CUET UG Applied Mathematics Booster Test 2

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CUET UG Applied Mathematics Booster Test 2 - Differentiation Basics

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 2 - Differentiation Basics

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QUESTION 1 OF 20

What is the derivative of the power function y = x^(-2) + 2x³?

Answer Review

1 What is the derivative of the power function y = x^(-2) + 2x³?

A -2x^(-3) + 6x²

B 2x^(-3) + 5x²

C -2x^(-1) + 6x

D -x^(-3) + 3x²

💡 Short Explanation

Power rule applied term-wise Derivative of x⁻² is -2x⁻³ Derivative of 2x³ is 6x²

✅ Correct Answer Explanation

Using d(xⁿ)/dx = nxⁿ⁻¹: x⁻² → -2x⁻³ and 2x³ → 6x², so total derivative is -2x⁻³ + 6x².

❓ Why Other Options Are Incorrect

Option B → Wrong sign in first term
Option C → Incorrect powers and coefficients
Option D → Incorrect differentiation of both terms

🎯 Strategy

Used: Term-wise Differentiation

Application: Apply power rule separately to each term

Final Logic: Add derivatives of each term

🧠 Memory Trick

"Differentiate each term separately"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Derivatives – Topic: Power Rule

2 Match the exponential and logarithmic functions in List I with their derivatives in List II:

List I	List II
1. y = 2^x	a. y' = e^x
2. y = log(x)	b. y' = 2^x log 2
3. y = e^(2x)	c. y' = 2e^(2x)
4. y = e^x	d. y' = 1/x

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

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

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

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

💡 Short Explanation

2^x → 2^x log2 log x → 1/x e^(2x) → 2e^(2x) e^x → e^x

✅ Correct Answer Explanation

Standard derivative rules match exactly with Option A pairing.

❓ Why Other Options Are Incorrect

Option B → Incorrect mapping of exponential/log forms
Option C → Completely mismatched derivatives
Option D → Wrong structure pairing

🎯 Strategy

Used: Formula Matching

Application: Apply known derivative formulas

Final Logic: Direct correspondence of functions

🧠 Memory Trick

"e stays same, 2^x gets log2"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Derivatives – Topic: Standard Derivatives

3 Which of the following statements about constant and sum rules are TRUE?
(1) d(c f(x))/dx = c f'(x)
(2) d(f(x) ± g(x))/dx = f'(x) ± g'(x)
(3) d(x^n)/dx = nx^(n-1) applies to any real number n.
(4) The derivative of a sum is the product of the derivatives.

A 1, 2, 4

B 1, 2

C 1, 2, 3

D 1, 2, 3, 4

💡 Short Explanation

Constant multiple rule is correct Sum/difference rule correct Power rule valid for real n Product claim is false

✅ Correct Answer Explanation

1, 2, and 3 are standard NCERT rules. → 4 is false because derivative of sum is not product.

❓ Why Other Options Are Incorrect

Option A → Omits correct statement 3
Option B → Incomplete
Option D → Includes incorrect statement 4

🎯 Strategy

Used: Rule Validation

Application: Verify each rule independently

Final Logic: Eliminate false statement 4

🧠 Memory Trick

"Sum stays sum, never product"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Derivatives – Topic: Differentiation Rules

4 Identify the incorrect statement regarding product and quotient differentiation rules:

A Product rule is correct

B Quotient rule denominator is square of numerator function

C Quotient rule has negative sign in numerator

D d(x e^x)/dx = e^x (x + 1)

💡 Short Explanation

Denominator should be g(x)² Not numerator squared Other rules correct

✅ Correct Answer Explanation

Correct quotient rule denominator is (g(x))², not numerator squared.

❓ Why Other Options Are Incorrect

Option A → Correct product rule
Option C → Correct sign structure
Option D → Correct derivative

🎯 Strategy

Used: Error Spotting

Application: Check rule structure

Final Logic: Denominator identification error

🧠 Memory Trick

"Denominator always squared"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Derivatives – Topic: Product & Quotient Rule

📄 Passage

The equation 3x² + xy + y = 0 can be written as y = (-3x²)/(1+x), i.e., y can be explicitly expressed in terms of x only. We have been finding derivatives in such cases directly. However, let us consider an equation like x³ + y³ - 9xy = 0. Here, y cannot be explicitly expressed as a single isolated function of x. Such functions are called implicit functions. We treat y as a differentiable function of x and differentiate both sides of the equation with respect to x using standard differentiation rules, specifically applying the chain rule to the y terms. The process by which we systematically isolate and find dy/dx is called implicit differentiation.

According to the passage, what defines the implicit differentiation concept?

A y can be isolated easily

B y cannot be explicitly expressed as a single function of x

C Derivative is always zero

D Function has only constants

💡 Short Explanation

y is not isolated Treat y as function Differentiate both sides

✅ Correct Answer Explanation

Implicit functions require differentiation without solving for y explicitly.

❓ Why Other Options Are Incorrect

Option A → Opposite of definition
Option C → False
Option D → Irrelevant

🎯 Strategy

Used: Definition Recall

Application: Identify implicit function meaning

Final Logic: No explicit isolation of y

🧠 Memory Trick

"Hidden y → implicit"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Derivatives – Topic: Implicit Differentiation

📄 Passage

When performing derivatives of implicit equations involving y, how is y treated structurally?

A Constant

B Independent variable

C Differentiable function of x

D Zero

💡 Short Explanation

y depends on x Chain rule applies dy/dx introduced

✅ Correct Answer Explanation

In implicit differentiation, y is assumed to be a function of x, so chain rule is applied.

❓ Why Other Options Are Incorrect

Option A → Incorrect assumption
Option B → Not independent
Option D → Invalid

🎯 Strategy

Used: Conceptual Classification

Application: Identify variable dependency

Final Logic: y = f(x) assumption

🧠 Memory Trick

"y depends on x always"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Derivatives – Topic: Implicit Functions

7 Apply the chain rule in implicit forms to the circle equation x² + y² = r² to find dy/dx:

A x/y

B -x/y

C y/x

D -y/x

💡 Short Explanation

Differentiate both sides 2x + 2y dy/dx = 0 Solve for dy/dx

✅ Correct Answer Explanation

dy/dx = -x/y after rearranging the differentiated equation.

❓ Why Other Options Are Incorrect

Option A → Missing negative sign
Option C → Wrong ratio
Option D → Incorrect inversion

🎯 Strategy

Used: Chain Rule Application

Application: Differentiate implicit equation

Final Logic: Rearrangement gives negative ratio

🧠 Memory Trick

"Circle slope = negative ratio"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Derivatives – Topic: Circle Differentiation

8 For the logarithmic implicit expression log(xy) = x + y, the implicit derivative dy/dx evaluates to:

A (y(1-x))/(x(y-1))

B (x(1-y))/(y(x-1))

C (y(x-1))/(x(1-y))

D (y(x+1))/(x(y+1))

💡 Short Explanation

Log differentiation Differentiate both sides Solve algebraically

✅ Correct Answer Explanation

Differentiating gives (1/x + 1/y dy/dx) = 1 + dy/dx, solving yields option C.

❓ Why Other Options Are Incorrect

Option A → Wrong sign structure
Option B → Incorrect variable placement
Option D → Incorrect derivation

🎯 Strategy

Used: Algebraic Rearrangement

Application: Collect dy/dx terms

Final Logic: Solve linear equation

🧠 Memory Trick

"Logs → rearrange → isolate y'"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Logarithmic Differentiation – Topic: Implicit Log Functions

9 A curve is given by parametric representation x = g(t) and y = f(t). The variable t is formally known as:

A Constant

B Independent domain

C Parameter

D Implicit variable

💡 Short Explanation

t defines both x and y Controls curve Not dependent variable

✅ Correct Answer Explanation

In parametric equations, t is called the parameter controlling the curve.

❓ Why Other Options Are Incorrect

Option A → Not constant
Option B → Incorrect term
Option D → Not implicit variable

🎯 Strategy

Used: Definition Recall

Application: Identify role of t

Final Logic: t defines parameterization

🧠 Memory Trick

"t = parameter"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Parametric Equations – Topic: Parameter Definition

10 For the parametric curve x = 1/(1+t) and y = t/(1+t), the derivative of the parametric curve dy/dx equals:

A -1

B 1

C t

D -t

💡 Short Explanation

Compute dy/dt Compute dx/dt Divide

✅ Correct Answer Explanation

dy/dt = 1/(1+t)² and dx/dt = -1/(1+t)², so dy/dx = -1.

❓ Why Other Options Are Incorrect

Option B → Wrong sign
Option C → Incorrect simplification
Option D → Extra variable

🎯 Strategy

Used: Parametric Differentiation

Application: Ratio of derivatives

Final Logic: Cancellation leads to -1

🧠 Memory Trick

"Opposites cancel → -1"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Parametric Differentiation – Topic: Basic Parametric Curve

11 By executing parameter elimination on x = t³ and y = t⁶, the resulting cartesian relation is:

A y = x²

B y = x³

C x = y²

D x² = y³

💡 Short Explanation

Eliminate parameter t Express both in same power Substitute relation

✅ Correct Answer Explanation

x = t³ ⇒ t = x^(1/3) → y = t⁶ = (t³)² = x²

❓ Why Other Options Are Incorrect

Option B → Incorrect power relation
Option C → Wrong rearrangement
Option D → Incorrect exponent structure

🎯 Strategy

Used: Parameter Elimination

Application: Express both variables in terms of t

Final Logic: Convert both to same base power

🧠 Memory Trick

"Cube to square relation"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Parametric Equations – Topic: Elimination of Parameter

12 The slope of the tangent from the parametric form x = at², y = 2at evaluated at t = 4 is:

A 1/2

B 1/4

C 4

D 2a

💡 Short Explanation

dy/dx = 1/t Substitute t = 4 Simplify

✅ Correct Answer Explanation

For x = at² and y = 2at, dy/dx = (2a)/(2at) = 1/t → At t = 4, slope = 1/4

❓ Why Other Options Are Incorrect

Option A → Incorrect substitution
Option C → Wrong inversion
Option D → Extra parameter a

🎯 Strategy

Used: Direct Formula Application

Application: Use known parametric derivative

Final Logic: Substitute value of t

🧠 Memory Trick

"Slope = 1/t always"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Derivatives – Topic: Parametric Differentiation

13 To differentiate functions of type f(x)^g(x) such as y = (1+x)^(1/x), the most appropriate first step is:

A Use the power rule directly

B Apply quotient rule on exponent

C Take logarithm of both sides

D Differentiate implicitly without logs

💡 Short Explanation

Variable exponent present Log simplifies expression Enables differentiation

✅ Correct Answer Explanation

Logarithmic differentiation is required when both base and exponent vary.

❓ Why Other Options Are Incorrect

Option A → Power rule invalid here
Option B → Not applicable
Option D → Inefficient method

🎯 Strategy

Used: Method Selection

Application: Identify mixed exponent structure

Final Logic: Convert using logarithm

🧠 Memory Trick

"Power + variable → take log"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Logarithmic Differentiation – Topic: Variable Exponent Functions

14 Differentiating using logarithms on log y = (1/x) log(1+x) via product rule yields:

A (1/y) y' = (1/x)(1/(1+x)) - (1/x²) log(1+x)

B y' = (1/(1+x))

C y' = -1/x²

D (1/y) y' = x / (1+x)

💡 Short Explanation

Product rule used Differentiate log form Apply chain rule

✅ Correct Answer Explanation

Differentiating log y = (1/x) log(1+x) gives product rule result combining derivatives of both factors.

❓ Why Other Options Are Incorrect

Option B → Incomplete derivative
Option C → Irrelevant simplification
Option D → Incorrect transformation

🎯 Strategy

Used: Product + Chain Rule

Application: Differentiate product of functions

Final Logic: Apply both differentiation rules

🧠 Memory Trick

"Log product → product rule inside log"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Logarithmic Differentiation – Topic: Product Rule in Logs

15 If y = x^x, what is the exact numerical value of dy/dx evaluated at x = e?

A e^e

B 2e^e

C e

D 2e

💡 Short Explanation

dy/dx = x^x(1 + log x) Substitute x = e log e = 1

✅ Correct Answer Explanation

dy/dx = e^e(1 + 1) = 2e^e

❓ Why Other Options Are Incorrect

Option A → Missing factor 2
Option C → Missing power structure
Option D → Incorrect scaling

🎯 Strategy

Used: Substitution

Application: Apply known derivative formula

Final Logic: Plug x = e into expression

🧠 Memory Trick

"log e = 1 doubles result"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Logarithmic Differentiation – Topic: x^x Evaluation

16 For mixed variable exponents xy + y^x = ab (constants), if u = xy and v = y^x, what is du/dx + dv/dx?

A ab

B 0

C 1

D a+b

💡 Short Explanation

Differentiate both terms RHS constant Sum derivative = 0

✅ Correct Answer Explanation

Since xy + y^x = constant (ab), derivative of RHS is 0, hence du/dx + dv/dx = 0.

❓ Why Other Options Are Incorrect

Option A → Misinterprets constant
Option C → Incorrect assumption
Option D → Not derived

🎯 Strategy

Used: Constant Rule

Application: Derivative of constant = 0

Final Logic: Sum of derivatives equals zero

🧠 Memory Trick

"Constant → derivative zero"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Implicit Differentiation – Topic: Mixed Functions

17 If y = x² + 3x + 4, what is the second order derivative d²y/dx²?

A 2x+3

B 2

C 3

D 0

💡 Short Explanation

First derivative: 2x + 3 Second derivative: constant Simplify

✅ Correct Answer Explanation

d²y/dx² = derivative of (2x + 3) = 2

❓ Why Other Options Are Incorrect

Option A → First derivative
Option C → Incorrect constant
Option D → Wrong

🎯 Strategy

Used: Sequential Differentiation

Application: Differentiate twice

Final Logic: Constant second derivative

🧠 Memory Trick

"Quadratic → second derivative constant"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Higher Order Derivatives – Topic: Quadratic Functions

18 Find the third and higher derivatives application for the function y = x⁴ + x³. The third derivative d³y/dx³ is:

A 12x² + 6x

B 24x + 6

C 4x³ + 3x²

D 24

💡 Short Explanation

Differentiate stepwise Third derivative reduces degree Apply power rule

✅ Correct Answer Explanation

y' = 4x³ + 3x² → y'' = 12x² + 6x → y''' = 24x + 6

❓ Why Other Options Are Incorrect

Option A → Second derivative
Option C → Original function structure
Option D → Not correct third derivative

🎯 Strategy

Used: Stepwise Differentiation

Application: Repeated differentiation

Final Logic: Three successive derivatives

🧠 Memory Trick

"Each derivative drops power"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Higher Order Derivatives – Topic: Polynomial Differentiation

19 If x = at² and y = 2at, what is the parametric second derivative d²y/dx² evaluated at t = 1?

A -1/(2a)

B 1/(2a)

C -1/a

D 1/a

💡 Short Explanation

dy/dx = 1/t Differentiate again Substitute t = 1

✅ Correct Answer Explanation

d²y/dx² = d/dt(1/t) × dt/dx = (-1/t²) × (1/2at) = -1/(2at³) → at t=1 gives -1/(2a)

❓ Why Other Options Are Incorrect

Option B → Wrong sign
Option C → Incorrect scaling
Option D → Missing negative sign

🎯 Strategy

Used: Chain Rule Extension

Application: Convert second derivative via t

Final Logic: Apply derivative of dy/dx

🧠 Memory Trick

"Second derivative introduces minus"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Parametric Differentiation – Topic: Second Derivative

20 In repeated differentiation problems, if y = e^(2x), the n-th derivative of y resolves to:

A 2^n e^(2x)

B 2 e^(nx)

C n e^(2x)

D e^(2x)

💡 Short Explanation

Each differentiation multiplies by 2 Repeated pattern Exponential remains same

✅ Correct Answer Explanation

Each derivative brings factor 2, so n-th derivative is 2^n e^(2x).

❓ Why Other Options Are Incorrect

Option B → Incorrect exponent structure
Option C → Wrong scaling
Option D → Only first derivative form

🎯 Strategy

Used: Pattern Recognition

Application: Identify repeated derivative pattern

Final Logic: Multiply factor 2 repeatedly

🧠 Memory Trick

"Each derivative doubles factor 2"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Higher Order Derivatives – Topic: Exponential Functions

CUET UG Applied Mathematics Booster Test 2 - Differentiation Basics

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