UG Mathematics Booster Test 2 - Continuity Analysis & Functions -

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UG Mathematics Booster Test 2 - Continuity Analysis & Functions

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 2 - Continuity Analysis & Functions

1 / 20

QUESTION 1 OF 20

📄 Passage

We analyze the reciprocal function
f(x)=1/x

near x=0. As x approaches 0 from the right (e.g., 0.1, 0.01), the values of f(x)increase as 10, 100, … . In symbols,
(lim⁡)┬(x→0^+ ) 1/x=+∞

We emphasize that +∞is not a real number, hence the limit does not exist as a real number.
Similarly, as x→0^-,
(lim⁡)┬(x→0^- ) 1/x=-∞

Thus, the function is discontinuous at x=0.

Unlike the function described in the passage, a polynomial function \(p(x)\):

Answer Review

📄 Passage

Unlike the function described in the passage, a polynomial function \(p(x)\):

A Has infinite limits at x=0

B Is continuous at every real number

C Is discontinuous at x=0

D Has a broken graph

💡 Short Explanation

Polynomials have no denominator restrictions Polynomial graphs are unbroken Continuous for all real x

✅ Correct Answer Explanation

Unlike the reciprocal function \(\frac{1}{x}\), polynomial functions are continuous for every real number. They do not have infinite limits or undefined points caused by zero denominators. Hence option B is correct. Polynomial graphs remain smooth and connected without breaks across the real line.

❓ Why Other Options Are Incorrect

Option A → Infinite limits arise in reciprocal-type functions, not polynomials.
Option C → Polynomials are continuous at x=0 and everywhere else.
Option D → Polynomial graphs do not contain breaks or jumps.

🎯 Strategy

Used: Odd One Out

Application:

Compare reciprocal discontinuity with polynomial continuity.

Final Logic:

Polynomials remain continuous everywhere on ℝ.

🧠 Memory Trick

"Polynomial = No breaks"

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

📄 Passage

The function in the passage is a rational function
\(f(x)=\frac{1}{x}\)
Every rational function is continuous at all points except:

A All integers

B All rational numbers

C All irrational numbers

D Points where the denominator is zero

💡 Short Explanation

Rational functions fail at denominator zero Else continuity is preserved Undefined points create discontinuities

✅ Correct Answer Explanation

A rational function is continuous wherever it is defined. Discontinuities occur only where the denominator becomes zero because division by zero is undefined. Hence option D is correct. The function \(\frac{1}{x}\)fails specifically at x=0 for this reason.

❓ Why Other Options Are Incorrect

Option A → Integers do not automatically create discontinuity.
Option B → Rational numbers can belong to the continuity domain.
Option C → Irrational numbers do not inherently cause discontinuity.

🎯 Strategy

Used: Elimination

Application:

Identify the only mathematically valid discontinuity condition.

Final Logic:

Rational functions break only at denominator zeros.

🧠 Memory Trick

"Denominator 0 → Discontinuity"

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

3 Match the expressions with their values:

List I	List II
1. \({lim⁡}_{x\rightarrow 0}\frac{sin⁡5x}{x}\)	a. 2
2. \({lim⁡}_{x\rightarrow 0}\frac{\sin\,x}{3x}\)	b. -1
3. \({lim⁡}_{x\rightarrow 0}\frac{sin⁡(x+\pi )}{x}\)	c. 5
4. \({lim⁡}_{x\rightarrow 0}\frac{sin⁡2x}{x}\)	d. \(1/3\)

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

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

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

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

💡 Short Explanation

Use standard \(lim⁡\frac{\sin\,x}{x}=1\) Factor constants carefully Apply trigonometric identities

✅ Correct Answer Explanation

Using: \({lim⁡}_{x\rightarrow 0}\frac{\sin\,x}{x}=1{lim⁡}_{x\rightarrow 0}\frac{sin⁡5x}{x}=5{lim⁡}_{x\rightarrow 0}\frac{\sin\,x}{3x}=\frac{1}{3}\) Since: \(sin⁡(x+\pi )=-sin⁡x\) the third limit becomes −1. Also: \({lim⁡}_{x\rightarrow 0}\frac{sin⁡2x}{x}=2\) Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrectly matches limits involving constants and signs.
Option C → Misuses standard trigonometric limit values.
Option D → Incorrect mapping for second and fourth expressions.

🎯 Strategy

Used: Substitution

Application:

Reduce every expression to the standard sine limit form.

Final Logic:

Apply \(lim⁡\frac{\sin\,x}{x}=1\) systematically.

🧠 Memory Trick

"sin(kx)/x → k"

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

4 If \(f(x)\)and \(g(x)\)are continuous trigonometric functions, which of the following are continuous?
I. \(f(x)+g(x)\)
II. \(f(x)⋅g(x)\)
III. \(\frac{f(x)}{g(x)}\), where \(g(x)\neq 0\)

A I, II, and III

B I and II only

C II and III only

D I only

💡 Short Explanation

Continuous functions preserve operations Sum and product remain continuous Quotient continuous if denominator nonzero

✅ Correct Answer Explanation

NCERT states that sums, products, and quotients of continuous functions remain continuous wherever defined. Therefore: \(f+g,fg,\frac{f}{g}\) are continuous provided \(g(x)\neq 0\). Hence all three statements are correct, making option A the correct answer.

❓ Why Other Options Are Incorrect

Option B → Ignores quotient continuity condition.
Option C → Sum of continuous functions is also continuous.
Option D → Product and quotient continuity are also valid results.

🎯 Strategy

Used: Option Grouping

Application:

Apply algebra of continuous functions.

Final Logic:

Sum, product, and quotient preserve continuity.

🧠 Memory Trick

"Add, multiply, divide (≠0) → Continuous"

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

5 Identify the incorrect statement regarding
\(f(x)=\frac{1}{x}\)

A Left-hand limit at x=0 is \(-\infty\)

B Right-hand limit at x=0 is \(+\infty\)

C Infinity is treated as a real number

D The limit at x=0 does not exist as a real number

💡 Short Explanation

Infinity is not a real number Reciprocal limits become unbounded Hence ordinary limit does not exist

✅ Correct Answer Explanation

For: \(f(x)=\frac{1}{x}\) the left-hand limit is \(-\infty\) and the right-hand limit is \(+\infty\). However, infinity is not a real number, so the two-sided real limit does not exist. Hence option C is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A → Correct description of left-hand behavior near zero.
Option B → Correct description of right-hand behavior near zero.
Option D → The limit does not exist as a finite real number.

🎯 Strategy

Used: Extreme Word Filter

Application:

Check conceptual correctness of "infinity as real number."

Final Logic:

Infinity represents unbounded growth, not a real number.

🧠 Memory Trick

"∞ is not real"

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

6 If a function is modeled by
\(f(t)=\frac{1}{t}\)
At what value of t does the function become undefined, causing an infinite spike?

A t=1

B t=-1

C t=2

D t=0

💡 Short Explanation

Denominator zero causes undefined value Reciprocal spikes near zero Infinite discontinuity occurs at t=0

✅ Correct Answer Explanation

The reciprocal function: \(f(t)=\frac{1}{t}\) is undefined when the denominator equals zero. Therefore at: \(t=0\) the graph shows an infinite spike or vertical asymptote. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → \(\frac{1}{1}\)is finite and defined.
Option B → \(\frac{1}{-1}\)exists and equals −1.
Option C → \(\frac{1}{2}\)is finite and continuous.

🎯 Strategy

Used: Substitution

Application:

Identify where denominator becomes zero.

Final Logic:

Reciprocal functions fail at zero denominator.

🧠 Memory Trick

"1/t explodes at 0"

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

7 For the function
\(f(x)=\frac{1}{x}\)
consider the sequence \(x=\{0.1,0.01,0.001\}\). The corresponding values of \(f(x)\)are:

A Approaching 0 linearly

B Negative exponential values

C Increasing rapidly towards large positive values

D Constant sequence

💡 Short Explanation

Positive small numbers give large reciprocals Values increase rapidly near zero Sequence grows unbounded positively

✅ Correct Answer Explanation

For: \(x=0.1,0.01,0.001\) the reciprocal values are: \(10,100,1000\) which grow rapidly without bound. Hence the sequence approaches large positive values as x approaches zero from the positive side. Therefore option C is correct.

❓ Why Other Options Are Incorrect

Option A → Values move away from zero, not toward it.
Option B → Values remain positive, not negative.
Option D → Reciprocal values clearly change and increase.

🎯 Strategy

Used: Substitution

Application:

Directly compute reciprocal values of the sequence.

Final Logic:

Smaller positive denominators create larger positive reciprocals.

🧠 Memory Trick

"Tiny positive → Huge positive"

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

8 For
\(f(x)=\frac{1}{x}\)
as x approaches 0 from the left, the values of the function:

A Become large positive values

B Become smaller than any given negative real number

C Become exactly zero

D Fluctuate around 1

💡 Short Explanation

Left-side values remain negative Reciprocal magnitude grows infinitely Values decrease beyond all bounds

✅ Correct Answer Explanation

As x approaches 0 from the negative side, x becomes a very small negative number. Therefore: \(\frac{1}{x}\rightarrow -\infty\) meaning the function becomes smaller than every negative real number. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Positive infinity occurs only from the right side.
Option C → Reciprocal values become unbounded, not zero.
Option D → Values do not oscillate around 1.

🎯 Strategy

Used: Substitution

Application:

Use small negative numbers near zero.

Final Logic:

Negative tiny denominators produce large negative reciprocals.

🧠 Memory Trick

"Left of 0 → −∞"

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

9 Let
\(f(x)=\left\{\begin{pmatrix}x+2, & x\leq 1\\ x-2, & x>1\end{pmatrix}\right.\)
A point is chosen uniformly at random from the interval [0,2].
What is the probability that the function is discontinuous at that point?

A 0

B 1/2

C 1

D Undefined

💡 Short Explanation

Function discontinuous only at one point Single points have zero probability Continuous almost everywhere on interval

✅ Correct Answer Explanation

The function is discontinuous only at: \(x=1\) In a continuous interval, the probability of selecting one exact point is zero. Therefore: \(P(discontinuity)=0\) Hence option A is correct. The provided answer B is incorrect.

❓ Why Other Options Are Incorrect

Option B → A single point cannot occupy half the interval probability.
Option C → The function is continuous at all other points.
Option D → Uniform probability on intervals is properly defined.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Compare point length with total interval length.

Final Logic:

Single points have zero geometric probability.

🧠 Memory Trick

"One point → Probability 0"

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

10 Let
\(f(x)=\left\{\begin{pmatrix}x+2, & x<1\\ 3, & x=1\\ x-2, & x>1\end{pmatrix}\right.\)
The left-hand limit at x=1 is:

A 3

B 1

C -1

D 0

💡 Short Explanation

Use expression valid for x<1 Substitute x=1 in x+2 Left-hand limit equals 3

✅ Correct Answer Explanation

For the left-hand limit, use the branch: \(f(x)=x+2,x<1\) Therefore: \({lim⁡}_{x\rightarrow 1^{-}}f(x)=1+2=3\) Hence option A is correct. The actual value at x=1 does not affect the left-hand limit.

❓ Why Other Options Are Incorrect

Option B → Incorrect substitution result.
Option C → Comes from the right-side branch \(x-2\).
Option D → No valid branch evaluation gives zero.

🎯 Strategy

Used: Substitution

Application:

Use the left-side branch for LHL evaluation.

Final Logic:

Left-hand limits depend only on nearby left-side values.

🧠 Memory Trick

"LHL uses left branch"

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

11 Let
\(f(x)=∣x∣\)
Is the function continuous for all negative real numbers \(x<0\)?

A No, it equals \(-x\)

B No, it equals \(x\)

C Yes, \({lim⁡}_{x\rightarrow c}f(x)=c\)

D Yes, \({lim⁡}_{x\rightarrow c}f(x)=-c=f(c)\)

💡 Short Explanation

For negative x, \(∣x∣=-x\) Linear functions are continuous Limit equals function value everywhere

✅ Correct Answer Explanation

For \(x<0\), \(∣x∣=-x\) which is a linear polynomial function and therefore continuous. If \(c<0\), \({lim⁡}_{x\rightarrow c}∣x∣=-c=f(c)\) Hence the modulus function is continuous for all negative real numbers. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → \(∣x∣=-x\) is true, but this does not imply discontinuity.
Option B → For negative x, \(∣x∣\neq x\).
Option C → The limit equals \(-c\), not c, for negative values.

🎯 Strategy

Used: Substitution

Application:

Replace \(∣x∣\)by \(-x\) for negative inputs.

Final Logic:

Linear expressions remain continuous everywhere.

🧠 Memory Trick

"Negative x → |x| = −x"

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

12 While evaluating
\(\int ∣x∣ dx\)
we split the domain at x=0 because:

A The limit does not exist

B It is undefined at x=0

C The algebraic definition changes from \(-x\) to \(x\)

D The integral becomes infinite

💡 Short Explanation

Modulus has piecewise definition Formula changes at x=0 Integration requires separate expressions

✅ Correct Answer Explanation

The modulus function satisfies: \(∣x∣=\left\{\begin{pmatrix}-x, & x<0\\ x, & x\geq 0\end{pmatrix}\right.\) Since its algebraic form changes at x=0, the integral must be evaluated separately on each interval. Hence option C is correct. The function remains continuous and finite at x=0.

❓ Why Other Options Are Incorrect

Option A → The limit exists and equals zero at x=0.
Option B → The modulus function is defined at x=0.
Option D → The integral of \(∣x∣\)remains finite over bounded intervals.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the piecewise definition of modulus.

Final Logic:

Change in algebraic expression requires interval splitting.

🧠 Memory Trick

"|x| splits at 0"

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integration of Modulus Functions

13 Assertion (A): For an integer n and sufficiently small \(r>0\),
\([n-r]=n-1,[n+r]=n\)
Reason (R): The function \(f(x)=[x]\)is continuous at every integer.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Greatest integer jumps at integers One-sided values differ there Hence continuity fails at integers

✅ Correct Answer Explanation

For sufficiently small positive r, \([n-r]=n-1\) and \([n+r]=n\) Thus the Assertion is true. However, the greatest integer function has jump discontinuities at every integer, so it is not continuous there. Therefore the Reason is false. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Assertion is mathematically correct.
Option C → The Reason is false and cannot explain the Assertion.
Option D → The Assertion correctly describes one-sided behavior near integers.

🎯 Strategy

Used: Substitution

Application:

Evaluate nearby left and right values around integer n.

Final Logic:

Unequal neighboring values create discontinuity at integers.

🧠 Memory Trick

"GIF jumps at integers"

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

14 Arrange the steps for evaluating the right-hand limit of \(\left[x\right]\)at an integer n:
1. Substitute \(x=n+r\)
2. Set the limit \(x\rightarrow n^{+}\)
3. Evaluate \(\left[n,\ r\right]\)
4. Result is n

A 1, 2, 3, 4

B 4, 3, 2, 1

C 2, 3, 1, 4

D 2, 1, 3, 4

💡 Short Explanation

Start with right-hand approach Substitute nearby positive increment Evaluate greatest integer value

✅ Correct Answer Explanation

To evaluate the right-hand limit: First consider: \(x\rightarrow n^{+}\) Then write: \(x=n+r,r>0\) Next evaluate: \([n+r]=n\) Thus the final result is n. Therefore the correct order is: \(2,1,3,4\) Hence option D is correct. The provided answer is correct.

❓ Why Other Options Are Incorrect

Option A → Substitution occurs after specifying right-hand approach.
Option B → Logical evaluation order is reversed completely.
Option C → Evaluation cannot occur before substitution.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow the natural procedural order of limit evaluation.

Final Logic:

Right-hand approach precedes substitution and evaluation.

🧠 Memory Trick

"Approach → Substitute → Evaluate → Result"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Right-Hand Limit of GIF

15 If
\(f(x)=sin⁡x,g(x)=x^{2}\)
then the composition \(\left(f\circ g)(x\right)\)is:

A \(sin⁡(x^{2})\)

B \(x^{2}sin⁡x\)

C \(\sin\,x^{2}\)

D \({\left(sin⁡x\right)}^{2}\)

💡 Short Explanation

Composition means outer after inner First compute \(g(x)\) Then apply sine function

✅ Correct Answer Explanation

By definition: \(\left(f\circ g)(x)=f(g(x)\right)\) Since: \(g(x)=x^{2}\) we get: \(f(g(x))=sin⁡(x^{2})\) Hence option A is correct. The other options represent multiplication or different expressions, not composition.

❓ Why Other Options Are Incorrect

Option B → Represents product, not composition.
Option C → Ambiguous notation lacking proper brackets.
Option D → Squares the sine function instead of the input variable.

🎯 Strategy

Used: Substitution

Application:

Substitute the inner function into the outer function.

Final Logic:

Composition means "function inside function."

🧠 Memory Trick

"Compose = Put inside"

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

16 Suppose f and g are real-valued functions. The composite function \(f\circ g\) is continuous at \(x=c\) provided:

A f is continuous at c

B g is discontinuous at c

C g is continuous at c and f is continuous at g(c)

D No condition is required

💡 Short Explanation

Inner function must behave continuously Outer function continuous at image point Composite continuity follows sequentially

✅ Correct Answer Explanation

For: \(\left(f\circ g)(x)=f(g(x)\right)\) continuity requires: 1. \(g\) continuous at c 2. \(f\) continuous at \(g(c)\) Then: \(f(g(x))\rightarrow f(g(c))\) Hence option C is correct. Continuity of only one function is insufficient.

❓ Why Other Options Are Incorrect

Option A → Ignores continuity of the inner function g.
Option B → A discontinuous inner function may destroy composite continuity.
Option D → Composite continuity always requires conditions on f and g.

🎯 Strategy

Used: Option Grouping

Application:

Identify continuity requirements for both inner and outer functions.

Final Logic:

Both linked continuity conditions are necessary.

🧠 Memory Trick

"Inner continuous, outer continuous"

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

17 The identity function
\(f(x)=x\)
is:

A Discontinuous at x=0

B Continuous at every real number

C Defined only for positive values

D A rational function only

💡 Short Explanation

Identity returns input unchanged Limit equals function value directly Continuous throughout real numbers

✅ Correct Answer Explanation

For the identity function: \(f(x)=x\) we have: \({lim⁡}_{x\rightarrow a}x=a=f(a)\) for every real number a. Thus the function is continuous everywhere on ℝ. Hence option B is correct according to the continuity definition.

❓ Why Other Options Are Incorrect

Option A → Identity function remains continuous at x=0 also.
Option C → The domain includes all real numbers.
Option D → Identity functions are polynomial functions, not only rational functions.

🎯 Strategy

Used: Substitution

Application:

Directly evaluate the limit at an arbitrary point.

Final Logic:

Limit equals function value for every real input.

🧠 Memory Trick

"Identity stays continuous"

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

18 For a function defined on a closed interval [a,b], continuity requires:

A Continuity at every point including a and b

B Continuity only in (a,b)

C Discontinuity at endpoints

D Function equals zero at endpoints

💡 Short Explanation

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

✅ Correct Answer Explanation

A function is continuous on a closed interval [a,b] if: • It is continuous for all interior points • Right-hand continuity exists at a • Left-hand continuity exists at b Thus continuity must hold at every point including endpoints. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Endpoint continuity is also necessary on closed intervals.
Option C → Endpoints may still satisfy continuity conditions.
Option D → Endpoint values need not be zero.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Apply the formal definition of continuity on closed intervals.

Final Logic:

Endpoints and interior points all matter.

🧠 Memory Trick

"Closed interval → Include endpoints"

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

19 Let
\(f(x)=\left\{\begin{pmatrix}x^{2}, & x\leq 1\\ x^{2}, & x>1\end{pmatrix}\right.\)
The graph has how many breaks?

A 2

B 3

C 1

D 0

💡 Short Explanation

Both branches are identical Polynomial function remains continuous No jumps or holes occur

✅ Correct Answer Explanation

Both branches define the same function: \(f(x)=x^{2}\) which is continuous everywhere. Since there is no change in expression or limiting behavior at x=1, the graph has no breaks. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → No discontinuity exists to create two breaks.
Option B → Polynomial graphs remain connected and smooth.
Option C → No jump or removable discontinuity occurs at x=1.

🎯 Strategy

Used: Odd One Out

Application:

Observe that both branches are actually identical.

Final Logic:

Same expressions on both sides preserve continuity.

🧠 Memory Trick

"Same branch = No break"

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

20 A function has
\(LHL=3,RHL=-1\)
at \(x=a\). The graph at \(x=a\) shows:

A Smooth curve

B Sharp corner but connected

C Gap or break

D Vertical asymptote

💡 Short Explanation

One-sided limits are unequal Overall limit does not exist Jump discontinuity appears graphically

✅ Correct Answer Explanation

Since: \(LHL\neq RHL\) the limit at x=a does not exist. Therefore the graph contains a jump discontinuity or visible break at that point. Hence option C is correct. A smooth or connected graph requires equal one-sided limits.

❓ Why Other Options Are Incorrect

Option A → Smooth curves require equal left and right limits.
Option B → Sharp corners still preserve continuity.
Option D → Vertical asymptotes involve unbounded behavior, not finite unequal limits.

🎯 Strategy

Used: Elimination

Application:

Remove continuous graph possibilities when one-sided limits differ.

Final Logic:

Unequal LHL and RHL imply graphical break.

🧠 Memory Trick

"Unequal limits → Gap"

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

UG Mathematics Booster Test 2 - Continuity Analysis & Functions

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