CUET UG Mathematics Booster Test 3 - Differentiability Concepts -

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CUET UG Mathematics Booster Test 3 - Differentiability Concepts

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 3 - Differentiability Concepts

1 / 20

QUESTION 1 OF 20

Using the limit definition of derivative, if a position vector is given by
\(\vec{r}(t)=t\hat{i}+t\hat{j}\)
evaluate the magnitude of
\({lim⁡}_{h\rightarrow 0}\frac{\vec{r}(t+h)-\vec{r}(t)}{h}.\)

Answer Review

1 Using the limit definition of derivative, if a position vector is given by
\(\vec{r}(t)=t\hat{i}+t\hat{j}\)
evaluate the magnitude of
\({lim⁡}_{h\rightarrow 0}\frac{\vec{r}(t+h)-\vec{r}(t)}{h}.\)

A 0

B \(\sqrt{2}\)

C 1

D −1

💡 Short Explanation

Differentiate vector components separately Velocity vector becomes \(\hat{i}+\hat{j}\) Magnitude equals \(\sqrt{2}\)

✅ Correct Answer Explanation

Given: \(\vec{r}(t)=t\hat{i}+t\hat{j}\) Then: \(\vec{r}(t+h)=(t+h)\hat{i}+(t+h)\hat{j}\) So, \(\frac{\vec{r}(t+h)-\vec{r}(t)}{h}=\hat{i}+\hat{j}\) Taking limit gives the same vector. Its magnitude is: \(\sqrt{1^{2}+1^{2}}=\sqrt{2}\) Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Magnitude is not zero because both vector components are non-zero.
Option C → Ignores contribution from second component.
Option D → Magnitude cannot be negative.

🎯 Strategy

Used: Substitution

Application:

Substitute \(t+h\) into vector expression and simplify.

Final Logic:

Resulting vector magnitude is \(\sqrt{2}\).

🧠 Memory Trick

"\(1^{2}+1^{2}=2\) → root 2"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Derivative Using First Principle

2 Assertion (A): The difference quotient for \(f(x)=kx\)(where \(k\) is constant) evaluates to \(k\) before taking the limit.
Reason (R):
\(\frac{f(x+h)-f(x)}{h}=\frac{k(x+h)-kx}{h}=k\)

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Linear function simplifies directly Difference quotient becomes constant Reason correctly proves assertion

✅ Correct Answer Explanation

For: \(f(x)=kx\) we get: \(\frac{f(x+h)-f(x)}{h}=\frac{k(x+h)-kx}{h}=\frac{kh}{h}=k\) Thus the quotient itself equals \(k\) before limit evaluation. Hence both Assertion and Reason are true, and Reason explains Assertion.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used: Substitution

Application:

Replace function directly into difference quotient.

Final Logic:

Simplification immediately gives \(k\).

🧠 Memory Trick

"Linear slope stays constant"

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

3 The derivative of a composite function \(y=f(g(x))\)can be expressed using which of the following notations?
\(\frac{dy}{dx}=\frac{dy}{du}⋅\frac{du}{dx}, where u=g(x)\)
\(f^{'}(g(x))⋅g^{'}(x)\)
\(\frac{d}{dx}[f(g(x))]=f^{'}(x)g^{'}(x)\)

A 1 and 2 only

B 1 and 3 only

C 2 and 3 only

D 1, 2, and 3

💡 Short Explanation

Chain rule uses inner and outer derivatives Correct notation needs \(f^{'}(g(x))\) Option 3 misses composition structure

✅ Correct Answer Explanation

Chain rule states: \(\frac{dy}{dx}=\frac{dy}{du}⋅\frac{du}{dx}\) or equivalently: \(\frac{d}{dx}[f(g(x))]=f^{'}(g(x))g^{'}(x)\) Option 3 incorrectly writes \(f^{'}(x)\)instead of \(f^{'}(g(x))\). Hence only statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

Option B → Includes incorrect statement 3.
Option C → Omits correct statement 1.
Option D → Statement 3 is mathematically incorrect.

🎯 Strategy

Used: Elimination

Application:

Verify exact chain rule notation carefully.

Final Logic:

Inner function must appear inside \(f^{'}\).

🧠 Memory Trick

"Outer at inner × inner derivative"

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

4 Consider the implicitly defined function:
\(x^{2}+y^{2}=1\)
Identify the incorrect statement:

A We can differentiate both sides with respect to \(x\)

B \(y\) cannot be treated as a function of \(x\)

C The derivative of a constant is 0

D The derivative of \(x\) with respect to \(x\) is 1

💡 Short Explanation

Implicit functions allow \(y=f(x)\)locally Differentiation possible on both sides Constants differentiate to zero

✅ Correct Answer Explanation

In implicit differentiation, \(y\) is treated as a function of \(x\). Thus: \(\frac{d}{dx}(y^{2})=2y\frac{dy}{dx}\) Hence statement B is incorrect. Statements A, C, and D follow standard differentiation rules.

❓ Why Other Options Are Incorrect

Option A → Correctly describes implicit differentiation.
Option C → Constant derivative is always zero.
Option D → \(\frac{d}{dx}(x)=1\) is fundamental.

🎯 Strategy

Used: Elimination

Application:

Check standard differentiation properties one by one.

Final Logic:

Only statement B contradicts implicit differentiation.

🧠 Memory Trick

"Implicit means \(y\) depends on \(x\)"

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

5 A function is differentiable at a point only if the limit
\({lim⁡}_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)
exists. This limit fails to exist when:

A \(f(x)\)is a polynomial

B \(f(x)\)is continuous and smooth

C Left-hand and right-hand limits are unequal

D \(f(x)\)is trigonometric

💡 Short Explanation

Derivative requires unique limit Unequal one-sided limits destroy existence Smooth functions remain differentiable

✅ Correct Answer Explanation

For differentiability, left-hand derivative and right-hand derivative must be equal. If these one-sided limits differ, the derivative limit fails to exist. Therefore option C correctly identifies the failure condition.

❓ Why Other Options Are Incorrect

Option A → Polynomials are differentiable everywhere.
Option B → Smooth continuous functions satisfy derivative conditions.
Option D → Standard trigonometric functions are differentiable.

🎯 Strategy

Used: Elimination

Application:

Test differentiability condition directly.

Final Logic:

Unequal one-sided limits imply no derivative.

🧠 Memory Trick

"LHD = RHD for derivative"

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

6 Match List I with List II:

List I	List II
1. Left derivative	a. \({lim⁡}_{h\rightarrow 0^{-}}\frac{f(x+h)-f(x)}{h}\)
2. Right derivative	b. \({lim⁡}_{h\rightarrow 0^{-}}f(x+h)\)
3. Derivative	c. Limit exists and is finite
4. Left-hand limit	d. \({lim⁡}_{h\rightarrow 0^{+}}\frac{f(x+h)-f(x)}{h}\)

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

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

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

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

💡 Short Explanation

Left derivative uses negative side Right derivative uses positive side Derivative must exist finitely

✅ Correct Answer Explanation

Left derivative uses: \(h\rightarrow 0^{-}\) while right derivative uses: \(h\rightarrow 0^{+}\) Derivative exists only when the limit is finite. Left-hand limit corresponds to: \({lim⁡}_{h\rightarrow 0^{-}}f(x+h)\) Thus option A gives the correct matching.

❓ Why Other Options Are Incorrect

Option B → Interchanges left and right derivative definitions.
Option C → Incorrectly matches derivative definition.
Option D → Mismatches all major definitions.

🎯 Strategy

Used: Option Grouping

Application:

Match symbols with standard derivative notation.

Final Logic:

Correct signs identify left and right derivatives.

🧠 Memory Trick

"Minus left, plus right"

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

7 Let
\(f(x)=\left\{\begin{pmatrix}x^{2}, & x<1\\ 2x-1, & x\geq 1\end{pmatrix}\right.\)
Find the left-hand derivative at \(x=1\).

A −2

B 0

C 1

D 2

💡 Short Explanation

Use left-side expression only Differentiate \(x^{2}\) Evaluate derivative at \(x=1\)

✅ Correct Answer Explanation

For \(x<1\), \(f(x)=x^{2}\) Hence: \(f^{'}(x)=2x\) At \(x=1\): \(f_{-}^{'}(1)=2(1)=2\) Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → Wrong derivative sign.
Option B → Derivative is not zero at \(x=1\).
Option C → Confuses derivative with function value.

🎯 Strategy

Used: Substitution

Application:

Differentiate left-side branch only.

Final Logic:

Left derivative equals \(2\).

🧠 Memory Trick

"Left branch controls LHD"

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

8 Arrange the correct sequence of steps to evaluate the right-hand derivative of \(f(x)=x^{2}\)at \(x=1\):
1. Evaluate the limit
2. Write the definition:
\({lim⁡}_{h\rightarrow 0^{+}}\frac{f(1+h)-f(1)}{h}\)
3. Substitute the function into the expression
4. Simplify the algebraic expression

A 1, 2, 3, 4

B 2, 3, 4, 1

C 3, 4, 1, 2

D 4, 2, 3, 1

💡 Short Explanation

Begin with derivative definition Substitute function expression Simplify before evaluating limit

✅ Correct Answer Explanation

Correct order: First write derivative definition, then substitute: \(f(x)=x^{2}\) After substitution simplify algebraically, then finally evaluate the limit. Hence sequence: \(2\rightarrow 3\rightarrow 4\rightarrow 1\) is correct.

❓ Why Other Options Are Incorrect

Option A → Evaluates limit before setup.
Option C → Misses proper starting definition.
Option D → Simplification cannot precede substitution.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow natural derivative-solving workflow.

Final Logic:

Definition → substitution → simplification → limit.

🧠 Memory Trick

"Define → substitute → simplify → limit"

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

9 For a function to be differentiable on a closed interval \(\left[a,\ b\right]\), which condition must be satisfied?

A Differentiable only on \(\left(a,\ b\right)\)

B Differentiable only at endpoints

C Continuous but not differentiable

D Differentiable on \(\left(a,\ b\right)\)and one-sided derivatives at \(a\) and \(b\)

💡 Short Explanation

Interior points need ordinary derivative Endpoints use one-sided derivatives Closed interval includes boundaries

✅ Correct Answer Explanation

A function is differentiable on: \(\left[a,\ b\right]\) if it is differentiable throughout: \(\left(a,\ b\right)\) and possesses right derivative at \(a\) and left derivative at \(b\). Thus option D gives the rigorous condition.

❓ Why Other Options Are Incorrect

Option A → Ignores endpoint conditions.
Option B → Interior differentiability is essential.
Option C → Differentiability requires continuity.

🎯 Strategy

Used: Elimination

Application:

Check closed interval endpoint conditions.

Final Logic:

One-sided derivatives are necessary at boundaries.

🧠 Memory Trick

"Closed interval needs side derivatives"

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

10 A function is differentiable on an open interval \(\left(a,\ b\right)\). Which feature is NOT allowed within this interval?

A Jump discontinuity

B Horizontal tangent

C Local maximum

D Inflection point

💡 Short Explanation

Differentiability implies continuity Jump discontinuity breaks continuity Smooth turning points are allowed

✅ Correct Answer Explanation

Differentiability guarantees continuity. A jump discontinuity destroys continuity, so differentiability cannot exist there. Horizontal tangents, maxima, and inflection points may still occur in differentiable functions. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Horizontal tangents occur in smooth curves.
Option C → Local maxima can exist in differentiable functions.
Option D → Inflection points may still be differentiable.

🎯 Strategy

Used: Elimination

Application:

Use theorem: differentiable implies continuous.

Final Logic:

Discontinuity prevents differentiability.

🧠 Memory Trick

"No jumps in differentiable graphs"

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

📄 Passage

Let \(y=u(x)⋅v(x)\)be the product of two functions. The Leibnitz (Product) Rule states that
\(\frac{d}{dx}[u(x)v(x)]=u(x)\frac{dv}{dx}+v(x)\frac{du}{dx}.\)
This rule is essential when dealing with algebraic operations on functions. Similarly, if
\(y=\frac{u(x)}{v(x)},\)
the Quotient Rule is applied as
\(\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=\frac{v(x)\frac{du}{dx}-u(x)\frac{dv}{dx}}{{\left[v(x)\right]}^{2}}.\)

Though the passage explicitly defines the multiplicative product and quotient rules, the sum rule operates distinctly. How does the sum rule structurally contrast?
\(\frac{d}{dx}[u(x)+v(x)]=\frac{du}{dx}+\frac{dv}{dx}.\)

A It involves multiplication of the respective derivatives.

B It requires calculating the square of the denominator.

C It is a direct linear operation evaluating strictly to

D It relies on the nested chain rule inherently.

💡 Short Explanation

Sum rule is linear Differentiate terms separately Add derivatives directly

✅ Correct Answer Explanation

The sum rule states: \(\frac{d}{dx}[u(x)+v(x)]=u^{'}(x)+v^{'}(x)\) Unlike product and quotient rules, no multiplication structure or denominator appears. Differentiation distributes directly across addition. Hence option C correctly describes the structural contrast.

❓ Why Other Options Are Incorrect

Option A → Multiplication belongs to product rule, not sum rule.
Option B → Squared denominator appears in quotient rule only.
Option D → Chain rule applies to compositions, not ordinary sums.

🎯 Strategy

Used: Option Grouping

Application:

Compare structures of derivative rules carefully.

Final Logic:

Sum rule simply adds derivatives termwise.

🧠 Memory Trick

"Sum → Differentiate and add"

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

📄 Passage

If applying the passage's quotient rule directly to
\(y=\frac{u(x)}{c}, where c is a numeric constant,\)
which rule effectively yields the identical outcome?

A Chain rule applied twice

B Product rule (treating \(\frac{1}{c}u(x)\)) and constant multiplier rule

C Additive difference rule

D Implicit differentiation

💡 Short Explanation

Constant denominator becomes scalar multiple Differentiate only variable part Constant factor remains unchanged

✅ Correct Answer Explanation

Since: \(\frac{u(x)}{c}=\frac{1}{c}u(x)\) the derivative becomes: \(\frac{1}{c}u^{'}(x)\) This follows from constant multiplier and product rule concepts. Hence option B correctly explains the equivalent method.

❓ Why Other Options Are Incorrect

Option A → Repeated chain rule unnecessary.
Option C → Difference rule irrelevant here.
Option D → No implicit relation exists.

🎯 Strategy

Used: Substitution

Application:

Rewrite quotient as scalar multiplication.

Final Logic:

Constant denominator simplifies differentiation directly.

🧠 Memory Trick

"Constant below = constant outside"

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

13 A transmitted data signal's amplitude varies as
\(y=xe^{x}.\)
Using the product rule, what is the instantaneous rate of change \(dy/dx\)?

A \(e^{x}\)

B \(xe^{x}\)

C \(e^{x}(1+x)\)

D \(e^{x}(x-1)\)

💡 Short Explanation

Apply product rule Differentiate each factor once Factor common exponential term

✅ Correct Answer Explanation

Given: \(y=xe^{x}\) Using product rule: \(\frac{dy}{dx}=x\frac{d}{dx}(e^{x})+e^{x}\frac{d}{dx}(x)=xe^{x}+e^{x}=e^{x}(x+1)\) Thus option C is correct.

❓ Why Other Options Are Incorrect

Option A → Misses derivative contribution of \(x\).
Option B → Original function, not derivative.
Option D → Incorrect sign after addition.

🎯 Strategy

Used: Substitution

Application:

Apply product rule termwise.

Final Logic:

Product differentiation gives \(e^{x}(x+1)\).

🧠 Memory Trick

"First keep second + second keep first"

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

14 Differentiate the quotient equation strictly using the quotient rule structure:
\(y=\frac{x^{2}+1}{x}.\)

A \(\frac{x^{2}-1}{x^{2}}\)

B \(\frac{x^{2}+1}{x^{2}}\)

C \(2x\)

D \(\frac{2x}{x^{2}}\)

💡 Short Explanation

Use quotient rule carefully Differentiate numerator and denominator Simplify final expression

✅ Correct Answer Explanation

Using quotient rule: \(y=\frac{u}{v}\) where \(u=x^{2}+1,v=x\) Then: \(u^{'}=2x,v^{'}=1y^{'}=\frac{x(2x)-(x^{2}+1)(1)}{x^{2}}=\frac{2x^{2}-x^{2}-1}{x^{2}}=\frac{x^{2}-1}{x^{2}}\) Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Numerator simplification incorrect.
Option C → Ignores denominator differentiation.
Option D → Incomplete quotient differentiation.

🎯 Strategy

Used: Substitution

Application:

Apply quotient rule formula directly.

Final Logic:

Simplification yields \(\frac{x^{2}-1}{x^{2}}\).

🧠 Memory Trick

"Low d-high minus high d-low"

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

15 The area of a physical square expands following
\(A=(x+1)^{2}.\)
First, expanding this polynomial yields
\(A=x^{2}+2x+1.\)
What is its explicit derivative \(dA/dx\)?

A \(2x+1\)

B \(2x+2\)

C \(x^{2}+2x+1\)

D \(2(x+1)^{2}\)

💡 Short Explanation

Differentiate polynomial termwise Constant derivative becomes zero Combine remaining terms

✅ Correct Answer Explanation

Given: \(A=x^{2}+2x+1\) Differentiate: \(\frac{dA}{dx}=2x+2+0=2x+2\) Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Misses derivative of \(2x\).
Option C → Original polynomial, not derivative.
Option D → Incorrect application of chain rule.

🎯 Strategy

Used: Substitution

Application:

Expand polynomial before differentiating.

Final Logic:

Polynomial derivative gives \(2x+2\).

🧠 Memory Trick

"Power comes down by one"

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

16 For an alternating electrical current case, dynamic voltage is
\(V=5sin⁡x.\)
What is the numerical magnitude of the derivative evaluated precisely at \(x=0\)?

A 0

B 1

C 5

D −5

💡 Short Explanation

Differentiate sine function Substitute \(x=0\) Use \(cos⁡0=1\)

✅ Correct Answer Explanation

Given: \(V=5sin⁡x\) Differentiate: \(\frac{dV}{dx}=5cos⁡x\) At \(x=0\): \(\frac{dV}{dx}=5cos⁡0=5(1)=5\) Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Derivative at zero is not zero.
Option B → Ignores coefficient 5.
Option D → Wrong sign for cosine derivative value.

🎯 Strategy

Used: Substitution

Application:

Use standard derivative then evaluate numerically.

Final Logic:

\(5cos⁡0=5\).

🧠 Memory Trick

"Sin becomes cos"

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

17 Identify the completely incorrect mathematical statement concerning the concept "Differentiable implies continuous".
\({lim⁡}_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\neq \frac{f(x+h)-f(x)}{h}.\)
\({lim⁡}_{h\rightarrow 0}[f(x+h)-f(x)]=0.\)

A The proof fundamentally relies on showing

B It uses the limit manipulation property

C If a function is starkly discontinuous at \(x=a\), it cannot possibly be differentiable there.

D Every strictly differentiable function evaluates as continuous.

💡 Short Explanation

Statement contradicts derivative definition Differentiability requires valid limit Continuous follows from differentiability

✅ Correct Answer Explanation

Option A is mathematically meaningless because differentiability requires: \({lim⁡}_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\) to exist finitely. The statement incorrectly compares a limit with its expression using inequality. Hence option A is completely incorrect.

❓ Why Other Options Are Incorrect

Option B → Correctly reflects continuity proof manipulation.
Option C → Discontinuous functions cannot be differentiable.
Option D → Standard theorem: differentiable implies continuous.

🎯 Strategy

Used: Elimination

Application:

Verify theorem statements conceptually.

Final Logic:

Only option A contradicts derivative logic.

🧠 Memory Trick

"No derivative without continuity"

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

18 The converse of the theorem "Differentiable ⇒ Continuous" is mathematically not true. Which statement accurately represents this false converse?

A Discontinuous functions are smoothly differentiable.

B Every continuous function is smoothly differentiable.

C Differentiable functions are mathematically not continuous.

D If a function is not differentiable, it is strictly discontinuous.

💡 Short Explanation

Converse reverses theorem direction Continuous need not imply differentiable Modulus function is counterexample

✅ Correct Answer Explanation

The original theorem states: \(Differentiable\Rightarrow Continuous\) Its converse is: \(Continuous\Rightarrow Differentiable\) which is false. Hence option B correctly states the false converse.

❓ Why Other Options Are Incorrect

Option A → Discontinuous functions cannot be smoothly differentiable.
Option C → Opposite of true theorem.
Option D → Non-differentiability does not always imply discontinuity.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Reverse theorem logically to identify converse.

Final Logic:

Converse claims continuity guarantees differentiability.

🧠 Memory Trick

"Continuous not always smooth"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Converse of Differentiability Theorem

19 Which of the following analytical functions exhibit a classic "sharp corner" (rendering a non-differentiable point) exactly at \(x=0\)?
1. \(f(x)=∣x∣\)
2. \(f(x)=x^{2/3}\)
3. \(f(x)=x^{2}\)

A 1 only

B 2 only

C 1 and 2 only

D 1 and 3 only

💡 Short Explanation

Modulus has corner at origin \(x^{2/3}\)has cusp-like behavior \(x^{2}\)remains smooth everywhere

✅ Correct Answer Explanation

\(f(x)=∣x∣\) has unequal side derivatives at 0. Also, \(f(x)=x^{2/3}\) has a cusp/non-smooth tangent behavior at 0. Both are non-differentiable there. But: \(x^{2}\) is smooth and differentiable everywhere. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores \(x^{2/3}\)singular behavior.
Option B → Ignores modulus sharp corner.
Option D → \(x^{2}\)is differentiable at 0.

🎯 Strategy

Used: Odd One Out

Application:

Separate smooth polynomial from non-smooth graphs.

Final Logic:

Only first two fail differentiability at 0.

🧠 Memory Trick

"Corner and cusp break derivative"

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

20 For the absolute modulus function
\(f(x)=∣x∣,\)
formally evaluate \(df/dx\) exactly at the origin point \(x=0\).

A 0

B 1

C Does not exist

D −1

💡 Short Explanation

LHD and RHD differ Unique tangent absent Hence derivative undefined

✅ Correct Answer Explanation

For: \(f(x)=∣x∣\) Left derivative at 0 is: \(-1\) Right derivative at 0 is: \(1\) Since both are unequal, derivative at: \(x=0\) does not exist. Therefore option C is correct.

❓ Why Other Options Are Incorrect

Option A → Derivative is not zero.
Option B → Only right derivative equals 1.
Option D → Only left derivative equals −1.

🎯 Strategy

Used: Elimination

Application:

Compare left and right derivatives.

Final Logic:

Unequal one-sided derivatives imply non-differentiability.

🧠 Memory Trick

"Left −1, Right +1 → No derivative"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Modulus Function at Origin

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