CUET UG Mathematics Booster Test 1 - Properties of Indefinite Integrals -

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CUET UG Mathematics Booster Test 1 - Properties of Indefinite Integrals

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Properties of Indefinite Integrals

1 / 20

QUESTION 1 OF 20

Match the fundamental differentiation of integrals with their simplified forms:

List I	List II
1. d/dx∫e^(2x) dx	a. √x
2. d/dx∫log x dx	b. log x
3. d/dx∫tan x dx	c. tan x
4. d/dx∫√x dx	d. e^(2x)

Answer Review

1 Match the fundamental differentiation of integrals with their simplified forms:

List I	List II
1. d/dx∫e^(2x) dx	a. √x
2. d/dx∫log x dx	b. log x
3. d/dx∫tan x dx	c. tan x
4. d/dx∫√x dx	d. e^(2x)

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

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

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

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

💡 Short Explanation

Differentiation and integration are inverse operations. d/dx[∫f(x)dx] = f(x). Each integral differentiates back to its original integrand.

✅ Correct Answer Explanation

By the inverse property of integration and differentiation, differentiating an indefinite integral returns the original integrand. Thus ∫e^(2x)dx → e^(2x), ∫log x dx → log x, ∫tan x dx → tan x, and ∫√x dx → √x. Therefore the matching sequence is 1-d, 2-b, 3-c, 4-a.

❓ Why Other Options Are Incorrect

Option B → Incorrect matching of log x, tan x, and e^(2x).
Option C → Several functions are mismatched with their original integrands.
Option D → Does not follow the inverse property of differentiation and integration.

🎯 Strategy

Used: Option Grouping

Application: Match each differentiated integral directly with its original integrand.

Final Logic: Differentiating an indefinite integral reproduces the integrand.

🧠 Memory Trick

"Diff of Integral = Original Function."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integration as Inverse Process of Differentiation

2 A data analyst tracks the rate of change of a moving average,
M'(t)=3t²−2t.
To find the moving average equation M(t) (integral of the derivative), which form represents it?

A 6t−2+C

B 3t³−t²+C

C t³−2t²+C

D t³−t²+C

💡 Short Explanation

Integrate term by term. ∫3t²dt = t³. ∫−2tdt = −t².

✅ Correct Answer Explanation

Using polynomial integration, ∫(3t²−2t)dt = 3(t³/3)−2(t²/2)+C = t³−t²+C. Option D satisfies differentiation back to 3t²−2t. The remaining options fail to produce the given derivative when differentiated.

❓ Why Other Options Are Incorrect

Option A → Derivative is 6, not 3t²−2t.
Option B → Derivative becomes 9t²−2t.
Option C → Derivative becomes 3t²−4t.

🎯 Strategy

Used: Substitution

Application: Apply standard power-rule integration term by term.

Final Logic: Integrating 3t²−2t yields t³−t²+C.

🧠 Memory Trick

"Increase power by 1, divide by new power."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integration of Polynomial Functions

3 Assertion (A): The functions F(x)=sin²x and G(x)=−cos²x have the same derivative.
Reason (R): Two anti-derivatives of the same function differ only by a constant.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

F'(x)=2sin x cos x. G'(x)=2sin x cos x. Functions differ by a constant.

✅ Correct Answer Explanation

Differentiating both functions gives 2sin x cos x. Also, sin²x + cos²x = 1, so F(x)−G(x)=1, a constant. NCERT states anti-derivatives of the same function differ only by a constant. Hence both statements are true and the reason explains the assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are actually true.
Option B → Reason is not false.
Option D → Assertion is true since both derivatives are identical.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Verify both statements independently and check explanatory relationship.

Final Logic: Same derivative and constant difference confirm both statements.

🧠 Memory Trick

"Same derivative ⇒ differ by constant."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Family of Anti-derivatives

4 If
∫f(x)dx = F(x)+C₁ and ∫f(x)dx = G(x)+C₂,
then the difference F(x)−G(x) is:

A f(x)

B A constant

C Zero strictly

D F'(x)−G'(x)

💡 Short Explanation

Both are anti-derivatives of the same function. Their derivatives are equal. Difference remains constant.

✅ Correct Answer Explanation

Since F'(x)=f(x) and G'(x)=f(x), we get F'(x)−G'(x)=0. Therefore F(x)−G(x) must be a constant. NCERT states that all anti-derivatives of a function differ only by an arbitrary constant.

❓ Why Other Options Are Incorrect

Option A → Difference is not f(x).
Option C → Difference need not be zero; any constant is possible.
Option D → F'(x)−G'(x)=0, not the function difference.

🎯 Strategy

Used: Elimination

Application: Use derivative equality to eliminate variable expressions.

Final Logic: Equal derivatives imply constant difference.

🧠 Memory Trick

"Same slope, constant gap."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: General Anti-derivative

5 The integral
∫(x²+eˣ)dx
evaluates an area-related function. What is the correct integrated form?

A 2x+eˣ+C

B x³/3+xeˣ+C

C x³/3+eˣ+C

D x³+eˣ+C

💡 Short Explanation

Integrate each term separately. ∫x²dx=x³/3. ∫eˣdx=eˣ.

✅ Correct Answer Explanation

Using the sum rule, ∫(x²+eˣ)dx = ∫x²dx + ∫eˣdx = x³/3 + eˣ + C. Option C follows standard integration formulas. Other options either differentiate incorrectly or omit required factors.

❓ Why Other Options Are Incorrect

Option A → Not an integral of x²+eˣ.
Option B → Derivative of xeˣ is not eˣ.
Option D → Integral of x² is x³/3, not x³.

🎯 Strategy

Used: Option Grouping

Application: Integrate each term independently and compare.

Final Logic: Sum rule gives x³/3+eˣ+C.

🧠 Memory Trick

"Power + Exponential = Separate Integrals."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Sum Rule of Integration

6 Evaluate the sum expression:
∫(2x²−3sin x+5√x)dx.

A (2x³)/3+3cos x+(5x^(3/2))/2+C

B (2x³)/3−3cos x+(10x^(3/2))/3+C

C (2x³)/3+3cos x+(10x^(3/2))/3+C

D 4x−3cos x+5/(2√x)+C

💡 Short Explanation

Apply integration term-wise. ∫sin x dx = −cos x. ∫√x dx = (2/3)x^(3/2).

✅ Correct Answer Explanation

Using linearity: ∫2x²dx=(2x³)/3, ∫−3sin x dx=3cos x, and ∫5√x dx=5·(2/3)x^(3/2)=(10/3)x^(3/2). Adding all terms gives Option C.

❓ Why Other Options Are Incorrect

Option A → Coefficient of x^(3/2) is incorrect.
Option B → Sign of cosine term is incorrect.
Option D → Does not represent proper integration.

🎯 Strategy

Used: Elimination

Application: Integrate each term and compare coefficients carefully.

Final Logic: Only Option C matches all standard formulas.

🧠 Memory Trick

"Sin gives minus cos; negative sin gives plus cos."

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

7 Which of the following properties are valid for the constant multiple rule of integrals?
1. ∫kf(x)dx = k∫f(x)dx
2. ∫f(kx)dx = k∫f(x)dx
3. It applies to any real number k

A 1 and 3 only

B 2 and 3 only

C 1 and 2 only

D 1, 2, and 3

💡 Short Explanation

Constant can be taken outside the integral. Rule works for real constants. ∫f(kx)dx requires substitution.

✅ Correct Answer Explanation

Property (1) is the standard constant multiple rule. Property (3) is true because k may be any real constant. Property (2) is false because changing the argument from x to kx requires substitution and introduces scaling factors. Hence only 1 and 3 are correct.

❓ Why Other Options Are Incorrect

Option B → Includes false statement (2).
Option C → Includes false statement (2).
Option D → Statement (2) violates substitution principles.

🎯 Strategy

Used: Elimination

Application: Test each statement against standard integration properties.

Final Logic: Only statements 1 and 3 are universally valid.

🧠 Memory Trick

"Constant Out, Variable Change Needs Substitution."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Constant Multiple Rule

8 Evaluate the scalar multiplied vector integral:
3∫(cos x î + eˣ ĵ)dx.

A 3sin x î +3eˣx ĵ +C⃗

B sin x î +eˣ ĵ +C⃗

C −3sin x î +3eˣ ĵ +C⃗

D 3sin x î +3eˣ ĵ +C⃗

💡 Short Explanation

Integrate each vector component. ∫cos x dx=sin x. Multiply result by 3.

✅ Correct Answer Explanation

Vector integration is performed component-wise. Thus ∫(cos x î + eˣ ĵ)dx = sin x î + eˣ ĵ + C⃗. Multiplying by 3 gives 3sin x î + 3eˣ ĵ + C⃗. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect integration of eˣ.
Option B → Omits factor 3.
Option C → Integral of cos x is not −sin x.

🎯 Strategy

Used: Option Grouping

Application: Integrate components separately and apply scalar multiplication.

Final Logic: Component-wise integration yields Option D.

🧠 Memory Trick

"Integrate First, Scale Later."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Linearity of Integration

9 Identify the INCORRECT statement regarding
∫[k₁f₁(x)+k₂f₂(x)]dx:

A It equals k₁∫f₁(x)dx+k₂∫f₂(x)dx.

B It equals (k₁+k₂)∫[f₁(x)f₂(x)]dx.

C It demonstrates the linearity property of integrals.

D The rule extends to any finite number of functions.

💡 Short Explanation

Integration is linear. Sum rule applies term-wise. Product form is not implied.

✅ Correct Answer Explanation

Linearity states ∫[k₁f₁+k₂f₂]dx = k₁∫f₁dx + k₂∫f₂dx. It does not convert a sum into the integral of a product. Hence Option B contradicts the fundamental linearity property and is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A → Correct statement of linearity.
Option C → Correct interpretation of the rule.
Option D → Linearity extends to finite sums of functions.

🎯 Strategy

Used: Odd One Out

Application: Compare each statement with the linearity theorem.

Final Logic: Only Option B violates linearity.

🧠 Memory Trick

"Integral Splits Sums, Not Products."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: General Linearity Property

10 Arrange the steps to integrate
∫(2−3sin x)/(cos²x) dx:
1. Split into ∫2/(cos²x)dx − ∫(3sin x)/(cos²x)dx.
2. Rewrite as 2∫sec²xdx − 3∫sec xtan xdx.
3. Apply standard formulas.
4. Obtain 2tan x − 3sec x + C.

A 1, 2, 3, 4

B 2, 1, 4, 3

C 1, 3, 2, 4

D 3, 1, 2, 4

💡 Short Explanation

Use linearity first. Convert into standard trigonometric forms. Apply known integrals.

✅ Correct Answer Explanation

The correct sequence begins by splitting the integral using linearity. Next rewrite using sec²x and sec xtan x identities. Then apply standard integration formulas ∫sec²x dx = tan x and ∫sec xtan x dx = sec x. Finally obtain 2tan x − 3sec x + C.

❓ Why Other Options Are Incorrect

Option B → Rewriting occurs after splitting, not before.
Option C → Standard formulas cannot be applied before rewriting.
Option D → Application of formulas must occur near the end.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Follow the natural procedural order of integration.

Final Logic: Split → Rewrite → Integrate → Simplify.

🧠 Memory Trick

"Split → Convert → Integrate → Result."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Trigonometric Integrals and Linearity

11 The graph of f(x)=sec²x represents a rate of change. By direct identification, the area bounded under it relates to which primitive anti-derivative function?

A cot x+C

B tan x+C

C −tan x+C

D sec x tan x+C

💡 Short Explanation

∫sec²x dx = tan x + C. sec²x is the derivative of tan x. Direct inspection gives the anti-derivative.

✅ Correct Answer Explanation

The inspection method uses known derivative-integral pairs. Since d/dx(tan x)=sec²x, the anti-derivative of sec²x is tan x+C. Option B directly matches the standard trigonometric integration formula. The remaining options differentiate to functions different from sec²x.

❓ Why Other Options Are Incorrect

Option A → d/dx(cot x)=−csc²x, not sec²x.
Option C → d/dx(−tan x)=−sec²x.
Option D → d/dx(sec x tan x)=sec x(tan²x+sec²x).

🎯 Strategy

Used: Elimination

Application: Recall standard derivative-integral pairs and eliminate mismatches.

Final Logic: sec²x is the derivative of tan x.

🧠 Memory Trick

"Sec-square gives Tan."

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

12 A numerical algorithm seeks the anti-derivative of
eˣ+1/x.
What is its numerical value at exactly x=1, given the constant of integration C=0?

A e+1

B e−1

C 1

D e

💡 Short Explanation

∫(eˣ+1/x)dx=eˣ+ln|x|+C. At x=1, ln1=0. C=0 is given.

✅ Correct Answer Explanation

Integrating gives eˣ+ln|x|+C. Substituting x=1 and C=0 yields e¹+ln(1)=e+0=e. Therefore Option D is correct. Option A incorrectly treats ln1 as 1, while the other options ignore part of the anti-derivative.

❓ Why Other Options Are Incorrect

Option A → ln1 equals 0, not 1.
Option B → No subtraction occurs.
Option C → Ignores the exponential contribution.

🎯 Strategy

Used: Substitution

Application: Integrate first and then substitute x=1.

Final Logic: e+ln1=e.

🧠 Memory Trick

"ln1 = 0."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Standard Integrals of eˣ and 1/x

13 Find the integral for the generic quadratic form:
∫(ax²+bx+c)dx.

A (ax³)/3+(bx²)/2+cx+C

B 2ax+b+C

C (ax³)/3+(bx²)/2+C

D ax³+bx²+cx+C

💡 Short Explanation

Apply power rule term-wise. Constant integrates to cx. Add constant of integration.

✅ Correct Answer Explanation

Using ∫xⁿdx=xⁿ⁺¹/(n+1), we obtain ∫ax²dx=(ax³)/3, ∫bxdx=(bx²)/2, and ∫cdx=cx. Adding the arbitrary constant gives Option A. This is the standard NCERT polynomial integration formula.

❓ Why Other Options Are Incorrect

Option B → This is a derivative expression.
Option C → Omits the integral of c.
Option D → Missing division by powers after integration.

🎯 Strategy

Used: Option Grouping

Application: Integrate each polynomial term independently.

Final Logic: Increase power by one and divide by the new power.

🧠 Memory Trick

"Power Up, Divide Down."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Polynomial Integration

14 A higher degree polynomial integral
∫(x³+3x+4)/√x dx
is evaluated in a statistics distribution. What is the fractional power of x in the integrated term originating from the x³ numerator?

A 7/2

B 5/2

C 3/2

D 1/2

💡 Short Explanation

x³/√x=x^(5/2). Integrating raises power by one. New power becomes 7/2.

✅ Correct Answer Explanation

The term from the numerator becomes x³/x^(1/2)=x^(5/2). Applying the power rule, ∫x^(5/2)dx=x^(7/2)/(7/2)+C. Therefore the integrated term contains x^(7/2), making Option A correct.

❓ Why Other Options Are Incorrect

Option B → Power before integration.
Option C → Not obtained from integrating x^(5/2).
Option D → Arises from different exponents entirely.

🎯 Strategy

Used: Elimination

Application: Simplify exponents first, then integrate.

Final Logic: 5/2 increases to 7/2 after integration.

🧠 Memory Trick

"5/2 → 7/2 after integration."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Integration of Algebraic Functions

📄 Passage

Substitution is a powerful technique in integration. When the integrand is of the form f(g(x))g'(x), substituting t=g(x) reduces the integral to ∫f(t)dt. For rational functions, substitution often involves identifying the derivative of the denominator in the numerator, leading to logarithmic results like
∫1/x dx = ln|x| + C.

From the passage, evaluate
∫2x/(x²+1)dx.

A tan⁻¹(x)+C

B 1/(x²+1)+C

C ln(x²+1)+C

D ln|2x|+C

💡 Short Explanation

Denominator is x²+1. Numerator is its derivative. Use logarithmic substitution.

✅ Correct Answer Explanation

Let u=x²+1. Then du=2x dx. The integral becomes ∫du/u=ln|u|+C. Since x²+1 is always positive, the result is ln(x²+1)+C. Thus Option C correctly applies the substitution principle described in the passage.

❓ Why Other Options Are Incorrect

Option A → Integral of 1/(1+x²), not 2x/(x²+1).
Option B → Differentiation does not yield the integrand.
Option D → Incorrect logarithmic argument.

🎯 Strategy

Used: Substitution

Application: Identify denominator and its derivative in the numerator.

Final Logic: u=x²+1 converts the integral into ∫du/u.

🧠 Memory Trick

"Derivative Above, Log Below."

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

📄 Passage

Evaluate
∫(ln x)²/x dx
using the substitution principle mentioned.

A ln x/3+C

B (ln x)³/3+C

C 2ln x+C

D 1/x²+C

💡 Short Explanation

Let t=ln x. Then dt=dx/x. Integrate t².

✅ Correct Answer Explanation

Using substitution t=ln x, we obtain dt=dx/x. The integral becomes ∫t²dt=t³/3+C. Replacing t by ln x gives (ln x)³/3+C. Therefore Option B correctly applies the substitution method.

❓ Why Other Options Are Incorrect

Option A → Integrates t² incorrectly.
Option C → Derivative does not reproduce the integrand.
Option D → Unrelated to logarithmic substitution.

🎯 Strategy

Used: Substitution

Application: Replace ln x by a simpler variable.

Final Logic: ∫t²dt=t³/3.

🧠 Memory Trick

"ln x becomes t."

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

17 Evaluate the direct trigonometric integral:
∫sec x(sec x+tan x)dx.

A sec²x+tan x+C

B ln|sec x+tan x|+C

C tan²x+sec x+C

D tan x+sec x+C

💡 Short Explanation

Expand the integrand. Integrate sec²x and sec x tan x. Add results.

✅ Correct Answer Explanation

sec x(sec x+tan x)=sec²x+sec x tan x. Using standard formulas, ∫sec²x dx=tan x and ∫sec x tan x dx=sec x. Therefore the integral equals tan x+sec x+C. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Represents neither integration result nor anti-derivative.
Option B → Integral of sec x only.
Option C → Differentiation does not yield the integrand.

🎯 Strategy

Used: Option Grouping

Application: Split into standard trigonometric integrals.

Final Logic: sec²x → tan x and sec x tan x → sec x.

🧠 Memory Trick

"Sec-square gives Tan; Sec-Tan gives Sec."

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

18 Evaluate the combined trigonometric form:
∫(sin²x−cos²x)/(sin²x cos²x)dx.

A tan x−cot x+C

B tan x+cot x+C

C −tan x+cot x+C

D sec x+csc x+C

💡 Short Explanation

Simplify into sec²x−csc²x. Integrate term-wise. Use standard formulas.

✅ Correct Answer Explanation

\(\frac{{sin⁡}^{2}x-{cos⁡}^{2}x}{{sin⁡}^{2}x{cos⁡}^{2}x}={sec⁡}^{2}x-{csc⁡}^{2}x\) Hence, \(\int ({sec⁡}^{2}x-{csc⁡}^{2}x)dx=tan⁡x+cot⁡x+C\) because ∫sec²x dx=tan x and ∫−csc²x dx=cot x. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Sign of cot x is incorrect.
Option C → Sign of tan x is incorrect.
Option D → Differentiation does not produce the integrand.

🎯 Strategy

Used: Elimination

Application: Rewrite the expression into standard trigonometric forms.

Final Logic: sec²x−csc²x integrates to tan x+cot x.

🧠 Memory Trick

"Minus csc² gives plus cot."

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Trigonometric Identities in Integration

19 If
∫(2x−3)dx=f(x)
and it is given that f(2)=1, what is the constant C?

A 1

B 2

C −1

D 3

💡 Short Explanation

Integrate first. Use the given condition. Solve for C.

✅ Correct Answer Explanation

f(x)=x²−3x+C. Using f(2)=1, 1=4−6+C 1=−2+C C=3. Therefore Option D is correct. This is a standard application of initial conditions used to determine the arbitrary constant after integration.

❓ Why Other Options Are Incorrect

Option A → Gives f(2)=−1.
Option B → Gives f(2)=0.
Option C → Gives f(2)=−3.

🎯 Strategy

Used: Substitution

Application: Substitute the given point after integration.

Final Logic: f(2)=1 leads directly to C=3.

🧠 Memory Trick

"Integrate → Substitute → Solve C."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Determination of Arbitrary Constant

20 The unique solution curve for the differential equation
dy/dx=eˣ
passing through the point (0,2) is:

A y=eˣ−1

B y=eˣ+1

C y=eˣ+2

D y=eˣ

💡 Short Explanation

Integrate dy/dx=eˣ. Obtain y=eˣ+C. Use the given point.

✅ Correct Answer Explanation

Integrating gives y=eˣ+C. Applying the condition (0,2), 2=e⁰+C 2=1+C C=1. Hence the unique solution is y=eˣ+1. The provided answer C is incorrect because y=eˣ+2 gives y(0)=3, not 2.

❓ Why Other Options Are Incorrect

Option A → Gives y(0)=0.
Option C → Gives y(0)=3.
Option D → Gives y(0)=1.

🎯 Strategy

Used: Substitution

Application: Integrate first and then apply the initial condition.

Final Logic: y=eˣ+C and y(0)=2 imply C=1.

🧠 Memory Trick

"Point fixes the constant."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Solution with Initial Conditions

CUET UG Mathematics Booster Test 1 - Properties of Indefinite Integrals

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