CUET UG Mathematics Booster Test 2 - Differentiability Concepts -

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CUET UG Mathematics Booster Test 2 - Differentiability Concepts

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 2 - Differentiability Concepts

1 / 20

QUESTION 1 OF 20

Derivative Definition (Integral Connection)
If
\(\frac{d}{dx}(x^{2})=2x\)
then
\(\int 2x dx=\)

Answer Review

1 Derivative Definition (Integral Connection)
If
\(\frac{d}{dx}(x^{2})=2x\)
then
\(\int 2x dx=\)

A \(x^{2}\)

B \(x^{2}+C\)

C \(2x+C\)

D \(x+C\)

💡 Short Explanation

Integration reverses differentiation Antiderivative of \(2x\) is \(x^{2}\) Include constant of integration

✅ Correct Answer Explanation

Since: \(\frac{d}{dx}(x^{2})=2x\) the antiderivative of \(2x\) is: \(x^{2}\) Indefinite integration always includes an arbitrary constant: \(\int 2x dx=x^{2}+C\) Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Missing constant of integration.
Option C → Derivative of \(2x\) equals 2, not \(2x\).
Option D → Derivative gives 1, not \(2x\).

🎯 Strategy

Used: Substitution

Application:

Reverse the differentiation process directly.

Final Logic:

Integral is antiderivative plus constant.

🧠 Memory Trick

"Integral = Reverse derivative + C"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Relation Between Derivative and Integral

2 Difference Quotient (Numerical)
Evaluate the difference quotient for
\(f(x)=3x\frac{f(x+h)-f(x)}{h}\)

A 5

B \(3+h\)

C 3

D \(3h\)

💡 Short Explanation

Substitute function into quotient Simplify numerator carefully Cancel common factor \(h\)

✅ Correct Answer Explanation

Given: \(f(x)=3x\) then: \(f(x+h)=3(x+h)=3x+3h\) Thus: \(\frac{f(x+h)-f(x)}{h}=\frac{3x+3h-3x}{h}=\frac{3h}{h}=3\) Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → No algebraic basis.
Option B → Incorrect simplification.
Option D → Extra factor \(h\) retained mistakenly.

🎯 Strategy

Used: Substitution

Application:

Insert \(x+h\) into function definition.

Final Logic:

Difference quotient simplifies to constant slope.

🧠 Memory Trick

"Linear function → constant derivative"

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

3 \(f^{'}(x)\)Notation
If
\(f(x)=cos⁡x\)
then
\(f^{'}(x)=\)

A \(\sin\,x\)

B \(\tan\,x\)

C \(\cos\,x\)

D \(-sin⁡x\)

💡 Short Explanation

Cosine derivative gives negative sine Standard trigonometric derivative formula Sign is important

✅ Correct Answer Explanation

The derivative rule for cosine is: \(\frac{d}{dx}(cos⁡x)=-sin⁡x\) Thus: \(f^{'}(x)=-sin⁡x\) Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Missing negative sign.
Option B → Tangent is unrelated derivative here.
Option C → Cosine is original function, not derivative.

🎯 Strategy

Used: Memory Recall

Application:

Recall standard trigonometric derivative identities.

Final Logic:

Cosine differentiates into negative sine.

🧠 Memory Trick

"Cos becomes minus sin"

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

4 \(dy/dx\) Notation (Vector Form)
If
\(\vec{r}(t)=t\hat{i}+t^{2}\hat{j}\)
then
\(\frac{d\vec{r}}{dt}=\)

A \(\hat{i}+2t\hat{j}\)

B \(t\hat{i}+2t\hat{j}\)

C \(\hat{i}+t^{2}\hat{j}\)

D \(2\hat{i}+t\hat{j}\)

💡 Short Explanation

Differentiate vector components separately Derivative of \(t\) is 1 Derivative of \(t^{2}\)is \(2t\)

✅ Correct Answer Explanation

Differentiate each component independently: \(\frac{d}{dt}(t)=1,\frac{d}{dt}(t^{2})=2t\) Hence: \(\frac{d\vec{r}}{dt}=\hat{i}+2t\hat{j}\) Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → First component differentiated incorrectly.
Option C → Second component not differentiated.
Option D → Both component derivatives incorrect.

🎯 Strategy

Used: Substitution

Application:

Apply differentiation to each vector component.

Final Logic:

Vector derivative equals componentwise derivative.

🧠 Memory Trick

"Differentiate each vector term separately"

📚 Class XII – NCERT Mathematics – Chapter: Vector Algebra – Topic: Vector Differentiation

5 For the derivative to exist at a point:
1. Left-hand derivative exists and is finite
2. Right-hand derivative exists and is finite
3. Both are equal

A 1 and 2 only

B 1, 2, and 3

C 3 only

D 1 and 3 only

💡 Short Explanation

Both side derivatives must exist Both must be finite Equality ensures differentiability

✅ Correct Answer Explanation

A derivative exists at a point only if: Left-hand derivative exists, Right-hand derivative exists, Both are equal and finite. If either side fails or values differ, differentiability does not exist. Hence all three conditions are necessary. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Equality condition missing.
Option C → Existence and finiteness also required.
Option D → Right derivative condition omitted.

🎯 Strategy

Used: Option Grouping

Application:

Check complete differentiability conditions systematically.

Final Logic:

Equal finite side derivatives define derivative existence.

🧠 Memory Trick

"LHD = RHD"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Existence of Derivative

6 If a limit evaluates to:
\(-\infty\)
then:

A Function is differentiable

B Function is continuous

C Function is not differentiable

D Right-hand limit must be finite

💡 Short Explanation

Infinite limits are not finite values Derivative must remain finite Infinite slope destroys differentiability

✅ Correct Answer Explanation

A derivative must exist as a finite real number. If a limit tends to: \(-\infty\) then the derivative is not finite and differentiability fails. Infinite limits indicate vertical behavior or asymptotic growth. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Infinite derivative is not valid differentiability.
Option B → Continuity not guaranteed by infinite limit.
Option D → No such requirement exists.

🎯 Strategy

Used: Elimination

Application:

Reject statements inconsistent with finite derivative definition.

Final Logic:

Infinite limit prevents derivative existence.

🧠 Memory Trick

"Infinite slope → No derivative"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Non-Differentiability

7 Left Derivative (Numerical)
Let
\(f(x)=∣x-1∣\)
Find the left derivative at \(x=1\).

A 5

B 1

C 0

D -1

💡 Short Explanation

For \(x<1, ∣x-1∣=1-x\) Derivative becomes \(-1\) Use left-side expression only

✅ Correct Answer Explanation

For: \(x<1\) we write: \(∣x-1∣=1-x\) Differentiating: \(\frac{d}{dx}(1-x)=-1\) Hence the left-hand derivative at: \(x=1\) equals: \(-1\) Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → No derivative calculation gives 5.
Option B → This is right derivative value.
Option C → Derivative is nonzero.

🎯 Strategy

Used: Substitution

Application:

Replace modulus by branch definition for left side.

Final Logic:

Left branch slope equals \(-1\).

🧠 Memory Trick

"Left branch falls → slope negative"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Left-Hand Derivative

8 Match functions with right derivatives at \(x=0\):

List I	List II
1. \(f(x)=x\)	a. 1
2. \(f(x)=x^{2}\)	b. 0
3. \(f(x)=x^{3}\)	c. 3
4. \(f(x)=∣x∣\)	d. 1

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

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

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

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

💡 Short Explanation

Differentiate each function at 0 Right derivative of modulus equals 1 Polynomial derivatives evaluated directly

✅ Correct Answer Explanation

At \(x=0\): \(\frac{d}{dx}(x)=1\frac{d}{dx}(x^{2})=2x=0\frac{d}{dx}(x^{3})=3x^{2}=0\) and right derivative of: \(∣x∣\) equals 1. The provided matching in option A contains an inconsistency for item 3. Correct matching should be: \(1-a, 2-b, 3-b, 4-d\) Hence none of the options are perfectly correct.

❓ Why Other Options Are Incorrect

Option A → \(x^{3}\)derivative at 0 equals 0, not 1.
Option B → Multiple derivative matches incorrect.
Option C → Polynomial derivative values mismatched.

🎯 Strategy

Used: Elimination

Application:

Compute each derivative independently.

Final Logic:

Correct derivative matching exposes option inconsistency.

🧠 Memory Trick

"Power derivatives vanish at 0"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Right-Hand Derivative

9 Differentiability on Closed Interval
Identify the incorrect statement for differentiability on \(\left[a,\ b\right]\):

A Differentiable at every point in \(\left(a,\ b\right)\)

B Right-hand derivative exists at \(a\)

C Right-hand derivative exists at \(b\)

D Left-hand derivative exists at \(b\)

💡 Short Explanation

Closed interval endpoints use one-sided derivatives Right derivative at left endpoint Left derivative at right endpoint

✅ Correct Answer Explanation

For differentiability on: \(\left[a,\ b\right]\) the function must be differentiable in: \(\left(a,\ b\right)\) and satisfy: Right-hand derivative at \(a\), Left-hand derivative at \(b\). Right-hand derivative at \(b\) is unnecessary because points to the right lie outside the interval. Hence option C is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct interior differentiability condition.
Option B → Correct endpoint derivative condition.
Option D → Required at right endpoint.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Match endpoint derivative direction with interval structure.

Final Logic:

Endpoint derivatives are one-sided.

🧠 Memory Trick

"Start → Right, End → Left"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Differentiability on Closed Intervals

10 Differentiability on Open Interval
For differentiability on \(\left(a,\ b\right)\):

A Sharp corners are allowed

B Function must be linear

C Only continuity required

D Unique tangent exists at every point

💡 Short Explanation

Differentiability means smooth tangent Sharp corners destroy derivative Continuity alone is insufficient

✅ Correct Answer Explanation

A function is differentiable on: \(\left(a,\ b\right)\) if it has a unique finite tangent (slope) at every interior point. Sharp corners or cusps prevent derivative existence. Differentiability is stronger than continuity. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → Sharp corners make derivatives fail.
Option B → Nonlinear functions can also be differentiable.
Option C → Continuous functions may still be nondifferentiable.

🎯 Strategy

Used: Elimination

Application:

Remove statements inconsistent with geometric meaning of derivative.

Final Logic:

Differentiability requires unique tangent slope.

🧠 Memory Trick

"Smooth curve → derivative exists"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Geometric Meaning of Derivative

11 Algebra of Derivatives (Sum Rule – Arrange in Order)
Arrange the steps to differentiate
\(f(x)=u(x)+v(x)\)
1. Add the resulting derivatives: \(f^{'}(x)=u^{'}(x)+v^{'}(x)\)
2. Recognize the terms as \(u(x)\)and \(v(x)\)
3. Differentiate \(u(x):u^{'}(x)\)
4. Differentiate \(v(x):v^{'}(x)\)

A 1, 2, 3, 4

B 2, 4, 3, 1

C 2, 1, 4, 3

D 4, 3, 2, 1

💡 Short Explanation

Identify component functions first Differentiate separately Combine using sum rule

✅ Correct Answer Explanation

To differentiate: \(f(x)=u(x)+v(x)\) first identify both functions. Then differentiate each separately: \(u^{'}(x), v^{'}(x)\) Finally apply the sum rule: \(f^{'}(x)=u^{'}(x)+v^{'}(x)\) Hence the logical order is: \(2,4,3,1\) So option B is correct.

❓ Why Other Options Are Incorrect

Option A → Addition cannot occur before differentiation.
Option C → Sum rule applied prematurely.
Option D → Reverses the correct logical sequence.

🎯 Strategy

Used: Arrange in Order

Application:

Follow natural differentiation workflow stepwise.

Final Logic:

Identify → Differentiate → Combine.

🧠 Memory Trick

"Split → Differentiate → Add"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Sum Rule of Derivatives

12 Algebra of Derivatives (Difference Rule)
Let
\(f(x)=x^{2},g(x)=x\)
Find
\(\frac{d}{dx}[f(x)-g(x)]\)

A \(2x-1\)

B \(2x+1\)

C \(x^{2}-x\)

D \(2x\)

💡 Short Explanation

Differentiate terms separately Apply difference rule Combine carefully with sign

✅ Correct Answer Explanation

Given: \(f(x)=x^{2},g(x)=x\) Then: \(f^{'}(x)=2x,g^{'}(x)=1\) Using difference rule: \(\frac{d}{dx}[f(x)-g(x)]=f^{'}(x)-g^{'}(x)=2x-1\) Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect sign during subtraction.
Option C → Original expression, not derivative.
Option D → Forgot derivative of \(x\).

🎯 Strategy

Used: Substitution

Application:

Compute derivatives individually before subtraction.

Final Logic:

Difference rule gives \(2x-1\).

🧠 Memory Trick

"Differentiate each, then subtract"

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

📄 Passage

"Let \(f(x)\)be a composite function defined as
\(f(x)=v(u(x))\)
If both \(\frac{du}{dx}\)and \(\frac{dv}{du}\)exist, then
\(\frac{df}{dx}=\frac{dv}{du}⋅\frac{du}{dx}\)
This is known as the Chain Rule."

For a product
\(y=u(x)v(x)\)
the product rule states:

A Differentiates inner function only

B Divides derivatives

C \(y^{'}=u^{'}v+uv^{'}\)

D Subtracts derivatives

💡 Short Explanation

Product differentiation uses two terms Differentiate one factor at a time Add both resulting products

✅ Correct Answer Explanation

For: \(y=u(x)v(x)\) the product rule is: \(\frac{dy}{dx}=u^{'}v+uv^{'}\) Differentiate first function keeping second unchanged, then vice versa, and add both products. Hence option C correctly represents the product rule.

❓ Why Other Options Are Incorrect

Option A → Describes chain rule partially, not product rule.
Option B → No derivative division occurs.
Option D → Product rule involves addition, not subtraction.

🎯 Strategy

Used: Memory Recall

Application:

Recall standard differentiation identities.

Final Logic:

Product rule always produces two additive terms.

🧠 Memory Trick

"First derive first + first derive second"

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

📄 Passage

If
\(y=\frac{u}{v}\)
and we write
\(y=u⋅v^{-1}\)
then deriving the quotient rule requires:

A Sum rule

B Difference rule

C Integral rule

D Product rule

💡 Short Explanation

Quotient rewritten as product Use reciprocal power form Apply product differentiation

✅ Correct Answer Explanation

The quotient: \(\frac{u}{v}\) can be written as: \(u⋅v^{-1}\) Differentiating this expression requires the product rule along with chain/power rules. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → Sum rule not applicable.
Option B → No subtraction structure involved.
Option C → Integration unrelated to quotient differentiation.

🎯 Strategy

Used: Substitution

Application:

Rewrite quotient into multiplicative form.

Final Logic:

Product rule becomes necessary after rewriting.

🧠 Memory Trick

"Quotient = Product with inverse"

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

15 Polynomial Derivative (Numerical)
Let
\(f(x)=x^{3}+5x^{2}\)
Find \(f^{'}(2)\).

A 60

B 40

C 20

D 15

💡 Short Explanation

Differentiate termwise Substitute \(x=2\) carefully Cross-check arithmetic

✅ Correct Answer Explanation

Given: \(f(x)=x^{3}+5x^{2}\) Differentiate: \(f^{'}(x)=3x^{2}+10x\) Now substitute \(x=2\): \(f^{'}(2)=3(4)+10(2)=12+20=32\) Thus the provided answer is incorrect. Correct value is 32, which is absent from the options.

❓ Why Other Options Are Incorrect

Option A → Arithmetic mistake.
Option B → Partial computation only.
Option D → Ignores cubic derivative contribution.

🎯 Strategy

Used: Substitution

Application:

Apply power rule before numerical evaluation.

Final Logic:

Correct derivative gives 32.

🧠 Memory Trick

"Differentiate first, substitute later"

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

16 Trigonometric Derivatives (Assertion–Reason)
Assertion (A):
\(\frac{d}{dx}(sin⁡x)=cos⁡x\)
Reason (R):
\(\frac{d}{dx}(cos⁡x)=-sin⁡x\)

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Both derivative identities are correct Standard trigonometric rules Related derivative relations explain behavior

✅ Correct Answer Explanation

The derivative identities: \(\frac{d}{dx}(sin⁡x)=cos⁡x\) and \(\frac{d}{dx}(cos⁡x)=-sin⁡x\) are both true. These paired trigonometric derivative relations are foundational results. Hence option C is correct. The provided answer was incorrect.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically true.
Option B → Reason is not false.
Option D → Assertion is true, not false.

🎯 Strategy

Used: Elimination

Application:

Verify each trigonometric derivative separately.

Final Logic:

Both standard derivative formulas are correct.

🧠 Memory Trick

"Sin → Cos, Cos → −Sin"

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

17 Differentiability ⇒ Continuity
Which theorem proves that every differentiable function is continuous?

A Chain Rule

B Product Rule

C Mean Value Theorem

D Theorem 3

💡 Short Explanation

Differentiability implies continuity Standard theorem in NCERT Converse is not always true

✅ Correct Answer Explanation

NCERT states explicitly in Theorem 3: "If a function is differentiable at a point, then it is continuous there." Thus differentiability guarantees continuity. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Chain rule concerns composite differentiation.
Option B → Product rule differentiates products only.
Option C → Mean Value Theorem requires differentiability but does not define this implication directly.

🎯 Strategy

Used: Memory Recall

Application:

Recall named NCERT theorem statement.

Final Logic:

Theorem 3 states differentiable ⇒ continuous.

🧠 Memory Trick

"Differentiable means automatically continuous"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Differentiability Implies Continuity

18 If a function is continuous, then:

A It may or may not be differentiable

B It is always differentiable

C It is never differentiable

D Differentiable except at 0

💡 Short Explanation

Continuity weaker than differentiability Sharp corners may exist Modulus function is classic example

✅ Correct Answer Explanation

Continuity does not guarantee differentiability. For example: \(f(x)=∣x∣\) is continuous at \(x=0\) but not differentiable there because left and right derivatives differ. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → False due to modulus counterexample.
Option C → Many continuous functions are differentiable.
Option D → No universal exception at 0 exists.

🎯 Strategy

Used: Counterexample Method

Application:

Use modulus function to test statements.

Final Logic:

Continuous functions can fail differentiability.

🧠 Memory Trick

"Continuous ≠ always smooth"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Continuity vs Differentiability

19 Non-Differentiable Case (Sharp Corner)
At a sharp corner point on a graph, the derivative:

A Equals zero

B Is finite and positive

C Does not exist

D Approaches infinity

💡 Short Explanation

Sharp corners change slope abruptly LHD and RHD differ Unique tangent absent

✅ Correct Answer Explanation

At a sharp corner, left-hand and right-hand derivatives are unequal. Since derivative requires equal finite side derivatives, differentiability fails. Therefore the derivative does not exist. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Derivative need not be zero.
Option B → Slope is not uniquely defined.
Option D → Infinite derivative is not necessary.

🎯 Strategy

Used: Graph Interpretation

Application:

Interpret geometric meaning of corner points.

Final Logic:

Unequal side slopes destroy derivative.

🧠 Memory Trick

"Corner means no tangent"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Non-Differentiability at Sharp Corners

20 Modulus Function at \(x=0\)
For
\(f(x)=∣x∣\)
the right-hand derivative at \(x=0\) is:

A -1

B 1

C 0

D Does not exist

💡 Short Explanation

For \(x>0, ∣x∣=x\) Derivative of \(x\) is 1 Right derivative uses positive branch

✅ Correct Answer Explanation

For: \(x>0\) the modulus function becomes: \(∣x∣=x\) Thus: \(\frac{d}{dx}(x)=1\) Hence the right-hand derivative at: \(x=0\) equals 1. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → This is left-hand derivative value.
Option C → Derivative is not zero.
Option D → Right derivative exists individually.

🎯 Strategy

Used: Substitution

Application:

Replace modulus using positive-side definition.

Final Logic:

Positive branch slope equals 1.

🧠 Memory Trick

"Right side rises → slope +1"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: One-Sided Derivatives

CUET UG Mathematics Booster Test 2 - Differentiability Concepts

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