CUET UG Mathematics Booster Test 2 - Special Integrals & Standard Forms -

🚀

CUET UG Mathematics Booster Test 2 - Special Integrals & Standard Forms

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 2 - Special Integrals & Standard Forms

1 / 20

QUESTION 1 OF 20

For
\(I=\int \frac{dx}{x^{2}-16},\)
which of the following are true?
1. The denominator factors into linear terms.
2. The domain excludes \(x=\pm 4\).
3. \(I=\frac{1}{8}log⁡∣\frac{x-4}{x+4}∣+C\)

Answer Review

1 For
\(I=\int \frac{dx}{x^{2}-16},\)
which of the following are true?
1. The denominator factors into linear terms.
2. The domain excludes \(x=\pm 4\).
3. \(I=\frac{1}{8}log⁡∣\frac{x-4}{x+4}∣+C\)

A 1 and 2 only

B 1, 2, and 3

C 2 and 3 only

D 1 and 3 only

💡 Short Explanation

\(x^{2}-16=(x-4)(x+4)\). Function undefined at ±4. Standard logarithmic result applies.

✅ Correct Answer Explanation

The denominator factors as \(x^{2}-16=(x-4)(x+4),\) so statement 1 is true. Since the denominator becomes zero at \(x=\pm 4\), statement 2 is true. Using the standard formula \(\int \frac{dx}{x^{2}-a^{2}}=\frac{1}{2a}log⁡∣\frac{x-a}{x+a}∣+C,\) with \(a=4\), statement 3 is also true. Therefore all three statements are correct.

❓ Why Other Options Are Incorrect

Option A → Omits statement 3, which is a valid standard result.
Option C → Ignores the factorization property.
Option D → Ignores the domain restriction at \(x=\pm 4\).

🎯 Strategy

Used: Option Grouping

Application:

Verify each statement independently using factorization, domain analysis and standard integration formulas.

Final Logic:

All three statements are valid.

🧠 Memory Trick

"Difference of squares → factor, exclude roots, log result."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Standard Integral \(\int dx/(x^{2}-a^{2})\)

2 Match the integrals with their corresponding results:

List I	List II
1. ∫dx/(x²−4)	a. 1/5 tan⁻¹(x/5)+C
2. ∫dx/(x²+25)	b. 1/8 log∣(x−4)/(x+4)∣+C
3. ∫dx/(x²+9)	c. 1/4 log∣(x−2)/(x+2)∣+C
4. ∫dx/(x²−16)	d. 1/3 tan⁻¹(x/3)+C

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

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

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

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

💡 Short Explanation

Use standard \(x^{2}-a^{2}\)logarithmic form. Use standard \(x^{2}+a^{2}\)inverse tangent form. Match coefficients carefully.

✅ Correct Answer Explanation

\(\int \frac{dx}{x^{2}-4}=\frac{1}{4}log⁡∣\frac{x-2}{x+2}∣\int \frac{dx}{x^{2}+25}=\frac{1}{5}{tan⁡}^{-1}(x/5)\int \frac{dx}{x^{2}+9}=\frac{1}{3}{tan⁡}^{-1}(x/3)\int \frac{dx}{x^{2}-16}=\frac{1}{8}log⁡∣\frac{x-4}{x+4}∣\) Hence 1-c, 2-a, 3-d, 4-b.

❓ Why Other Options Are Incorrect

Option B → Swaps logarithmic results.
Option C → Matches logarithmic and inverse tangent forms incorrectly.
Option D → Ignores the standard coefficient values.

🎯 Strategy

Used: Option Grouping

Application:

Separate plus-square and minus-square integrals before matching.

Final Logic:

Minus-square → log; plus-square → tan⁻¹.

🧠 Memory Trick

"Minus gives log, plus gives tan⁻¹."

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

3 Evaluate
\(\int_{0}^{2\sqrt{2}}\,\frac{dx}{\sqrt{16-x^{2}}}.\)

A π/4

B π/3

C π/6

D π/2

💡 Short Explanation

Use inverse sine standard form. Evaluate between limits. Simplify using known trigonometric values.

✅ Correct Answer Explanation

\(\int \frac{dx}{\sqrt{16-x^{2}}}={sin⁡}^{-1}(x/4)+C\) Thus \({\left[{sin⁡}^{-1}(x/4)\right]}_{0}^{2\sqrt{2}}={sin⁡}^{-1}\left(\frac{\sqrt{2}}{2}\right)-{sin⁡}^{-1}(0)=\frac{\pi }{4}\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Corresponds to \({sin⁡}^{-1}(\sqrt{3}/2)\).
Option C → Corresponds to \({sin⁡}^{-1}(1/2)\).
Option D → Would require upper limit 4.

🎯 Strategy

Used: Substitution

Application:

Recognize the standard inverse sine integral.

Final Logic:

Upper limit gives \({sin⁡}^{-1}(\sqrt{2}/2)=\pi /4\).

🧠 Memory Trick

"Root \(a^{2}-x^{2}\)→ sin⁻¹(x/a)."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Standard Integral \(\int dx/\sqrt{a^{2}-x^{2}}\)

4 Evaluate
\(\int_{5}^{6}\,\frac{dx}{\sqrt{x^{2}-16}}.\)

A log((3+√5)/4)

B log(2)

C log((3−√5)/4)

D log(5)

💡 Short Explanation

Apply logarithmic standard form. Substitute upper and lower limits. Simplify the logarithmic ratio.

✅ Correct Answer Explanation

\(\int \frac{dx}{\sqrt{x^{2}-16}}=log⁡∣x+\sqrt{x^{2}-16}∣+C\) Therefore \(=log⁡(6+\sqrt{20})-log⁡(5+\sqrt{9})=log⁡(6+2\sqrt{5})-log⁡(8)\) Since \(6+2\sqrt{5}=2(3+\sqrt{5})\), \(=log⁡\left(\frac{3+\sqrt{5}}{4}\right)\) Hence the provided answer A is correct, not B.

❓ Why Other Options Are Incorrect

Option B → Does not result from the exact limit evaluation.
Option C → Uses the reciprocal-related expression incorrectly.
Option D → Ignores subtraction of logarithms.

🎯 Strategy

Used: Substitution

Application:

Use the standard logarithmic antiderivative and evaluate limits exactly.

Final Logic:

Area equals \(\log\,\left(\frac{3+\sqrt{5}}{4}\right)\).

🧠 Memory Trick

"√(x²−a²) ⇒ log(x+root)."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integral involving \(\sqrt{x^{2}-a^{2}}\)

5 Find the average value of
\(f(x)=\frac{1}{\sqrt{x^{2}+16}}\)
over the interval [0,3].

A 1/3 log(3)

B 1/4 log(2)

C 1/3 log(2)

D 1/2 log(3)

💡 Short Explanation

Average value = area ÷ interval length. Use logarithmic standard integral. Evaluate and divide by 3.

✅ Correct Answer Explanation

\(Average=\frac{1}{3}\int_{0}^{3}\,\frac{dx}{\sqrt{x^{2}+16}}\) Using \(\int \frac{dx}{\sqrt{x^{2}+16}}=log⁡∣x+\sqrt{x^{2}+16}∣\) gives \(\frac{1}{3}[log⁡(8)-log⁡(4)]=\frac{1}{3}log⁡2\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect logarithmic value.
Option B → Uses wrong interval length.
Option D → Incorrect coefficient and logarithm.

🎯 Strategy

Used: Substitution

Application:

Compute area first and then divide by interval length.

Final Logic:

Average value \(=\frac{1}{3}log⁡2\).

🧠 Memory Trick

"Average = Integral ÷ Length."

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Integrals – Topic: Average Value of a Function

6 A PDF is given by
\(f(x)=\frac{c}{x^{2}+2x+2},x\in [0,1].\)
Find \(c\).

A tan⁻¹(3)

B π/4

C 1/(tan⁻¹(1/3))

D 1/(tan⁻¹(3))

💡 Short Explanation

Total probability must equal 1. Complete the square in denominator. Use inverse tangent evaluation over [0,1].

✅ Correct Answer Explanation

For a PDF, \(\int_{0}^{1}\,f(x) dx=1\) Thus \(c\int_{0}^{1}\,\frac{dx}{x^{2}+2x+2}=1\) Since \(x^{2}+2x+2=(x+1)^{2}+1,\int_{0}^{1}\,\frac{dx}{\left(x\ +\ 1)^{2}\ +\ 1\right.}={tan⁡}^{-1}(2)-{tan⁡}^{-1}(1)={tan⁡}^{-1}\left(\frac{1}{3}\right)\) Therefore \(c=\frac{1}{{tan⁡}^{-1}(1/3)}\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Gives the integral value incorrectly, not the normalization constant.
Option B → Does not satisfy the probability condition.
Option D → Uses \({tan⁡}^{-1}(3)\)instead of \({tan⁡}^{-1}(1/3)\).

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Apply the normalization condition \(\int f(x) dx=1\).

Final Logic:

\(c⋅{tan⁡}^{-1}(1/3)=1\Rightarrow c=1/{tan⁡}^{-1}(1/3)\).

🧠 Memory Trick

"PDF ⇒ total area = 1."

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Integrals – Topic: Probability Density Functions

7 Evaluate
\(\int \frac{dx}{2x^{2}+8x+10}.\)

A 1/4 tan⁻¹(x+2)+C

B tan⁻¹(x+2)+C

C 1/2 tan⁻¹((x+2)/2)+C

D 1/2 tan⁻¹(x+2)+C

💡 Short Explanation

Factor the quadratic. Complete the square. Apply the standard inverse tangent form.

✅ Correct Answer Explanation

\(2x^{2}+8x+10=2[(x+2)^{2}+1]\) Therefore \(\int \frac{dx}{2[(x+2)^{2}+1]}=\frac{1}{2}\int \frac{dx}{\left(x\ +\ 2)^{2}\ +\ 1\right.}=\frac{1}{2}{tan⁡}^{-1}(x+2)+C\) Hence the integral equals \(\frac{1}{2}{tan⁡}^{-1}(x+2)+C\) The provided answer C is incorrect.

❓ Why Other Options Are Incorrect

Option A → Coefficient should be \(1/2\), not \(1/4\).
Option B → Missing factor \(1/2\).
Option C → Uses an unnecessary scaling inside tan⁻¹.

🎯 Strategy

Used: Substitution

Application:

Reduce the quadratic to \(\left(x\ +\ a)^{2}\ +\ 1\right.\).

Final Logic:

Factor \(2\) first, then apply the standard formula.

🧠 Memory Trick

"Factor first, square later."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Completing the Square

8 Assertion (A):
\(\int \frac{e^{x}}{\sqrt{e^{2x}-1}}dx\)
cannot be evaluated analytically.
Reason (R):
Substitute \(t=e^{x}\)to get
\(\int \frac{dt}{\sqrt{t^{2}-1}}.\)

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Substitution converts the integral to a standard form. Standard logarithmic antiderivative exists. Hence the assertion is false.

✅ Correct Answer Explanation

Let \(t=e^{x},dt=e^{x} dx\) Then \(\int \frac{e^{x}}{\sqrt{e^{2x}-1}}dx=\int \frac{dt}{\sqrt{t^{2}-1}}\) which is a standard integral: \(=log⁡∣t+\sqrt{t^{2}-1}∣+C\) Thus the integral can be evaluated analytically. Assertion is false, while the reason is true. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reason is definitely true.
Option B → Assertion is false, not true.
Option C → Assertion is false.

🎯 Strategy

Used: Extreme Word Filter

Application:

The phrase "cannot be evaluated analytically" should be tested carefully.

Final Logic:

A direct substitution immediately solves the integral.

🧠 Memory Trick

"eˣ dx ⇒ try \(t=e^{x}\)."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Substitution Method and Standard Forms

9 Let
\(I(x)=\int \frac{dx}{x^{2}+4x+8}.\)
Given \(I(0)=k\pi\), find \(k\).

A 1/4

B 1/8

C 1/2

D 1/16

💡 Short Explanation

Complete the square. Use the condition \(I(0)\). Evaluate the antiderivative at zero.

✅ Correct Answer Explanation

\(x^{2}+4x+8=(x+2)^{2}+4\) Thus \(I(x)=\frac{1}{2}{tan⁡}^{-1}\left(\frac{x+2}{2}\right)+C\) Taking the standard antiderivative with \(C=0\), \(I(0)=\frac{1}{2}{tan⁡}^{-1}(1)=\frac{1}{2}⋅\frac{\pi }{4}=\frac{\pi }{8}\) Therefore \(k=\frac{1}{8}.\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Twice the required value.
Option C → Corresponds to \(\pi /2\), too large.
Option D → Half the required value.

🎯 Strategy

Used: Substitution

Application:

Reduce the denominator to \(\left(x\ +\ a)^{2}\ +\ b^{2}\right.\).

Final Logic:

\(I(0)=\frac{\pi }{8}\), so \(k=\frac{1}{8}\).

🧠 Memory Trick

"Complete square → tan⁻¹ → substitute."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integrals Reducible to Standard Forms

10 Evaluate
\(\int_{0}^{2}\,\frac{x+1}{x^{2}+2x+5} dx.\)

A 1/2 log(13/5)

B log(13/5)

C 1/2 log(5/13)

D log(5/13)

💡 Short Explanation

Numerator is proportional to derivative of denominator. Use logarithmic integration formula. Apply the limits.

✅ Correct Answer Explanation

Let \(u=x^{2}+2x+5\) Then \(du=2(x+1) dx\) Hence \(\int_{0}^{2}\,\frac{x+1}{x^{2}+2x+5} dx=\frac{1}{2}\int_{0}^{2}\,\frac{du}{u}=\frac{1}{2}log⁡∣u∣∣_{0}^{2}=\frac{1}{2}[log⁡(13)-log⁡(5)]=\frac{1}{2}log⁡\left(\frac{13}{5}\right)\) Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Missing factor \(1/2\).
Option C → Reverses the logarithmic ratio.
Option D → Wrong ratio and missing factor.

🎯 Strategy

Used: Substitution

Application:

Recognize that the numerator is half the derivative of the denominator.

Final Logic:

Direct \(u\)-substitution yields \(\frac{1}{2}log⁡(13/5)\).

🧠 Memory Trick

"Derivative on top ⇒ log below."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution

11 For
\(x^{2}+bx+c=(x+b/2)^{2}+k^{2},\)
arrange the values of \(k^{2}\)in increasing order:
I. \(x^{2}+2x+2\)
II. \(x^{2}+4x+8\)
III. \(x^{2}+6x+18\)

A III, II, I

B I, III, II

C I, II, III

D II, I, III

💡 Short Explanation

Complete the square for each expression. Find the corresponding \(k^{2}\). Arrange numerically.

✅ Correct Answer Explanation

For I: \(x^{2}+2x+2=(x+1)^{2}+1\) so \(k^{2}=1\). For II: \(x^{2}+4x+8=(x+2)^{2}+4\) so \(k^{2}=4\). For III: \(x^{2}+6x+18=(x+3)^{2}+9\) so \(k^{2}=9\). Therefore \(1<4<9\) giving the order I, II, III. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Gives descending order.
Option B → Places \(k^{2}=9\) before \(k^{2}=4\).
Option D → Incorrectly places II before I.

🎯 Strategy

Used: Option Grouping

Application:

Complete the square and compare the resulting constants.

Final Logic:

\(1<4<9\Rightarrow\) I, II, III.

🧠 Memory Trick

"\(k^{2}=c-\frac{b^{2}}{4}\)"

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Completing the Square

12 Identify the INCORRECT identity:

A \(x^{2}+2x+5=(x+1)^{2}+4\)

B \(x^{2}-6x+13=(x-3)^{2}+4\)

C \(2x^{2}+4x+10=2[(x+1)^{2}+4]\)

D \(x^{2}+bx+c=(x+b/2)^{2}+(c-b^{2})\)

💡 Short Explanation

Use the standard completing-square formula. Check each identity algebraically. Option D has an incorrect constant term.

✅ Correct Answer Explanation

The correct identity is \(x^{2}+bx+c={\left(x\ +\ \frac{b}{2}\right)}^{2}+\left(c\ −\ \frac{b^{2}}{4}\right)\) Option D incorrectly writes \(c-b^{2}\)instead of \(c-\frac{b^{2}}{4}\). Expanding A, B, and C verifies that they are all correct. Hence D is the incorrect identity.

❓ Why Other Options Are Incorrect

Option A → Expands correctly to \(x^{2}+2x+5\).
Option B → Expands correctly to \(x^{2}-6x+13\).
Option C → Expands correctly after factoring out 2.

🎯 Strategy

Used: Odd One Out

Application:

Compare each identity with the standard completing-square formula.

Final Logic:

Only Option D violates the standard formula.

🧠 Memory Trick

"Always divide \(b^{2}\)by 4."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Completing the Square

13 Evaluate
\(\int \frac{\cos\,x}{{sin⁡}^{2}x+4sin⁡x+5} dx.\)

A \(\frac{1}{2}{tan⁡}^{-1}(sin⁡x+2)+C\)

B \({tan⁡}^{-1}(sin⁡x-2)+C\)

C \({tan⁡}^{-1}(sin⁡x+2)+C\)

D \(log⁡∣sin⁡x+2∣+C\)

💡 Short Explanation

Let \(t=sin⁡x\). Denominator becomes \(\left(t\ +\ 2)^{2}\ +\ 1\right.\). Apply inverse tangent formula.

✅ Correct Answer Explanation

Using \(t=sin⁡x,dt=cos⁡x dx\) the integral becomes \(\int \frac{dt}{t^{2}+4t+5}=\int \frac{dt}{\left(t\ +\ 2)^{2}\ +\ 1\right.}={tan⁡}^{-1}(t+2)+C={tan⁡}^{-1}(sin⁡x+2)+C\) Thus Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Introduces an unnecessary factor \(1/2\).
Option B → Uses an incorrect shift.
Option D → Logarithmic forms arise when the derivative of the denominator appears.

🎯 Strategy

Used: Substitution

Application:

Recognize \(cos⁡x dx\) as \(d(sin⁡x)\).

Final Logic:

The integral reduces to \(\int dt/[(t+2)^{2}+1]\).

🧠 Memory Trick

"cos x dx ⇒ sin x substitution."

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

14 Evaluate
\(\int_{0}^{1}\,\frac{2x+3}{x^{2}+3x+2} dx.\)

A log(2)

B log(4)

C log(6)

D log(3)

💡 Short Explanation

Numerator equals derivative of denominator. Use logarithmic integration. Apply limits.

✅ Correct Answer Explanation

Let \(u=x^{2}+3x+2\) Then \(du=(2x+3) dx\) Therefore \(\int_{0}^{1}\,\frac{2x+3}{x^{2}+3x+2} dx=log⁡∣u∣∣_{0}^{1}=log⁡(6)-log⁡(2)=log⁡(3)\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Uses only part of the evaluation.
Option B → Incorrect ratio.
Option C → Ignores the lower-limit contribution.

🎯 Strategy

Used: Substitution

Application:

Identify numerator as the derivative of the denominator.

Final Logic:

\(log⁡(6)-log⁡(2)=log⁡(3)\).

🧠 Memory Trick

"Derivative on top ⇒ log below."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution

📄 Passage

Recognizing substitutions like \(x^{n}\rightarrow t\) simplifies rational integrals where the numerator is proportional to the derivative of a power in the denominator.

Evaluate
\(\int \frac{x^{2}}{x^{6}+a^{6}} dx.\)

A \(\frac{1}{3a^{3}}{tan⁡}^{-1}(x^{3}/a^{3})+C\)

B \(\frac{1}{a^{3}}{tan⁡}^{-1}(x^{3}/a^{3})+C\)

C \(\frac{1}{3a}{tan⁡}^{-1}(x^{3}/a)+C\)

D \(\frac{1}{a}{tan⁡}^{-1}(x^{3}/a^{3})+C\)

💡 Short Explanation

Use \(t=x^{3}\). Convert to standard inverse tangent form. Apply scaling factors carefully.

✅ Correct Answer Explanation

Let \(t=x^{3}\) Then \(dt=3x^{2}dx\) Thus \(\int \frac{x^{2}}{x^{6}+a^{6}}dx=\frac{1}{3}\int \frac{dt}{t^{2}+a^{6}}=\frac{1}{3}⋅\frac{1}{a^{3}}{tan⁡}^{-1}\left(\frac{t}{a^{3}}\right)+C=\frac{1}{3a^{3}}{tan⁡}^{-1}\left(\frac{x^{3}}{a^{3}}\right)+C\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Missing factor \(1/3\).
Option C → Incorrect scaling inside tan⁻¹.
Option D → Missing both coefficient adjustments.

🎯 Strategy

Used: Substitution

Application:

Choose \(t=x^{3}\)because its derivative appears in the numerator.

Final Logic:

\(dt=3x^{2}dx\) leads directly to the standard formula.

🧠 Memory Trick

"Power in denominator ⇒ same power substitution."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Reduction to Standard Forms

📄 Passage

Recognizing substitutions like \(x^{n}\rightarrow t\) simplifies rational integrals where the numerator is proportional to the derivative of a power in the denominator.

Evaluate
\(\int \frac{1}{x(x^{4}-1)} dx.\)

A \(\frac{1}{4}log⁡∣\frac{x^{4}+1}{x^{4}}∣+C\)

B \(\frac{1}{4}log⁡∣\frac{x^{4}-1}{x^{4}}∣+C\)

C \(\frac{1}{2}log⁡∣\frac{x^{4}-1}{x^{4}}∣+C\)

D \(\frac{1}{4}log⁡∣\frac{x^{4}}{x^{4}-1}∣+C\)

💡 Short Explanation

Multiply by \(x^{3}/x^{3}\). Use \(t=x^{4}\). Integrate the resulting rational function.

✅ Correct Answer Explanation

\(\int \frac{1}{x(x^{4}-1)}dx=\int \frac{x^{3}}{x^{4}(x^{4}-1)}dx\) Let \(t=x^{4},dt=4x^{3}dx\) Then \(=\frac{1}{4}\int \frac{dt}{t(t-1)}=\frac{1}{4}\int \left(\frac{1}{t-1}\ −\ \frac{1}{t}\right)dt=\frac{1}{4}log⁡∣\frac{t-1}{t}∣+C=\frac{1}{4}log⁡∣\frac{x^{4}-1}{x^{4}}∣+C\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Uses \(x^{4}+1\) instead of \(x^{4}-1\).
Option C → Coefficient should be \(1/4\).
Option D → Gives the negative of the required answer.

🎯 Strategy

Used: Substitution

Application:

Transform the quartic expression into a rational function in \(t\).

Final Logic:

Partial fractions give the logarithmic result.

🧠 Memory Trick

"Multiply by \(x^{3}\), then use \(x^{4}\)."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Rational Integrals by Substitution

17 Normalize \(f(x)=c\frac{x}{\sqrt{x^{4}+1}},x\in [0,1].\)

A \(\frac{2}{log⁡(1+\sqrt{2})}\)

B \(\frac{1}{log⁡(1+\sqrt{2})}\)

C \(\frac{2}{log⁡(\sqrt{2})}\)

D \(\frac{1}{log⁡(\sqrt{2})}\)

💡 Short Explanation

Total probability must equal 1. Use substitution \(t=x^{2}\). Apply the standard logarithmic integral.

✅ Correct Answer Explanation

For normalization, \(\int_{0}^{1}\,c\frac{x}{\sqrt{x^{4}+1}}dx=1\) Let \(t=x^{2},dt=2x dx\) Then \(\int_{0}^{1}\,\frac{x}{\sqrt{x^{4}+1}}dx=\frac{1}{2}\int_{0}^{1}\,\frac{dt}{\sqrt{t^{2}+1}}\) Using \(\int \frac{dt}{\sqrt{t^{2}+1}}=log⁡∣t+\sqrt{t^{2}+1}∣\) gives \(\frac{1}{2}log⁡(1+\sqrt{2})\) Hence \(c⋅\frac{1}{2}log⁡(1+\sqrt{2})=1c=\frac{2}{log⁡(1+\sqrt{2})}\) Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Misses the factor 2 arising from substitution.
Option C → Uses the wrong logarithmic value.
Option D → Incorrect logarithmic expression and coefficient.

🎯 Strategy

Used: Substitution

Application:

Transform \(x^{4}\)into a quadratic form using \(t=x^{2}\).

Final Logic:

Normalization gives \(c=2/log⁡(1+\sqrt{2})\).

🧠 Memory Trick

"PDF ⇒ area = 1."

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Integrals – Topic: Probability Density Functions

18 Evaluate displacement \(\int_{0}^{\ln\,4}\,\frac{e^{t}}{\sqrt{e^{2t}+9}}dt.\)

A log(4/(1+√10))

B log(9/(1+√10))

C log(9/√10)

D log(4/√10)

💡 Short Explanation

Use \(u=e^{t}\). Convert to standard logarithmic form. Apply limits carefully.

✅ Correct Answer Explanation

Let \(u=e^{t},du=e^{t}dt\) Then \(\int_{1}^{4}\,\frac{du}{\sqrt{u^{2}+9}}\) Using \(\int \frac{du}{\sqrt{u^{2}+9}}=log⁡∣u+\sqrt{u^{2}+9}∣\) gives \(log⁡(4+\sqrt{25})-log⁡(1+\sqrt{10})=log⁡(9)-log⁡(1+\sqrt{10})=log⁡\left(\frac{9}{1+\sqrt{10}}\right)\) Therefore the provided answer B is correct.

❓ Why Other Options Are Incorrect

Option A → Uses 4 instead of the correct upper-limit value 9.
Option C → Omits the lower-limit logarithmic term.
Option D → Incorrect evaluation of both limits.

🎯 Strategy

Used: Substitution

Application:

Recognize \(e^{t}dt\) as \(du\).

Final Logic:

Upper limit contributes \(\log\,9\), lower limit contributes \(log⁡(1+\sqrt{10})\).

🧠 Memory Trick

"eᵗdt ⇒ let \(u=e^{t}\)."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Substitution and Standard Forms

19 Find the average value of \(f(x)=\frac{1}{e^{x}+e^{-x}}\)
over \(\left[0,ln⁡\sqrt{3}\right]\).

A \(\frac{\pi }{6ln⁡\sqrt{3}}\)

B \(\frac{\pi }{4ln⁡\sqrt{3}}\)

C \(\frac{\pi }{12ln⁡\sqrt{3}}\)

D \(\frac{\pi }{3ln⁡\sqrt{3}}\)

💡 Short Explanation

Convert the denominator using \(e^{x}\). Apply substitution \(t=e^{x}\). Divide by interval length.

✅ Correct Answer Explanation

Average value: \(\frac{1}{\ln\,\sqrt{3}}\int_{0}^{\ln\,\sqrt{3}}\,\frac{dx}{e^{x}+e^{-x}}\) Multiply numerator and denominator by \(e^{x}\): \(=\frac{1}{\ln\,\sqrt{3}}\int_{0}^{\ln\,\sqrt{3}}\,\frac{e^{x}}{e^{2x}+1}dx\) Let \(t=e^{x}\): \(=\frac{1}{\ln\,\sqrt{3}}\int_{1}^{\sqrt{3}}\,\frac{dt}{t^{2}+1}=\frac{1}{\ln\,\sqrt{3}}{\left[{tan⁡}^{-1}t\right]}_{1}^{\sqrt{3}}=\frac{1}{\ln\,\sqrt{3}}\left(\frac{\pi }{3}\ −\ \frac{\pi }{4}\right)=\frac{\pi }{12ln⁡\sqrt{3}}\) Therefore the provided answer C is correct.

❓ Why Other Options Are Incorrect

Option A → Twice the actual average value.
Option B → Uses an incorrect inverse tangent difference.
Option D → Four times the correct value.

🎯 Strategy

Used: Substitution

Application:

Rewrite the denominator and convert to the standard inverse tangent integral.

Final Logic:

Average value \(=\frac{\pi }{12ln⁡\sqrt{3}}\).

🧠 Memory Trick

"Average = Integral ÷ Interval length."

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Integrals – Topic: Average Value of a Function

20 Which methods evaluate \(\int \frac{dx}{sin⁡x+cos⁡x}?\)
1. Half-angle substitution \(\left(tan⁡(x/2)\right)\)
2. Transform to \(csc⁡(x+\pi /4)\)
3. Use
\(\int csc⁡\theta d\theta =log⁡∣tan⁡\frac{\theta }{2}∣+C\)

A 1 and 2 only

B 2 and 3 only

C 1 and 3 only

D 1, 2, and 3

💡 Short Explanation

Multiple valid methods exist. Trigonometric transformation simplifies the denominator. Half-angle substitution also works.

✅ Correct Answer Explanation

Using \(sin⁡x+cos⁡x=\sqrt{2} sin⁡\left(x\ +\ \frac{\pi }{4}\right)\) the integral becomes a constant multiple of \(\int csc⁡\left(x\ +\ \frac{\pi }{4}\right)dx.\) The standard result \(\int csc⁡\theta d\theta =log⁡∣tan⁡\frac{\theta }{2}∣+C\) can then be applied. Alternatively, the Weierstrass substitution \(t=tan⁡(x/2)\)also evaluates the integral. Hence all three methods are valid.

❓ Why Other Options Are Incorrect

Option A → Omits the useful \(\csc\,\theta\) integration formula.
Option B → Ignores the half-angle substitution method.
Option C → Omits the direct transformation step.

🎯 Strategy

Used: Option Grouping

Application:

Check whether each listed technique independently leads to a valid evaluation.

Final Logic:

All three methods correctly evaluate the integral.

🧠 Memory Trick

"\(sin⁡x+cos⁡x\rightarrow \sqrt{2}sin⁡(x+\pi /4)\)."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Trigonometric Integrals and Substitutions

CUET UG Mathematics Booster Test 2 - Special Integrals & Standard Forms

Test Complete!

Answer Review