CUET UG Applied Mathematics Booster Test 1

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CUET UG Applied Mathematics Booster Test 1 - Differentiation Basics

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 1 - Differentiation Basics

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QUESTION 1 OF 20

What is the derivative of the power function y = x^(3/2) with respect to x?

Answer Review

1 What is the derivative of the power function y = x^(3/2) with respect to x?

A (3/2) x^(1/2)

B (2/3) x^(1/2)

C x^(1/2)

D (3/2) x^(-1/2)

💡 Short Explanation

Power rule: d(xⁿ)/dx = nxⁿ⁻¹ n = 3/2 Result simplifies

✅ Correct Answer Explanation

Applying NCERT power rule, exponent comes down and is reduced by 1: d(x^(3/2))/dx = (3/2)x^(1/2)

❓ Why Other Options Are Incorrect

Option B → Incorrect coefficient
Option C → Missing multiplier
Option D → Wrong exponent reduction

🎯 Strategy

Used: Power Rule Application

Application: Apply exponent rule directly

Final Logic: Multiply exponent and reduce power

🧠 Memory Trick

"Power down, exponent minus one"

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

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

List I	List II
1. d(e^(2x))/dx	a. 2/x
2. d(log x²)/dx	b. 2e^(2x)
3. d(2^x)/dx	c. 2^x log 2
4. d(5)/dx	d. 0

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

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

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

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

💡 Short Explanation

e^(2x) → 2e^(2x) log x² → 2/x 2^x → 2^x log2 constant → 0

✅ Correct Answer Explanation

Standard differentiation rules applied term-wise match Option A exactly.

❓ Why Other Options Are Incorrect

Option B → Incorrect pairing of derivatives
Option C → Wrong matching order
Option D → Completely mismatched

🎯 Strategy

Used: Direct Matching

Application: Apply standard derivative formulas

Final Logic: Correct formula pairing

🧠 Memory Trick

"e doubles, log simplifies, constant dies"

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

3 Which of the following differentiation rule applications are correct?
(1) The derivative of 3x² + 5x is 6x + 5.
(2) d(f(x) - g(x))/dx = d(f(x))/dx - d(g(x))/dx
(3) Constants can be factored out, i.e., d(k f(x))/dx = k d(f(x))/dx.
(4) If y = 7, dy/dx = 7.

A 1, 2, 4

B 1, 2

C 1, 2, 3

D 1, 2, 3, 4

💡 Short Explanation

Polynomial derivative is correct Sum/difference rule correct Constant rule correct

✅ Correct Answer Explanation

1, 2, and 3 follow NCERT differentiation rules. → 4 is incorrect because derivative of constant 7 is 0, not 7.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used: Concept Elimination

Application: Verify each rule validity

Final Logic: Only 1, 2, 3 are correct

🧠 Memory Trick

"Constant always becomes zero"

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

4 Identify the incorrect application of the quotient and product rules:

A d/dx [1/x] = -1/x²

B Quotient rule subtracts f(x)g'(x) in numerator

C d/dx [e^x / x] = (x e^x - e^x)/x²

D d/dx [x / e^x] = (e^x + x e^x)/(e^x)²

💡 Short Explanation

Quotient rule requires subtraction Option D has incorrect sign Others are correct

✅ Correct Answer Explanation

Correct derivative is (e^x - x e^x)/(e^x)², so option D is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct
Option B → Correct rule description
Option C → Correct application

🎯 Strategy

Used: Error Detection

Application: Check sign in quotient rule

Final Logic: Sign error identifies incorrect option

🧠 Memory Trick

"Quotient = subtract, not add"

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

5 Implicit differentiation is used because:

A Different derivatives for each function

B We find derivatives without explicit formula

C Implicit equations never form curves

D Explicit formulas undefined

💡 Short Explanation

No need to isolate y Treat y as function Apply chain rule

✅ Correct Answer Explanation

NCERT defines implicit differentiation as differentiation without explicitly solving for y.

❓ Why Other Options Are Incorrect

Option A → Incorrect reasoning
Option C → False statement
Option D → Irrelevant claim

🎯 Strategy

Used: Concept Identification

Application: Recognize definition of method

Final Logic: Works without explicit expression

🧠 Memory Trick

"No solving needed"

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

6 For 3x² + xy + y = 0, implicit differentiation gives dy/dx forms:

A (-6x - 3x²)/(1+x)²

B (-6x - y)/(1+x)

C Both A and B correct

D Neither A nor B

💡 Short Explanation

Different forms of same result Using substitution or simplification Equivalent expressions

✅ Correct Answer Explanation

Both expressions reduce to same dy/dx after algebraic substitution using original equation.

❓ Why Other Options Are Incorrect

Option A → Correct form
Option B → Equivalent form
Option D → Incorrect since both are valid

🎯 Strategy

Used: Equivalence Check

Application: Verify algebraic equivalence

Final Logic: Both expressions match

🧠 Memory Trick

"Different form, same value"

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

7 Differentiating e^(xy) = 2x gives LHS as:

A e^(xy)(x + y)

B e^(xy)(y + x(dy/dx))

C e^(xy)(dy/dx)

D xy e^(xy-1)

💡 Short Explanation

Chain rule applied Product rule inside exponent dy/dx appears

✅ Correct Answer Explanation

d/dx[e^(xy)] = e^(xy)·d(xy)/dx = e^(xy)(y + x dy/dx).

❓ Why Other Options Are Incorrect

Option A → Missing dy/dx term
Option C → Incomplete differentiation
Option D → Wrong formula

🎯 Strategy

Used: Chain Rule

Application: Differentiate exponential composite function

Final Logic: Derivative of exponent multiplied

🧠 Memory Trick

"Exponential stays, inner differentiates"

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

8 If x^m y^n = (x+y)^(m+n), then dy/dx simplifies to:

A x/y

B y/x

C -y/x

D -x/y

💡 Short Explanation

Take logarithm Differentiate both sides Simplify symmetry

✅ Correct Answer Explanation

After log differentiation and simplification, dy/dx = y/x.

❓ Why Other Options Are Incorrect

Option A → Inverted ratio
Option C → Wrong sign
Option D → Incorrect direction

🎯 Strategy

Used: Log Differentiation

Application: Convert product powers to logs

Final Logic: Symmetry yields y/x

🧠 Memory Trick

"Swap powers → flip ratio"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Logarithmic Differentiation – Topic: Applications

9 Which defines a composite function in parametric representation?

A y = f(x)

B x² + y² = r²

C x = g(t), y = f(t)

D xy = c²

💡 Short Explanation

Parameter defines both variables t is independent parameter Forms curve representation

✅ Correct Answer Explanation

Parametric form expresses x and y as functions of parameter t.

❓ Why Other Options Are Incorrect

Option A → Not parametric
Option B → Implicit circle
Option D → Not parametric

🎯 Strategy

Used: Definition Recall

Application: Identify parametric structure

Final Logic: Both variables depend on t

🧠 Memory Trick

"t controls both x and y"

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

10 If x = 2at² and y = at⁴, dy/dx is:

A 2at

B t²

C 4at³

D a

💡 Short Explanation

dy/dt = 4at³ dx/dt = 4at Divide

✅ Correct Answer Explanation

dy/dx = (4at³)/(4at) = t².

❓ Why Other Options Are Incorrect

Option A → Incorrect simplification
Option C → dy/dt only
Option D → No derivation basis

🎯 Strategy

Used: Parametric Differentiation

Application: Apply ratio rule

Final Logic: Simplify derivative ratio

🧠 Memory Trick

"Cancel 4a → t squared remains"

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

11 To eliminate the parameter t and verify that x = at², y = 2at represents y² = 4ax, we:

A Add x and y directly.

B Substitute x and y into the equation to show (2at)² = 4a(at²).

C Differentiate x and y.

D Multiply x and y.

💡 Short Explanation

Use elimination method Substitute parametric forms Verify identity

✅ Correct Answer Explanation

From x = at² and y = 2at, substituting gives y² = (2at)² = 4a²t² = 4a(at²) = 4ax, confirming the relation.

❓ Why Other Options Are Incorrect

Option A → Irrelevant operation
Option C → Differentiation not required
Option D → Does not eliminate parameter

🎯 Strategy

Used: Substitution Method

Application: Directly substitute parameter expressions

Final Logic: Identity verified through substitution

🧠 Memory Trick

"Plug and prove"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Conic Sections – Topic: Parametric Form of Parabola

12 If a tangent from parametric form is given by x = at² and y = 2at, what is its slope at t = 3?

A 1/3

B 3

C 2a/3

D 1/6

💡 Short Explanation

dy/dx = 1/t Substitute t = 3 Result = 1/3

✅ Correct Answer Explanation

For x = at², y = 2at, dy/dx = (2a)/(2at) = 1/t. → At t = 3, slope = 1/3.

❓ Why Other Options Are Incorrect

Option B → Incorrect inversion
Option C → Extra parameter a
Option D → Incorrect simplification

🎯 Strategy

Used: Direct Substitution

Application: Use known parametric derivative formula

Final Logic: Substitute t into dy/dx = 1/t

🧠 Memory Trick

"Slope = 1/t always"

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

📄 Passage

Logarithmic differentiation is extremely useful not only for functions of the type [f(x)]^g(x) but also when differentiating complex expressions that involve products, quotients, and fractional powers. For example, if a function is given as y = (u(x)v(x))/w(x), taking the natural logarithm of both sides yields log y = log u(x) + log v(x) - log w(x). This property of logarithms converts cumbersome multiplication and division into simpler addition and subtraction. Differentiating both sides then gives (1/y)(dy/dx) = u'(x)/u(x) + v'(x)/v(x) - w'(x)/w(x). Multiplying the entire right-hand side by y gives the final derivative dy/dx safely.

According to the passage, applying logarithmic differentiation to a product of functions converts it into:

A A sum of logarithmic functions

B A quotient of logarithmic functions

C A single polynomial

D An exponential function

💡 Short Explanation

log transforms product into sum log(ab) = log a + log b Simplifies differentiation

✅ Correct Answer Explanation

Logarithmic properties convert multiplication into addition, making differentiation easier.

❓ Why Other Options Are Incorrect

Option B → Incorrect transformation
Option C → Not related to logs
Option D → No exponential conversion

🎯 Strategy

Used: Log Identity Recognition

Application: Apply log multiplication rule

Final Logic: Product becomes sum in logs

🧠 Memory Trick

"Multiply → becomes add in logs"

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

📄 Passage

After taking the logarithm on y = u(x)v(x) and differentiating, what is the proper expression isolated for dy/dx?

A y [u'(x)/u(x) + v'(x)/v(x)]

B u'(x) + v'(x)

C u'(x)v'(x)

D [u'(x)/u(x)] × [v'(x)/v(x)]

💡 Short Explanation

log y = log u + log v Differentiate both sides Multiply by y

✅ Correct Answer Explanation

From log y = log u + log v, differentiation gives (1/y) dy/dx = u'/u + v'/v, hence dy/dx = y[ u'/u + v'/v ].

❓ Why Other Options Are Incorrect

Option B → Missing logarithmic structure
Option C → Incorrect multiplication
Option D → Incorrect product of ratios

🎯 Strategy

Used: Log Differentiation

Application: Use log derivative formula

Final Logic: Multiply both sides by y

🧠 Memory Trick

"Log split → sum → multiply by y"

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

15 To differentiate y = xˣ, taking logarithm gives log y = x log x. The derivative dy/dx is:

A xˣ(log x)

B xˣ(1 + log x)

C 1 + log x

D x^(x−1)

💡 Short Explanation

Apply log differentiation Differentiate x log x Multiply back by y

✅ Correct Answer Explanation

log y = x log x ⇒ derivative gives y'/y = log x + 1 ⇒ dy/dx = xˣ(1 + log x).

❓ Why Other Options Are Incorrect

Option A → Missing +1 term
Option C → Missing multiplication by xˣ
Option D → Wrong rule

🎯 Strategy

Used: Log Differentiation

Application: Convert exponential form using logs

Final Logic: Combine log derivative and multiply by y

🧠 Memory Trick

"xˣ → multiply by (1 + log x)"

📚 Class XI – NCERT Mathematics (Part II) – Chapter: Logarithmic Differentiation – Topic: xˣ derivative

16 For mixed variable exponents, if xy = e^(x−y), taking logs gives log x + log y = x − y. Differentiating with respect to x yields dy/dx as:

A (x − y)/(x + y)

B (y(x−1))/(x(y+1))

C (y(1−x))/(x(y+1))

D (x(y−1))/(y(x+1))

💡 Short Explanation

Differentiate log equation Collect dy/dx terms Solve algebraically

✅ Correct Answer Explanation

Differentiating gives 1/x + (1/y)dy/dx = 1 − dy/dx. Solving yields dy/dx = y(x−1)/x(y+1).

❓ Why Other Options Are Incorrect

Option A → Incorrect simplification
Option C → Wrong sign arrangement
Option D → Variable inversion error

🎯 Strategy

Used: Algebraic Rearrangement

Application: Collect dy/dx terms on one side

Final Logic: Solve linear equation in dy/dx

🧠 Memory Trick

"Logs → rearrange → isolate y'"

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

17 If y = log x, find the second order derivative d²y/dx².

A 1/x

B 1/x²

C -1/x²

D -1/x

💡 Short Explanation

First derivative = 1/x Differentiate again Apply power rule

✅ Correct Answer Explanation

d/dx(1/x) = d(x⁻¹)/dx = −x⁻² = −1/x².

❓ Why Other Options Are Incorrect

Option A → First derivative only
Option B → Missing negative sign
Option D → Incorrect power

🎯 Strategy

Used: Direct Differentiation

Application: Apply power rule twice

Final Logic: Second derivative introduces negative sign

🧠 Memory Trick

"log → 1/x → -1/x²"

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

18 The third order derivative of y = f(x) is written correctly as:

A f''(x)

B d³y/dx³

C (dy/dx)³

D 3(dy/dx)

💡 Short Explanation

Third derivative notation Standard calculus form Represents rate of change of second derivative

✅ Correct Answer Explanation

The third derivative is denoted as d³y/dx³ or f'''(x) in NCERT notation.

❓ Why Other Options Are Incorrect

Option A → Second derivative
Option C → Not valid notation
Option D → Incorrect expression

🎯 Strategy

Used: Notation Recall

Application: Identify correct derivative order

Final Logic: Standard third derivative form

🧠 Memory Trick

"3rd power → triple d"

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

19 The second order derivative of a parametric function requires:

A Simply dividing second derivatives

B Differentiating dy/dx w.r.t t and multiplying by dt/dx

C Differentiating y once

D Multiplying second derivatives

💡 Short Explanation

Use chain rule extension Convert parameter derivative Apply dt/dx factor

✅ Correct Answer Explanation

NCERT formula: d²y/dx² = d/dt(dy/dx) × dt/dx.

❓ Why Other Options Are Incorrect

Option A → Incorrect method
Option C → Incomplete
Option D → No rule basis

🎯 Strategy

Used: Chain Rule Extension

Application: Convert second derivative via parameter t

Final Logic: Differentiate in t then convert to x

🧠 Memory Trick

"Differentiate in t, convert using dt/dx"

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

20 In repeated differentiation problems, if y = √(x² + 1), finding the second derivative primarily involves initially:

A Applying the chain rule to the power function

B Applying the product rule twice

C Using logarithmic differentiation only

D Integrating the function

💡 Short Explanation

Rewrite as (x² + 1)^(1/2) Apply chain rule Then differentiate again

✅ Correct Answer Explanation

The function is a composite function, so chain rule is the first required step before further differentiation.

❓ Why Other Options Are Incorrect

Option B → Not required structure
Option C → Not necessary here
Option D → Irrelevant operation

🎯 Strategy

Used: Functional Identification

Application: Recognize composite structure

Final Logic: Chain rule is the starting step

🧠 Memory Trick

"Root → always chain first"

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

CUET UG Applied Mathematics Booster Test 1 - Differentiation Basics

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