CUET UG Mathematics Booster Test 1 - Special Integrals & Standard Forms -

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CUET UG Mathematics Booster Test 1 - 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 1 - Special Integrals & Standard Forms

1 / 20

QUESTION 1 OF 20

Calculate the area of the region bounded by
\(y=\frac{1}{x^{2}-4},\)
the x-axis, and the ordinates \(x=3\) and \(x=4\).

Answer Review

1 Calculate the area of the region bounded by
\(y=\frac{1}{x^{2}-4},\)
the x-axis, and the ordinates \(x=3\) and \(x=4\).

A 1/4 log(3)

B 1/2 log(5/3)

C 1/2 log(3)

D 1/4 log(5/3)

💡 Short Explanation

Area equals definite integral. Use standard form \(\int dx/(x^{2}-a^{2})\). Apply limits 3 and 4 carefully.

✅ Correct Answer Explanation

Using \(\int \frac{dx}{x^{2}-4}=\frac{1}{4}log⁡∣\frac{x-2}{x+2}∣+C\) Area \(=\frac{1}{4}{\left[log\ ⁡\ \left(\frac{x-2}{x+2}\right)\right]}_{3}^{4}=\frac{1}{4}log⁡\left(\frac{\left(2/6\right)}{\left(1/5\right)}\right)=\frac{1}{4}log⁡\left(\frac{5}{3}\right)\) Hence Option D is correct. The other options arise from incorrect coefficient or limit evaluation.

❓ Why Other Options Are Incorrect

Option A → Uses incorrect logarithmic ratio.
Option B → Correct logarithmic argument but coefficient should be \(1/4\).
Option C → Neither coefficient nor logarithmic argument is correct.

🎯 Strategy

Used: Substitution

Application:

Recall the standard integral and substitute limits directly.

Final Logic:

Standard formula plus limits gives \(\frac{1}{4}log⁡(5/3)\).

🧠 Memory Trick

"x²−a² ⇒ log ratio ÷ 2a"

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

2 Consider a proposed probability density function
\(f(x)=\frac{C}{x^{2}-1},x\in (0,2).\)
Why is this mathematically invalid?

A The integral evaluates to a negative number.

B The log function is undefined for negative bounds.

C There is an infinite discontinuity (asymptote) at x=1, making the integral improper.

D C becomes imaginary.

💡 Short Explanation

PDF must be integrable. Denominator becomes zero at \(x=1\). Infinite discontinuity breaks normalization.

✅ Correct Answer Explanation

A valid probability density function must have a finite integral over its domain. Here \(f(x)=\frac{C}{x^{2}-1}\) has a vertical asymptote at \(x=1\). Since the interval \(\left(0\ ,\ 2\right)\)contains this singularity, the integral becomes improper and diverges. Therefore Option C is correct. Options A, B, and D do not explain the actual mathematical issue.

❓ Why Other Options Are Incorrect

Option A → Divergence occurs before any finite negative value is obtained.
Option B → Logarithmic evaluation is not the fundamental problem.
Option D → Constant \(C\) need not be imaginary.

🎯 Strategy

Used: Elimination

Application:

Check whether the function satisfies PDF requirements.

Final Logic:

Presence of a vertical asymptote inside the interval invalidates the PDF.

🧠 Memory Trick

"PDF ⇒ finite area required."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Improper Integrals and Applications

3 Evaluate: \(\int \frac{dx}{x^{2}+9}\)

A 1/9 tan⁻¹(x/3)+C

B 1/3 tan⁻¹(x/3)+C

C 3 tan⁻¹(3x)+C

D 1/3 tan⁻¹(3x)+C

💡 Short Explanation

Compare with \(a^{2}+x^{2}\). Here \(a=3\). Apply standard inverse tangent formula.

✅ Correct Answer Explanation

Using \(\int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}{tan⁡}^{-1}\left(\frac{x}{a}\right)+C\) with \(a=3\), \(\int \frac{dx}{x^{2}+9}=\frac{1}{3}{tan⁡}^{-1}\left(\frac{x}{3}\right)+C\) Thus Option B is correct. The remaining options contain incorrect coefficients or arguments.

❓ Why Other Options Are Incorrect

Option A → Coefficient should be \(1/3\), not \(1/9\).
Option C → Incorrect coefficient and argument.
Option D → Argument should be \(x/3\), not \(3x\).

🎯 Strategy

Used: Option Grouping

Application:

Identify \(a\) and compare options with the standard formula.

Final Logic:

\(a=3\Rightarrow \frac{1}{3}{tan⁡}^{-1}(x/3)\).

🧠 Memory Trick

"Plus square ⇒ tan⁻¹(x/a)/a"

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

4 Assertion (A): \(\int \frac{dx}{16+x^{2}}=4{tan⁡}^{-1}(4x)+C\)
Reason (R): \(\int \frac{dx}{a^{2}+x^{2}}=a{tan⁡}^{-1}(ax)+C\)

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Standard formula is misquoted. Assertion uses incorrect coefficient. Reason also states a wrong formula.

✅ Correct Answer Explanation

The correct formula is \(\int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}{tan⁡}^{-1}\left(\frac{x}{a}\right)+C\) For \(a=4\), \(\int \frac{dx}{16+x^{2}}=\frac{1}{4}{tan⁡}^{-1}(x/4)+C\) Therefore Assertion (A) is false. Reason (R) is also false because it incorrectly states \(a{tan⁡}^{-1}(ax)\). Hence Option A is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Compare both statements with the exact NCERT standard result.

Final Logic:

Both formulas differ from the standard form.

🧠 Memory Trick

"Divide by a, not multiply by a."

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

5 Arrange the following integrals in descending order:
I. \(\int_{0}^{1/2}\,\frac{dx}{\sqrt{1-x^{2}}}\)
II. \(\int_{0}^{1/\sqrt{2}}\,\frac{dx}{\sqrt{1-x^{2}}}\)
III. \(\int_{0}^{\sqrt{3}/2}\,\frac{dx}{\sqrt{1-x^{2}}}\)

A I, II, III

B I, III, II

C II, III, I

D III, II, I

💡 Short Explanation

Use inverse sine antiderivative. Compare upper limits. Larger upper limit gives larger value.

✅ Correct Answer Explanation

Using \(\int \frac{dx}{\sqrt{1-x^{2}}}={sin⁡}^{-1}(x)\) I = \({sin⁡}^{-1}(1/2)=\pi /6\) II = \({sin⁡}^{-1}(1/\sqrt{2})=\pi /4\) III = \({sin⁡}^{-1}(\sqrt{3}/2)=\pi /3\) Since \(\pi /3>\pi /4>\pi /6\) the descending order is III, II, I. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Gives ascending order.
Option B → Incorrect placement of II and I.
Option C → Places II above III incorrectly.

🎯 Strategy

Used: Option Grouping

Application:

Convert each integral into a known inverse trigonometric value.

Final Logic:

\(\pi /3>\pi /4>\pi /6\).

🧠 Memory Trick

"½, 1/√2, √3/2 ⇒ 30°,45°,60°."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integral involving \(\sqrt{1-x^{2}}\)

6 Which of the following are valid anti-derivatives of \(\int \frac{dx}{\sqrt{16-x^{2}}}?\)
1. \({sin⁡}^{-1}(x/4)+C_{1}\)
2. \(-{cos⁡}^{-1}(x/4)+C_{2}\)
3. \(\frac{1}{4}log⁡∣x+\sqrt{16-x^{2}}∣+C_{3}\)

A 1 only

B 2 and 3 only

C 1 and 2 only

D 1, 2, and 3

💡 Short Explanation

Standard result is inverse sine. Negative inverse cosine differs only by a constant. Logarithmic form belongs to another integral.

✅ Correct Answer Explanation

\(\int \frac{dx}{\sqrt{16-x^{2}}}={sin⁡}^{-1}(x/4)+C\) Also, \(-{cos⁡}^{-1}(x/4)+C\) has the same derivative. Therefore statements 1 and 2 are valid antiderivatives. Statement 3 is associated with logarithmic integrals involving \(\sqrt{x^{2}\pm a^{2}}\), not \(\sqrt{a^{2}-x^{2}}\). Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores valid antiderivative 2.
Option B → Includes invalid logarithmic expression.
Option D → Statement 3 is not an antiderivative.

🎯 Strategy

Used: Elimination

Application:

Differentiate each proposed antiderivative.

Final Logic:

Only statements 1 and 2 produce \(1/\sqrt{16-x^{2}}\).

🧠 Memory Trick

"a²−x² ⇒ sin⁻¹, not log."

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

7 Match the integrals with their corresponding results:

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

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

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

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

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

💡 Short Explanation

Both √(x²−a²) and √(x²+a²) forms give logarithmic results. Match the radical expression exactly. Compare integrand and antiderivative carefully.

✅ Correct Answer Explanation

The standard results are: \(\int \frac{dx}{\sqrt{x^{2}-a^{2}}}=log⁡∣x+\sqrt{x^{2}-a^{2}}∣+C\) and \(\int \frac{dx}{\sqrt{x^{2}+a^{2}}}=log⁡∣x+\sqrt{x^{2}+a^{2}}∣+C\) Therefore: 1→b, 2→a, 3→c, 4→d. Each result preserves the same expression under the square root. Hence Option A is the correct matching. All other options interchange radicals incorrectly.

❓ Why Other Options Are Incorrect

Option B → Swaps the results for \(x^{2}-1\) and \(x^{2}+1\).
Option C → Incorrectly matches every radical expression.
Option D → Does not preserve the original form of the square root.

🎯 Strategy

Used: Option Grouping

Application:

Match each integrand directly with the logarithmic antiderivative containing the identical radical.

Final Logic:

The expression under the square root remains unchanged in the final logarithmic result.

🧠 Memory Trick

"Root stays root inside the log."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Standard Integrals involving √(x²±a²)

8 A velocity component is \(v_{x}(x)=\frac{1}{\sqrt{x^{2}-9}}\)
Find displacement from \(x=4\) to \(x=5\).

A log(9/(4+√7))

B log(5/4)

C log(9)

D log(4+√7)

💡 Short Explanation

Displacement equals definite integral. Use the logarithmic standard form. Apply upper and lower limits.

✅ Correct Answer Explanation

\(\int \frac{dx}{\sqrt{x^{2}-9}}=log⁡∣x+\sqrt{x^{2}-9}∣+C\) Evaluating from 4 to 5: \(log⁡(5+\sqrt{16})-log⁡(4+\sqrt{7})=log⁡(9)-log⁡(4+\sqrt{7})=log⁡\left(\frac{9}{4+\sqrt{7}}\right)\) Thus Option A is correct. The other options represent incomplete evaluations of the definite integral.

❓ Why Other Options Are Incorrect

Option B → Does not arise from the logarithmic evaluation.
Option C → Uses only the upper-limit contribution.
Option D → Uses only the lower-limit contribution.

🎯 Strategy

Used: Substitution

Application:

Apply the standard antiderivative and substitute the limits directly.

Final Logic:

\(log⁡(9)-log⁡(4+\sqrt{7})=log⁡\left(\frac{9}{4+\sqrt{7}}\right)\).

🧠 Memory Trick

"√(x²−a²) ⇒ log(x+root)."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Integral involving √(x²−a²)

9 Identify the INCORRECT formula:

A ∫dx/(a²+x²)=1/a tan⁻¹(x/a)+C

B ∫dx/√(x²+a²)=log∣x+√(x²+a²)∣+C

C ∫dx/√(a²−x²)=sin⁻¹(x/a)+C

D ∫dx/√(x²+a²)=1/a sin⁻¹(x/a)+C

💡 Short Explanation

A, B and C are standard NCERT formulas. D incorrectly uses inverse sine. √(x²+a²) gives a logarithmic result.

✅ Correct Answer Explanation

The correct standard formulas are: \(\int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}{tan⁡}^{-1}(x/a)+C\int \frac{dx}{\sqrt{x^{2}+a^{2}}}=log⁡∣x+\sqrt{x^{2}+a^{2}}∣+C\int \frac{dx}{\sqrt{a^{2}-x^{2}}}={sin⁡}^{-1}(x/a)+C\) Therefore A, B and C are correct. Option D incorrectly assigns an inverse sine antiderivative to the \(\sqrt{x^{2}+a^{2}}\)form.

❓ Why Other Options Are Incorrect

Option A → Correct NCERT standard result.
Option B → Correct logarithmic standard form.
Option C → Correct inverse trigonometric standard result.

🎯 Strategy

Used: Odd One Out

Application:

Compare every formula with standard NCERT integral tables.

Final Logic:

Only Option D does not correspond to a valid standard integral formula.

🧠 Memory Trick

"Plus root ⇒ log, minus root ⇒ sin⁻¹."

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

10 The moving average value of \(y=\frac{1}{\sqrt{x^{2}+9}}\)
over the interval [0,4] is:

A log(3)

B 1/4 log(2)

C 1/4 log(3)

D 4log(3)

💡 Short Explanation

Average value = \(\frac{1}{b-a}\int_{a}^{b}\,f(x) dx\). Use logarithmic standard integral. Evaluate limits and divide by interval length.

✅ Correct Answer Explanation

Average value: \(\frac{1}{4}\int_{0}^{4}\,\frac{dx}{\sqrt{x^{2}+9}}\) Using \(\int \frac{dx}{\sqrt{x^{2}+a^{2}}}=log⁡∣x+\sqrt{x^{2}+a^{2}}∣+C=\frac{1}{4}{\left[log⁡(x+\sqrt{x^{2}+9})\right]}_{0}^{4}=\frac{1}{4}\left(log⁡(9)-log⁡(3)\right)=\frac{1}{4}log⁡(3)\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores division by interval length 4.
Option B → Incorrect logarithmic evaluation.
Option D → Multiplies by 4 instead of dividing by 4.

🎯 Strategy

Used: Substitution

Application:

First compute the definite integral, then divide by \(b-a\) to obtain the average value.

Final Logic:

Average value \(=\frac{1}{4}log⁡(3)\).

🧠 Memory Trick

"Average value = Area ÷ Interval length."

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

11 Evaluate \(I(x)=\int \frac{dx}{x^{2}+6x+13}\)
Given that \(I(-3)=0\), find \(I(-1)\).

A π/4

B π/8

C π/2

D π/6

💡 Short Explanation

Complete the square. Convert to standard tan⁻¹ form. Use the given condition to determine the constant.

✅ Correct Answer Explanation

\(x^{2}+6x+13=(x+3)^{2}+4\) Hence \(I(x)=\frac{1}{2}{tan⁡}^{-1}\left(\frac{x+3}{2}\right)+C\) Using \(I(-3)=0\), \(C=0\) Therefore \(I(-1)=\frac{1}{2}{tan⁡}^{-1}(1)=\frac{1}{2}⋅\frac{\pi }{4}=\frac{\pi }{8}\) Thus the provided answer is incorrect.

❓ Why Other Options Are Incorrect

Option A → Equals \(\pi /4\), twice the required value.
Option C → Much larger than the evaluated integral.
Option D → Does not arise from the inverse tangent calculation.

🎯 Strategy

Used: Substitution

Application:

Complete the square and apply the initial condition.

Final Logic:

\(I(-1)=\frac{1}{2}{tan⁡}^{-1}(1)=\frac{\pi }{8}\).

🧠 Memory Trick

"Complete square → apply condition."

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

12 Find the area under \(f(x)=\frac{1}{x^{2}-4x+8}\)
between \(x=2\) and \(x=4\).

A π/8

B π/4

C π/2

D π/6

💡 Short Explanation

Complete the square. Use inverse tangent formula. Evaluate between limits.

✅ Correct Answer Explanation

\(x^{2}-4x+8=(x-2)^{2}+4\) Therefore \(\int_{2}^{4}\,\frac{dx}{\left(x-2)^{2}+4\right.}=\frac{1}{2}{\left[{tan⁡}^{-1}\left(\frac{x-2}{2}\right)\right]}_{2}^{4}=\frac{1}{2}\left(\frac{\pi }{4}\ −\ 0\right)=\frac{\pi }{8}\) Thus the provided answer is correct.

❓ Why Other Options Are Incorrect

Option B → Omits the factor \(1/2\).
Option C → Overestimates the area.
Option D → Does not arise from limit evaluation.

🎯 Strategy

Used: Substitution

Application:

Convert the denominator into \(\left(x-a)^{2}+b^{2}\right.\).

Final Logic:

Area = \(\frac{1}{2}\times \frac{\pi }{4}=\frac{\pi }{8}\).

🧠 Memory Trick

"Square completed ⇒ tan⁻¹ appears."

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

13 Evaluate \(\int \frac{dx}{2x^{2}+4x+10}\)

A 1/2 tan⁻¹(x+1)+C

B 1/4 tan⁻¹(x+1)+C

C 1/2 tan⁻¹((x+1)/2)+C

D 1/4 tan⁻¹((x+1)/2)+C

💡 Short Explanation

Factor out 2. Complete the square. Use standard tan⁻¹ form.

✅ Correct Answer Explanation

\(2x^{2}+4x+10=2[(x+1)^{2}+4]\) Thus \(\int \frac{dx}{2[(x+1)^{2}+4]}=\frac{1}{2}\int \frac{dx}{\left(x\ +\ 1)^{2}\ +\ 4\right.}=\frac{1}{2}⋅\frac{1}{2}{tan⁡}^{-1}\left(\frac{x+1}{2}\right)+C=\frac{1}{4}{tan⁡}^{-1}\left(\frac{x+1}{2}\right)+C\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Missing both scaling factors.
Option B → Incorrect argument.
Option C → Missing factor \(1/2\).

🎯 Strategy

Used: Substitution

Application:

Reduce the quadratic to a standard inverse tangent form.

Final Logic:

Two factors of \(1/2\) produce coefficient \(1/4\).

🧠 Memory Trick

"Factor first, then square."

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

14 Evaluate \(\int \frac{e^{x}}{e^{2x}+4}dx\)

A tan⁻¹(e^x)+C

B 1/4 tan⁻¹(e^x/2)+C

C 1/2 tan⁻¹(e^x/2)+C

D tan⁻¹(e^x/2)+C

💡 Short Explanation

Let \(t=e^{x}\). Convert to standard form. Apply \(a=2\) formula.

✅ Correct Answer Explanation

Let \(t=e^{x},dt=e^{x}dx\) Then \(\int \frac{dt}{t^{2}+4}\) Using \(\int \frac{dt}{t^{2}+a^{2}}=\frac{1}{a}{tan⁡}^{-1}\left(\frac{t}{a}\right)\) with \(a=2\), \(=\frac{1}{2}{tan⁡}^{-1}(t/2)+C=\frac{1}{2}{tan⁡}^{-1}(e^{x}/2)+C\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Uses denominator \(1+t^{2}\).
Option B → Coefficient should be \(1/2\).
Option D → Missing the factor \(1/2\).

🎯 Strategy

Used: Substitution

Application:

Replace \(e^{x}\)by a single variable.

Final Logic:

Standard \(1/(t^{2}+4)\)formula yields coefficient \(1/2\).

🧠 Memory Trick

"eˣ dx ⇒ let t=eˣ."

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

📄 Passage

When evaluating integrals of the form \(\int \frac{\left(px\ +\ q\right)}{\left(ax^{2}\ +\ bx\ +\ c\right)}dx,\)
we write px+q=A(2ax+b)+B,
so the integral splits into a logarithmic term and an inverse tangent term.

For \(\frac{4x+1}{2x^{2}+x+1},\)
find the value of \(A\).

A 1/2

B 1

C 2

D 4

💡 Short Explanation

Compare coefficients of \(x\). Use derivative of denominator. Solve directly for \(A\).

✅ Correct Answer Explanation

\(2ax+b=4x+1\) for denominator \(2x^{2}+x+1\). Let \(4x+1=A(4x+1)+B\) Comparing coefficients: \(4=4A\) Therefore \(A=1\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Produces coefficient \(2x\).
Option C → Produces coefficient \(8x\).
Option D → Produces coefficient \(16x\).

🎯 Strategy

Used: Option Grouping

Application:

Match coefficients of \(x\) on both sides.

Final Logic:

\(4=4A\Rightarrow A=1\).

🧠 Memory Trick

"Match x-term first."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Rational Functions \(\left(px+q)/(ax^{2}+bx+c\right)\)

📄 Passage

For the same function, find the value of \(B\).

A 0

B 1

C -1

D 2

💡 Short Explanation

Use value of \(A\). Compare constant terms. Solve for \(B\).

✅ Correct Answer Explanation

From Question 15, \(A=1\) and \(4x+1=1(4x+1)+B\) Therefore \(B=0\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Adds an extra constant.
Option C → Makes both sides unequal.
Option D → Overestimates the constant term.

🎯 Strategy

Used: Option Grouping

Application:

Substitute the previously found value of \(A\).

Final Logic:

\(4x+1=(4x+1)+B\Rightarrow B=0\).

🧠 Memory Trick

"After A, compare constants."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Rational Function Decomposition

17 A probability density function is \(f(x)=K\frac{x}{x^{4}+1},x\in [0,\infty ).\)
Find \(K\).

A π/2

B 2/π

C π/4

D 4/π

💡 Short Explanation

Total probability equals 1. Evaluate improper integral. Solve for normalization constant.

✅ Correct Answer Explanation

\(1=K\int_{0}^{\infty }\,\frac{x}{x^{4}+1}dx\) Let \(t=x^{2}\), \(dt=2x dx\): \(=\frac{1}{2}\int_{0}^{\infty }\,\frac{dt}{t^{2}+1}=\frac{1}{2}{\left[{tan⁡}^{-1}t\right]}_{0}^{\infty }=\frac{\pi }{4}\) Hence \(K⋅\frac{\pi }{4}=1K=\frac{4}{\pi }\) Therefore the provided answer is correct.

❓ Why Other Options Are Incorrect

Option A → Does not normalize the density.
Option B → Produces total probability \(1/2\).
Option C → Gives probability less than 1.

🎯 Strategy

Used: Substitution

Application:

Transform the integral into the standard inverse tangent form.

Final Logic:

Integral equals \(\pi /4\), so \(K=4/\pi\).

🧠 Memory Trick

"PDF ⇒ area equals 1."

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

18 A particle has velocity \(v(t)=\frac{t}{t^{4}+16}.\)
If \(s(0)=0\), find \(s(t)\).

A 1/4 tan⁻¹(t²/4)

B 1/2 tan⁻¹(t²/4)

C 1/8 tan⁻¹(t²/4)

D 1/4 tan⁻¹(t²)

💡 Short Explanation

Integrate velocity. Use \(u=t^{2}\). Apply the condition \(s(0)=0\).

✅ Correct Answer Explanation

\(s(t)=\int \frac{t}{t^{4}+16}dt\) Let \(u=t^{2},du=2t dt\) Then \(=\frac{1}{2}\int \frac{du}{u^{2}+16}=\frac{1}{2}⋅\frac{1}{4}{tan⁡}^{-1}(u/4)=\frac{1}{8}{tan⁡}^{-1}(t^{2}/4)+C\) Using \(s(0)=0\), \(C=0\). Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Missing factor \(1/2\).
Option B → Coefficient too large.
Option D → Incorrect argument and coefficient.

🎯 Strategy

Used: Substitution

Application:

Use \(u=t^{2}\)to convert the integral into standard form.

Final Logic:

Two scaling factors produce coefficient \(1/8\).

🧠 Memory Trick

"t dt ⇒ use t²."

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

19 Assertion (A): \(\int \frac{\sin\,x}{1+{cos⁡}^{2}x} dx\)
is solved using substitution \(t=cos⁡x\).
Reason (R):
The derivative of \(\cos\,x\) is \(\sin\,x\).

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Numerator resembles derivative of denominator expression. Substitution simplifies the integral. Reason directly justifies the method.

✅ Correct Answer Explanation

Let \(t=cos⁡x\) Then \(dt=-sin⁡x dx\) and the integral becomes a rational function: \(-\int \frac{dt}{1+t^{2}}\) Therefore the assertion is true. The reason is also true because the derivative relationship motivates the substitution. Hence Option C is correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Check truth values and whether the reason explains the assertion.

Final Logic:

Derivative linkage makes the substitution natural.

🧠 Memory Trick

"Numerator ≈ derivative ⇒ substitute."

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

20 The area function \(A(x)=\int_{0}^{x}\,\frac{t}{t^{4}+a^{4}}dt\)
is equal to:

A 1/(2a²) tan⁻¹(x²/a²)

B 1/a² tan⁻¹(x²/a²)

C 1/2a tan⁻¹(x²/a)

D 1/a tan⁻¹(x²/a²)

💡 Short Explanation

Use substitution \(u=t^{2}\). Convert to standard inverse tangent form. Apply lower and upper limits.

✅ Correct Answer Explanation

Let \(u=t^{2},du=2t dt\) Then \(A(x)=\frac{1}{2}\int_{0}^{x^{2}}\,\frac{du}{u^{2}+a^{4}}\) Using \(\int \frac{du}{u^{2}+a^{4}}=\frac{1}{a^{2}}{tan⁡}^{-1}\left(\frac{u}{a^{2}}\right)\) gives \(A(x)=\frac{1}{2a^{2}}{tan⁡}^{-1}\left(\frac{x^{2}}{a^{2}}\right)\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Missing factor \(1/2\).
Option C → Incorrect scaling and argument.
Option D → Coefficient should be \(1/(2a^{2})\).

🎯 Strategy

Used: Substitution

Application:

Convert the quartic denominator to a quadratic expression.

Final Logic:

\(u=t^{2}\)immediately yields the standard inverse tangent form.

🧠 Memory Trick

"Quartic + tdt ⇒ use t²."

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

CUET UG Mathematics Booster Test 1 - Special Integrals & Standard Forms

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