CUET UG Mathematics Booster Test 2 - Introduction & Basics -

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CUET UG Mathematics Booster Test 2 - Introduction & Basics

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 2 - Introduction & Basics

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QUESTION 1 OF 20

A stock's moving growth rate data over time t is modeled by
g(t)=1/t.
Using integration, find the total accumulated growth data from t=1 to t=e².

Answer Review

1 A stock's moving growth rate data over time t is modeled by
g(t)=1/t.
Using integration, find the total accumulated growth data from t=1 to t=e².

A e²

B 1

C 2

D e

💡 Short Explanation

Total accumulation equals a definite integral. Anti-derivative of 1/t is ln t. Evaluate between 1 and e².

✅ Correct Answer Explanation

The accumulated growth is ∫₁ᵉ² (1/t)dt = [ln t]₁ᵉ² = ln(e²)−ln(1)=2−0=2. Therefore Option C is correct. Options A, B, and D result from confusing the logarithm with the exponential value or incomplete evaluation.

❓ Why Other Options Are Incorrect

Option A → Uses the upper limit directly.
Option B → Ignores the upper-limit contribution.
Option D → Confuses ln(e²) with e.

🎯 Strategy

Used: Substitution

Application: Use the standard logarithmic integral and evaluate limits.

Final Logic: ln(e²)−ln(1)=2.

🧠 Memory Trick

"1/x → Log."

📚 Class XII – NCERT Mathematics – Chapter: Definite Integrals – Topic: Evaluation Using Standard Integrals

2 Evaluate the area of the region represented by the integral
∫₀^π sin x dx.

A 0

B π/2

C 1

D 2

💡 Short Explanation

Integral of sin x is −cos x. Evaluate from 0 to π. Area equals 2.

✅ Correct Answer Explanation

∫₀^π sin x dx = [−cos x]₀^π = (−(−1))−(−1)=1+1=2. Since sin x is positive on [0,π], the integral equals the area under the curve. Thus Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Integral is not zero because area is positive.
Option B → Incorrect evaluation of limits.
Option C → Represents only half the actual area.

🎯 Strategy

Used: Substitution

Application: Apply the anti-derivative and evaluate limits.

Final Logic: −cosπ+cos0=2.

🧠 Memory Trick

"Half-wave sine area = 2."

📚 Class XII – NCERT Mathematics – Chapter: Definite Integrals – Topic: Area Interpretation

3 Assertion (A): The anti-derivative of e²ˣ is 2e²ˣ+C.
Reason (R): Integration is identical to differentiation, thus the chain rule applies via multiplication.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

∫e²ˣdx=(1/2)e²ˣ+C. Assertion is incorrect. Reason is also conceptually incorrect.

✅ Correct Answer Explanation

The correct anti-derivative is ∫e²ˣdx=(1/2)e²ˣ+C, not 2e²ˣ+C. Hence the assertion is false. The reason is also false because integration is the inverse of differentiation, not identical to it. Therefore both statements are false.

❓ Why Other Options Are Incorrect

Option B → Assertion is not true.
Option C → Both statements are false.
Option D → Reason is also false.

🎯 Strategy

Used: Substitution

Application: Differentiate proposed answers to verify.

Final Logic: d/dx[(1/2)e²ˣ]=e²ˣ.

🧠 Memory Trick

"Reverse Chain = Divide by Inner Derivative."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Exponential Integrals

4 Consider the family of curves given by
y=∫1/(1+x²) dx.
Graphically, all curves in this family are bounded horizontally by which asymptotes?

A y=0 and y=π

B y=−π/2 and y=π/2 (ignoring C shifting for bounds relative to themselves)

C y=−1 and y=1

D y=−e and y=e

💡 Short Explanation

Integral equals tan⁻¹x+C. arctan x approaches ±π/2. Horizontal asymptotes occur at those values.

✅ Correct Answer Explanation

Since ∫1/(1+x²)dx = tan⁻¹x + C, and tan⁻¹x approaches π/2 as x→∞ and −π/2 as x→−∞, the family has horizontal asymptotes at these values relative to each vertical shift. Thus Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Not the asymptotic limits of arctan.
Option C → These are not asymptotes of tan⁻¹x.
Option D → No exponential relationship exists.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recognize the standard anti-derivative.

Final Logic: Integral gives arctan x with limits ±π/2.

🧠 Memory Trick

"arctan → ±π/2."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Standard Integrals

5 If ∫f(x)dx=F(x)+C and ∫g(x)dx=G(x)+K, then which are valid properties?
1. ∫[k·f(x)]dx=kF(x)+C′
2. ∫[f(x)·g(x)]dx=F(x)·G(x)+C′
3. ∫[f(x)−g(x)]dx=F(x)−G(x)+C′

A 1 and 2 only

B 2 and 3 only

C 1 and 3 only

D 1, 2, and 3

💡 Short Explanation

Integration is linear. Product rule for integration does not exist. Statements 1 and 3 are valid.

✅ Correct Answer Explanation

Constant multiples and differences distribute through integration, making statements 1 and 3 correct. However, ∫(fg)dx ≠ (∫fdx)(∫gdx), so statement 2 is false. Therefore the correct choice is Option C.

❓ Why Other Options Are Incorrect

Option A → Includes false statement 2.
Option B → Omits valid statement 1.
Option D → Statement 2 is invalid.

🎯 Strategy

Used: Option Grouping

Application: Check each property independently.

Final Logic: Integration is linear, not multiplicative.

🧠 Memory Trick

"Sum Works, Product Doesn't."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Properties of Indefinite Integrals

6 Arrange the correct steps to evaluate the integral
∫(2x sin(x²+1)) dx
via substitution:
1. Substitute x²+1=t, yielding ∫sin t dt.
2. Integrate to get −cos t+C.
3. Resubstitute t to get −cos(x²+1)+C.
4. Determine 2x dx=dt.

A 1, 2, 4, 3

B 4, 2, 1, 3

C 1, 3, 2, 4

D 4, 1, 2, 3

💡 Short Explanation

Differentiate substitution first. Replace variables. Integrate and resubstitute.

✅ Correct Answer Explanation

First let t=x²+1, then dt=2xdx. Substituting gives ∫sin t dt. Integrating yields −cos t+C. Finally replace t by x²+1. Hence the correct order is 4→1→2→3, corresponding to Option D.

❓ Why Other Options Are Incorrect

Option A → Begins before finding dt.
Option B → Integrates before substitution.
Option C → Resubstitutes before integration.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Follow the standard substitution method.

Final Logic: Define t, convert, integrate, restore variable.

🧠 Memory Trick

"Substitute → Integrate → Back Substitute."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Substitution Method

7 A drone's vertical velocity vector is
v⃗(t)=(t−2)k̂.
If it starts at height 0k̂ at t=0, at what time t>0 does it return to the ground (position 0k̂)?

A t=4

B t=2

C t=3

D t=1

💡 Short Explanation

Integrate velocity to obtain position. Apply initial height condition. Solve position equal to zero.

✅ Correct Answer Explanation

Position is s(t)=∫(t−2)dt=t²/2−2t+C. Since s(0)=0, C=0. Setting s(t)=0 gives t(t/2−2)=0. The positive solution is t=4. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Velocity becomes zero, not position.
Option C → Position remains negative.
Option D → Position is not zero.

🎯 Strategy

Used: Substitution

Application: Integrate velocity and solve for position.

Final Logic: t²/2−2t=0 ⇒ t=4.

🧠 Memory Trick

"Velocity Integral = Position."

📚 Class XII – NCERT Mathematics – Chapter: Applications of Integration – Topic: Motion Problems

8 Calculate the definite integral utilized in a physics problem:
∫₀¹ xeˣ dx
using integration by parts.

A e

B 1

C e−1

D 0

💡 Short Explanation

Apply integration by parts. Integral becomes eˣ(x−1). Evaluate between 0 and 1.

✅ Correct Answer Explanation

Using integration by parts: ∫xeˣdx=eˣ(x−1)+C. Evaluating from 0 to 1 gives [eˣ(x−1)]₀¹ = 0−(−1)=1. Thus Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores lower-limit contribution.
Option C → Represents an intermediate expression.
Option D → Integral is positive.

🎯 Strategy

Used: Substitution

Application: Use the standard integration-by-parts result.

Final Logic: [eˣ(x−1)]₀¹=1.

🧠 Memory Trick

"xeˣ → eˣ(x−1)."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integration by Parts

9 Match the indefinite integrals to their standard forms:

List I	List II
1. ∫dx/(x²−a²)	a. ln\|x+√(x²−a²)\|+C
2. ∫dx/(a²−x²)	b. 1/2a ln\|(a+x)/(a−x)\|+C
3. ∫dx/√(x²−a²)	c. 1/2a ln\|(x−a)/(x+a)\|+C
4. ∫dx/√(a²−x²)	d. sin⁻¹(x/a)+C

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

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

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

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

💡 Short Explanation

These are standard integral formulas. Match each expression directly. Option C gives the correct correspondence.

✅ Correct Answer Explanation

Using standard NCERT results: ∫dx/(x²−a²)=c, ∫dx/(a²−x²)=b, ∫dx/√(x²−a²)=a, ∫dx/√(a²−x²)=d. Hence the correct matching is Option C.

❓ Why Other Options Are Incorrect

Option A → Several formulas are interchanged.
Option B → Incorrectly matches logarithmic forms.
Option D → Assigns inverse trigonometric form incorrectly.

🎯 Strategy

Used: Option Grouping

Application: Recall standard integration identities.

Final Logic: Match each integral with its standard result.

🧠 Memory Trick

"Root Minus a² → Log, Root a² Minus → Sin⁻¹."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Standard Forms

10 If the probability of an error distribution over symmetric limits is given by the definite integral
∫₋ₐᵃ x³cos x dx,
evaluate this exact probability.

A 2a³

B a⁴/4

C a³sin a

D 0

💡 Short Explanation

x³ is odd. cos x is even. Odd × even = odd function.

✅ Correct Answer Explanation

The integrand x³cos x is an odd function because x³ is odd and cos x is even. The integral of an odd function over symmetric limits [−a,a] is zero. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Does not follow odd-function symmetry.
Option B → Unrelated polynomial area result.
Option C → Not obtained from symmetry properties.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Identify parity of the integrand.

Final Logic: Odd function over symmetric limits integrates to zero.

🧠 Memory Trick

"Odd + Symmetric = Zero."

📚 Class XII – NCERT Mathematics – Chapter: Definite Integrals – Topic: Properties of Definite Integrals

11 By the Fundamental Theorem of Calculus, what is the derivative of the area function
A(x)=∫₀ˣ ln(1+t²) dt
with respect to x?

A ln(1+x²)

B 2x/(1+x²)

C xln(1+x²)

D ln x

💡 Short Explanation

Fundamental Theorem connects integration and differentiation. Differentiate an area function by evaluating the integrand at x. Replace t with x.

✅ Correct Answer Explanation

The Fundamental Theorem of Calculus states that if A(x)=∫ₐˣf(t)dt, then A'(x)=f(x). Here f(t)=ln(1+t²). Therefore A'(x)=ln(1+x²). Option A is correct. Options B and C arise from unnecessary differentiation or multiplication, while D ignores the actual integrand.

❓ Why Other Options Are Incorrect

Option B → Derivative of ln(1+x²), not the area function itself.
Option C → Incorrect multiplication by x.
Option D → Omits the term (1+x²).

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Directly apply the Fundamental Theorem of Calculus.

Final Logic: Derivative of an area function equals its integrand.

🧠 Memory Trick

"Derivative cancels integral."

📚 Class XII – NCERT Mathematics – Chapter: Definite Integrals – Topic: Fundamental Theorem of Calculus

12 Which of the following is an INCORRECT property connecting definite and indefinite integrals?

A ∫ₐᵇf(x)dx = ∫ₐᶜf(x)dx + ∫ᶜᵇf(x)dx

B ∫ₐᵇf(x)dx = ∫ᵇₐf(x)dx

C ∫ₐᵇf(x)dx = F(b)−F(a), where F'(x)=f(x)

D ∫₀ᵃf(x)dx = ∫₀ᵃf(a−x)dx

💡 Short Explanation

Reversing limits changes the sign. Integral-addition property is valid. Fundamental Theorem remains correct.

✅ Correct Answer Explanation

For definite integrals, ∫ₐᵇf(x)dx = −∫ᵇₐf(x)dx. Hence Option B is incorrect because it omits the negative sign. Options A, C, and D are standard properties of definite integrals given in NCERT.

❓ Why Other Options Are Incorrect

Option A → Correct additive property of definite integrals.
Option C → Fundamental Theorem of Calculus.
Option D → Standard substitution property.

🎯 Strategy

Used: Elimination

Application: Recall standard properties of definite integrals.

Final Logic: Reversing limits reverses the sign.

🧠 Memory Trick

"Flip Limits, Flip Sign."

📚 Class XII – NCERT Mathematics – Chapter: Definite Integrals – Topic: Properties of Definite Integrals

13 Arrange the steps to split the integrand
x/((x+1)(x+2))
using partial fractions:
1. Solve for A=−1 and B=2.
2. Set x/((x+1)(x+2)) = A/(x+1) + B/(x+2).
3. Equate x=A(x+2)+B(x+1).
4. Rewrite the integrand as −1/(x+1)+2/(x+2).

A 2, 1, 3, 4

B 1, 2, 4, 3

C 2, 3, 1, 4

D 3, 2, 1, 4

💡 Short Explanation

Begin with partial fraction assumption. Equate numerators. Solve constants and rewrite.

✅ Correct Answer Explanation

The standard procedure is: assume the partial fraction form, clear denominators to obtain the numerator equation, solve for A and B, then rewrite the expression. Thus the order is 2 → 3 → 1 → 4, giving Option C.

❓ Why Other Options Are Incorrect

Option A → Solves A and B before forming equations.
Option B → Starts with unknown values already found.
Option D → Equates terms before decomposition setup.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Follow the logical sequence of partial fractions.

Final Logic: Assume → Equate → Solve → Rewrite.

🧠 Memory Trick

"Assume, Equate, Solve, Rewrite."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Partial Fractions Method

14 According to the property
∫ₐᵇf(x)dx = ∫ₐᵇf(t)dt,
what does this imply about the variable of integration in a definite integral?

A It affects the numerical result depending on the letter chosen.

B It fundamentally changes the function's nature.

C It restricts the limits of integration.

D It acts solely as a dummy variable.

💡 Short Explanation

Variable name has no effect on value. Only limits and function matter. Integration variable is a placeholder.

✅ Correct Answer Explanation

The symbols x, t, u, etc., used inside a definite integral are dummy variables. Changing the letter does not alter the numerical value of the integral. Therefore Option D is correct. The remaining options incorrectly assign mathematical significance to the variable name.

❓ Why Other Options Are Incorrect

Option A → Result is unchanged by renaming the variable.
Option B → Function meaning remains identical.
Option C → Limits are unaffected.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Compare equivalent definite integral forms.

Final Logic: Integration variables are placeholders.

🧠 Memory Trick

"x, t, u — Same Value."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integral Notation

15 The integration technique "Integration by Parts" involves selecting a first function (u) and second function (v). Which principles typically guide this selection?
1. The second function must be easily integrable.
2. The derivative of the first function should ideally simplify the remaining integral.
3. Both functions must always be trigonometric.

A 1 and 2 only

B 2 and 3 only

C 1 and 3 only

D 1, 2, and 3

💡 Short Explanation

Choose u whose derivative simplifies. Choose the remaining part easy to integrate. Trigonometric functions are not mandatory.

✅ Correct Answer Explanation

Integration by Parts uses ∫u dv = uv − ∫v du. A good choice makes du simpler and v easy to obtain. Statement 3 is false because the method applies to many function types, not only trigonometric functions. Hence statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

Option B → Includes false statement 3.
Option C → Omits valid statement 2.
Option D → Statement 3 is incorrect.

🎯 Strategy

Used: Option Grouping

Application: Test each statement against the integration-by-parts rule.

Final Logic: Simpler derivative + easy integration.

🧠 Memory Trick

"Choose u to simplify."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integration by Parts

16 Assertion (A): Evaluating
∫sec x tan x dx
by reverse differentiation yields sec x + C.
Reason (R): The derivative of tan x is sec x tan x.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Derivative of sec x is sec x tan x. Hence the integral equals sec x + C. Reason uses the wrong derivative.

✅ Correct Answer Explanation

Since d/dx(sec x)=sec x tan x, we have ∫sec x tan x dx = sec x + C. Thus the assertion is true. However, the derivative of tan x is sec²x, not sec x tan x. Therefore the reason is false, making Option B correct.

❓ Why Other Options Are Incorrect

Option A → Assertion is true.
Option C → Reason is incorrect.
Option D → Assertion is not false.

🎯 Strategy

Used: Substitution

Application: Differentiate sec x and tan x separately.

Final Logic: sec x differentiates to sec x tan x.

🧠 Memory Trick

"sec → sec tan."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Standard Integrals

17 Given an anti-derivative F(x) of f(x)=4x³−6 such that F(0)=3. Determine the exact value of the arbitrary constant C when
F(x)=x⁴−6x+C.

A 6

B −3

C 3

D 0

💡 Short Explanation

Substitute x=0 into F(x). Use the given condition. Solve directly for C.

✅ Correct Answer Explanation

Given F(x)=x⁴−6x+C and F(0)=3, 0−0+C=3. Therefore C=3. Option C satisfies the condition exactly. All other values violate the initial condition.

❓ Why Other Options Are Incorrect

Option A → Gives F(0)=6.
Option B → Gives F(0)=−3.
Option D → Gives F(0)=0.

🎯 Strategy

Used: Substitution

Application: Insert the given point into the anti-derivative.

Final Logic: F(0)=C=3.

🧠 Memory Trick

"Plug x=0, Get C."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Arbitrary Constant

18 Match the initial conditions to their specific constants C for the family of curves
y=∫2x dx=x²+C:

List I	List II
1. Curve passes through (0,5)	a. C=1
2. Curve passes through (1,2)	b. C=−1
3. Curve passes through (2,0)	c. C=−4
4. Curve passes through (0,−1)	d. C=5

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

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

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

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

💡 Short Explanation

Substitute each point into y=x²+C. Solve for C individually. Match values accordingly.

✅ Correct Answer Explanation

(0,5) ⇒ C=5 (d) (1,2) ⇒ 2=1+C ⇒ C=1 (a) (2,0) ⇒ 0=4+C ⇒ C=−4 (c) (0,−1) ⇒ C=−1 (b) Hence the matching is 1-d, 2-a, 3-c, 4-b, which corresponds to Option D.

❓ Why Other Options Are Incorrect

Option A → Multiple constants are mismatched.
Option B → Incorrect assignment of C values.
Option C → Does not satisfy the given points.

🎯 Strategy

Used: Substitution

Application: Use each coordinate pair in y=x²+C.

Final Logic: Solve C separately for each condition.

🧠 Memory Trick

"Point Fixes C."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Family of Curves

19 Utilizing definite integration to find bounds, compute the numerical area represented by
∫₀^(π/2) sin(2x) dx.

A 1

B 2

C 1/2

D π

💡 Short Explanation

Integrate sin(2x). Apply limits 0 and π/2. Result simplifies to 1.

✅ Correct Answer Explanation

\(\int_{0}^{\pi /2}\,sin⁡(2x) dx={\left[\ −\ \frac{cos⁡(2x)}{2}\right]}_{0}^{\pi /2}=\frac{1}{2}+\frac{1}{2}=1.\) Therefore Option A is correct. The other values result from incorrect evaluation or omitted factors.

❓ Why Other Options Are Incorrect

Option B → Ignores the factor 1/2.
Option C → Uses only part of the evaluation.
Option D → Not related to the integral value.

🎯 Strategy

Used: Substitution

Application: Apply the standard trigonometric integration rule.

Final Logic: [−cos(2x)/2]₀^(π/2)=1.

🧠 Memory Trick

"sin(2x) → −cos(2x)/2."

📚 Class XII – NCERT Mathematics – Chapter: Definite Integrals – Topic: Trigonometric Integrals

20 In scientific kinetics, tracking accumulation requires definite integrals. Evaluate
∫₋(π/2)^(π/2) sin⁷x dx
using properties of odd functions.

A 14/15

B 0

C π/2

D 1

💡 Short Explanation

sin x is an odd function. Odd power preserves oddness. Integral over symmetric limits is zero.

✅ Correct Answer Explanation

Since sin⁷x is an odd function, sin⁷(−x)=−sin⁷x. The integral of any odd function over symmetric limits [−a,a] equals zero. Therefore ∫₋(π/2)^(π/2) sin⁷x dx = 0. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Not obtained from odd-function symmetry.
Option C → Unrelated to the integral value.
Option D → Ignores symmetry about the origin.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Identify the parity of the integrand.

Final Logic: Odd function over symmetric limits integrates to zero.

🧠 Memory Trick

"Odd + Symmetric = Zero."

📚 Class XII – NCERT Mathematics – Chapter: Definite Integrals – Topic: Properties of Definite Integrals

CUET UG Mathematics Booster Test 2 - Introduction & Basics

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