → \(\ln\,x\) is not defined for nonpositive numbers.

🎯 Strategy

Used: Elimination

Application:

Check the domain of the logarithmic function carefully.

Final Logic:

Continuity holds only on \(\left(0,\ \infty \right)\).

🧠 Memory Trick

"Log lives only for positives"

📚 Class XII – NCERT Mathematics – Chapter: Exponential and Logarithmic Functions – Topic: Continuity of Logarithmic Functions

18 If a function is defined on a discrete set (e.g., \(\left\{1,\ 2,\ 3\right\}\)), then under the topological definition of continuity it is:

A Continuous at all points in its domain

B Discontinuous at every point

C Impossible to analyze

D Not a valid function

💡 Short Explanation

Discrete domains isolate points No nearby domain points exist Continuity holds automatically

✅ Correct Answer Explanation

In a discrete domain, every point is isolated. Since there are no nearby domain points approaching a chosen point, continuity conditions become automatically satisfied. Therefore every function on a discrete set is continuous at all domain points. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Discontinuity requires neighboring approach behavior, absent in discrete sets.
Option C → Such functions are mathematically valid and analyzable.
Option D → Functions on finite or discrete domains are perfectly valid.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply the topological idea of isolated points.

Final Logic:

Isolated domain points imply automatic continuity.

🧠 Memory Trick

"Discrete points → Automatic continuity"

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

19 A function continuous on a closed interval [a,b] guarantees which property?

A It must cross the x-axis

B It is bounded and attains absolute maxima and minima (Extreme Value Theorem)

C It must be strictly increasing

D It has infinite area

💡 Short Explanation

Continuous closed-interval functions are bounded Maximum and minimum values exist Standard Extreme Value Theorem applies

✅ Correct Answer Explanation

The Extreme Value Theorem states that every continuous function on a closed interval [a,b] is bounded and attains both absolute maximum and minimum values. Hence option B is correct. Crossing the x-axis or monotonicity is not guaranteed.

❓ Why Other Options Are Incorrect

Option A → Continuous functions need not intersect the x-axis.
Option C → Continuity does not imply increasing behavior.
Option D → Continuous functions on closed intervals have finite bounded area behavior.

🎯 Strategy

Used: Option Grouping

Application:

Identify the theorem specifically associated with closed-interval continuity.

Final Logic:

EVT guarantees bounded extrema for continuous functions.

🧠 Memory Trick

"Closed + Continuous → Max & Min"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Extreme Value Theorem

20 If f is defined on [a,b], what is required for continuity at the right endpoint x=b?

A Check the right-hand limit at b

B Check both LHL and RHL at b

C \({lim⁡}_{x\rightarrow b^{-}}f(x)=f(b)\)

D The function is automatically discontinuous at endpoints

💡 Short Explanation

Endpoint continuity uses one-sided limits Right endpoint allows left approach only Function value must match LHL

✅ Correct Answer Explanation

At the right endpoint b of the interval [a,b], domain points exist only to the left. Therefore continuity is checked using: \({lim⁡}_{x\rightarrow b^{-}}f(x)=f(b)\) Hence option C is correct. Right-hand limits are unnecessary because no domain points lie beyond b.

❓ Why Other Options Are Incorrect

Option A → Right-hand limits are outside the interval domain.
Option B → Two-sided limits are not required at endpoints.
Option D → Endpoints can satisfy continuity using one-sided limits.

🎯 Strategy

Used: Elimination

Application:

Remove options requiring unavailable right-side domain points.

Final Logic:

Right endpoint continuity depends on 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 3 - Continuity Fundamentals

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