AP Inter 2nd Year

Reviewing AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India Questions and Answers can help students prepare confidently for exams.

AP Inter 2nd Year Civics 1st Lesson The Constitution of India Questions and Answers

Very Short Answer Questions

Question 1.
Cite any two major committees and their chairmen of the Constituent Assembly.
Answer:
Two major committees of the Constituent Assembly:

1. Union Powers Committee -Jawaharlal Nehru
   Drafting Committee -B. R. Ambedkar.

Question 2.
Mention Chief Commissioners’ provinces.
Answer:
Chief Commissioners provinces in the Indian Constituent Assembly of 1946 were:

1. Delhi
2. Ajmer-Merwara
3. Coorg
4. British Baluchistan.

Question 3.
How many women representatives were members of the Constituent Assembly indicate four such representatives?
Answer:
There were 15 women members in the Constituent Assembly of 1946,

1. Sarojini Naidu,
2. Durgabai Deshmukh,
3. Vijaya Lakshmi Pandit,
4. Ammu Swaminathan.

Question 4.
Which Supreme Court judgments have held that the Preamble is a part of the Constitution of India?
Answer:
The Supreme Court judgments that have held Preamble is a part of the Constitution of India:

1. Berubari Union Case (1960)
2. Kesavananda Bharati case (1973).
3. LIC of India case (1995).

Question 5.
Write about the Seventh Schedule of the Indian Constitution.
Answer:
The Seventh Schedule of the Indian Constitution is about the division of powers between the Union and the States. It contains three lists.

• List I (Union List),
• List II (State List),
• List III (Concurrent List).

Short Answer Questions

Question 1.
Describe the composition of the Drafting Committee of the Constituent Assembly.
Answer:

1. The Constituent Assembly was the special body formed in Nov-1946 to make Indian Constitution.
2. It contained 389 members (296 seats for British India and 93 for Princely States.)
3. Drafting committee of the constituent Assembly was formed on August 291947.
4. It was entrusted with the crucial task of preparing a draft of the new Constitution of India.
5. This committee comprised seven members and was headed by Dr.B.R.Ambedkar.
6. The Drafting Committee, after considering the proposals of various committees, prepared the First Draft of the Constitution of India, which was published in February 1948.
7. The people of India were given eight months to examine the draft and suggest amendments.
8. In the light of public comments, criticisms, and suggestions, the Committee prepared a Second Draft, which was published in October 1948.

Question 2.
Write any four sources of the Indian Constitution.
Answer:

• The main source of the Constitution was the Government of India Act, 1935.
• Indian Constitution was made after studying many known constitutions of the world.
• It borrowed many ideas from different countries and existing systems.

Four sources of the Indian Constitution:

1. Government of India Act, 1935 : Federal structure, Integrated Judiciary, Public Service Commission, President’s Rule, Office of Governor
2. British Constitution: Parliamentary system, Legislative Procedure, Rule of law, Single Citizenship.
3. USA Constitution : Fundamental Rights, Independence of Judiciary, Judicial review, Office of the Vice President, Impeachment of the President, Removal of Judges of Supreme court.
4. Canada Constitution : Federation with a Strong Centre, residuary powers in the Centre.

Question 3.
Mention any eight parts of the Indian Constitution.
Answer:
Originally, the Constitution of India contained 22 Parts, 395 Articles and 8 Schedules As of 2026, there are 25 Parts, 473 Articles and 12 Schedules.

Major Parts of the Indian Constitution :

1. Part I – Union and its Territory (Articles 1 to 4.
2. Part II – Citizenship (Articles 5 to 11.
3. Part III – Fundamental Rights (Articles 12 to 35)
4. Part IV – Directive Principles of State Policy (Articles 36 to 51.
5. Part IV-A – Fundamental Duties (Article 51 A)
6. Part V – Union Government (Articles 52 to 151.
7. Part VI – State Governments (Articles 152 to 237)
8. Part VII – Deleted by 7th Amendment Act-1956 (Articles 238 deleted)
9. Part VIII – Union Territories (Articles 239 to 242.
10. Part IX – Panchayats (Articles 243 – 243(0))

Long Answer Questions

Question 1.
Explain any six salient features of the Indian Constitution.
Answer:

• Indian Constitution starts with the Preamble.
• It reflects the aims, aspirations, and objectives of the people of India.

Salient Features of the Indian Constitution:

1. Lengthiest Written constitution:

• The constitution of India is the lengthiest written constitution in the world.
• Originally, the Indian constitution contained 22 Parts with 395 Articles and 8 schedules.
• As of 2026, it consists of 25 parts with 473 Articles and 12 schedules.

2. Combination of rigidity and flexibility:
The Constitution of India combines features of both rigid and flexible constitutions.

1. A rigid constitution requires a special procedure for its amendment,
2. A flexible constitution can be amended like ordinary laws.
   Indian Constitution is neither wholly rigid nor wholly flexible, but a unique blend of both.

3. Federal System with Unitary Bias:
The constitution of India includes features of both Unitary and Federal systems. It works as a federal system during normal times and Unitary system in emergencies.

Federal system: Two Governments (Central and State), division of powers, Supreme Constitution, Independent Judiciary. [ Notably the term Federation does not appear anywhere in the constitution instead, it uses “Union of States”]

Unitary system: Strong Centre, Single Constitution, Single Citizenship, Integrated Judiciary.

4. Parliamentary form of Government:

• The Constitution of India has adopted the British parliamentary system of government.
• The President of India is the nominal executive head.
• The Union Council of Ministers, headed by the Prime Minister, is the real executive authority. Majority party rule the legislature.

5. Fundamental Rights:
The basic human rights are as 6 Fundamental Rights given in Part III (Articles 12 to 35).
They are:

1. Right to equality
2. Right to freedom
3. Right against exploitation
4. Right to religion
5. Right to culture and educational right
6. Right to constitutional remedies.

6. Directive Principles:

• Directive Principles of State Policy are given in Part IV (Articles 36 to 51..
• They are not enforceable by courts, but they guide the government in making laws.
• They are divided into three types: Socialistic, Gandhian, and Liberal-Intellectual principles.

7. Fundamental Duties:
Originally, the Constitution did not include Fundamental Duties. They were added by the 42nd Constitutional Amendment Act, 1976. At first, ten duties were included, and later one more was added, making a total of eleven duties. These duties include respecting the Constitution, the National Flag and National Anthem, safeguarding public property etc.

Question 2.
Write the Preamble of the Indian Constitution and explain its significance.
Answer:

• The term “Preamble” refers to the introduction or preface to the Constitution.
• It reflects the aims, aspirations, and objectives of the people of India.
• The Preamble of the Indian Constitution is based on the Objectives Resolution, drafted and moved by Pandit Jawaharlal Nehru, and adopted by the Constituent Assembly in 1947.
• It has been amended only once, by the 42nd Constitutional Amendment Act, 1976, Which added three new words: Socialist, Secular and Integrity.

TEXT OF THE PREAMBLE

We, THE PEOPLE OF INDIA, having solemnly resolved to constitute India into a SOVEREIGN SOCIALIST SECULAR DEMOCRATIC REPUBLIC and to secure to all its citizens:

JUSTICE, Social, Economic and Political;
LIBERTY of thought, expression, belief, faith and worship;
EQUALITY of status and of opportunity; and to promote among them all;
FRATERNITY assuring the dignity of the individual and the unity and integrity of the Nation;
IN OUR CONSTITUENT ASSEMBLY this twenty-sixth day of November, 1949, do HEREBY ADOPT, ENACT AND GIVE TO OURSELVES THIS CONSTITUTION”.

Significance of the Preamble:
The Preamble contains the basic philosophy and fundamental values – political, moral, and social- on which the Constitution is founded. It reflects the grand vision of the Constituent Assembly and the dreams and aspirations of the founding fathers of the Constitution.

Admiring Comments on the Preamble by Eminent members and jurists:

1. “Preamble is the soul of the Constitution”. – Justice Hidayatullah
2. “Preamble is like the identity card of the Constitution.” – N. A. Palkhlvala
3. “Preamble is the “horoscope of our sovereign democratic republic”.- K.M.Munshi
4. “The Preamble to our Constitution expresses what we had thought or dreamt so long”. – Sir Alladi Krishnaswami Iyer

Multiple Choice Questions

Question 1.
Which Act separated the legislative functions of the Governor-General from the executive functions?
1. Charter Act of 1853
2. Charter Act of 1833
3. Indian Council Act of 1861
4. Indian Council Act of 1909
Answer:
1. Charter Act of 1853

Question 2.
Which among the following Indians were nominated by Lord Canning in 1862 to the Legislative Council?
a) Raja of Benares
b) Maharaja of Patiala
c) Sir Dinkar Rao
d)Dadabhai Naoroji
1. a, b, c
2. b, c, d
3. a & d
4. a, c, d
Answer:
1. a, b, c

Question 3.
Under the 1892 Act, members of councils were allowed to
1. Vote on budgetary demands
2. Ask questions to the executive (with restrictions)
3. Initiate bills independently
4. Elect the Viceroy
Answer:
2. Ask questions to the executive (with restrictions)

Question 4.
The Government of India Act of 1919 introduced, for the first time, in India
1. Federalism and Provincial Autonomy
2. Bicameralism and Direct Elections
3. Separate Electorates for Muslims
4. Complete Responsible Government
Answer:
2. Bicameralism and Direct Elections

Question 5.
The Government of India Act of 1935 contained
1. 250 sections and 8 schedules
2. 300 sections and 12 schedules
3. 321 sections and 10 schedules
4. 350 sections and 15 schedules
Answer:
3. 321 sections and 10 schedules

Question 6.
The Constituent Assembly was constituted under
1. Wavell Plan
2. Cripps Mission
3. August Offer
4. Cabinet Mission Plan
Answer:
4. Cabinet Mission Plan

Question 7.
Who chaired the Constituent Assembly when it met as a Legislative body?
1. Dr. Rajendra Prasad
2. H.C. Mukherjee
3. GV. Mavalankar
4. V.T. Krishnamachari
Answer:
3. GV. Mavalankar

Question 8.
Which of the following three words were added to the Preamble by the 42nd Amendment Act, 1976?
1. Socialist, Secular, Integrity
2. Socialist, Justice, Liberty
3. Secular, Republic Equality
4. Unity, Fraternity, Dignity
Answer:
1. Socialist, Secular, Integrity

Fill in the Blanks

Question 1.
Dual system of government (Double Government) was introduced by ___________
Answer:
Pitt’s India Act of 1784

Question 2.
The ‘portfolio system recognized under the Act of 1861 was introduced by ___________
Answer:
Lord Canning

Question 3.
The Indian Councils Act of 1909 is also known as ___________
Answer:
Morley-Minto Reforms

Question 4.
The first meeting of the Constituent Assembly of India was held on ___________
Answer:
9 December 1946

Question 5.
The permanent President of the Constituent Assembly was ___________
Answer:
Dr. Babu Rajendra Prasad

Question 6.
After Independence, the total strength of the Constituent Assembly was reduced from 389 to ___________
Answer:
299

One Word Answers

Question 1.
Which Act permitted the Christian missionaries to enter India?
Answer:
Charter Act of 1813

Question 2.
Who is the first Governor-General of India?
Answer:
Lord William Bentinck

Question 3.
Who was elected as the temporary President of the Constituent Assembly in December 1946?
Answer:
Dr. Sachidananda Sinha

Question 4.
Who moved the historic Objectives Resolution in the Constituent Assembly on December 13, 1946?
Answer:
Jawahartal Nehru

Question 5.
Who was the Chairman of the Drafting Committee of Indian Constitution?
Answer:
Dr. B.R. Ambedkar

Question 6.
Which Schedule of the Indian Constitution recognised the Languages?
Answer:
Eighth Schedule

AP Intermediate 2nd Year Civics Textbook Solutions

Civics Inter 2nd Year

• Chapter 1 The Constitution of India Questions and Answers
• Chapter 2 Fundamental Rights and Directive Principles of State Policy Questions and Answers
• Chapter 3 The Union Executive Questions and Answers
• Chapter 4 Union Legislature Questions and Answers
• Chapter 5 Union Judiciary Questions and Answers
• Chapter 6 The State Executive Questions and Answers
• Chapter 7 The State Legislature Questions and Answers
• Chapter 8 State Judiciary Questions and Answers
• Chapter 9 Union State Relations Questions and Answers
• Chapter 10 Local Governments in India Questions and Answers
• Chapter 11 Representation, Elections and Political Parties Questions and Answers

AP Inter 2nd Year Civics Weightage BluePrint 2026-2027

AP Inter 2nd Year Study Material

AP Board Solutions Inter 2nd Year – AP Board Solutions 12th Class

AP Intermediate 2nd Year Textbooks Solutions

AP Inter 2nd Year Study Material Pdf Download

• AP Inter 2nd Year Maths Study Material
• AP Inter 2nd Year Physics Study Material
• AP Inter 2nd Year Chemistry Study Material
• AP Inter 2nd Year Botany Study Material
• AP Inter 2nd Year Zoology Study Material
• AP Inter 2nd Year Telugu Study Material
• AP Inter 2nd Year English Study Material
• AP Inter 2nd Year Sanskrit Study Material
• AP Inter 2nd Year Hindi Study Material
• AP Inter 2nd Year Civics Study Material
• AP Inter 2nd Year Commerce Study Material
• AP Inter 2nd Year Accountancy Study Material
• AP Inter 2nd Year Economics Study Material

Practice AP Inter 2nd Year Maths Study Material Chapter 7 Integrals MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Integrals MCQ

Indefinite Integrals

Question 1.
The anti derivative of \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\) =
1) \(\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+C\)
2) \(\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^2+C\)
3) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)
4) \(\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+C\)
Solution:
3) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)
Anti derivative of \(\sqrt{x}+\frac{1}{\sqrt{x}}=\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x\)
⇒ I = \(\int x^{\frac{1}{2}} d x+\int x^{\frac{1}{2}} d x=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+c\)

Question 2.
The anti derivative of e^2logcotx
1) cot x – 1
2) tan x – cot x
3) -cot x – x
4) – 1 – cot x
Solution:
3) -cot x – x
I = \(\int e^{2 \log \cot x} d x=\int e^{\log _e \cot ^2 x} d x=\int \cot ^2 x d x=\int\left({cosec}^2 x-1\right) d x\) = -cot x – x + c

Question 3.
If \(\frac{d}{d x}\)f(x) = 4x³ – \(\frac{3}{x^4}\) such that f(2) = 0. Then f(x) is
1) \(x^4+\frac{1}{x^3}-\frac{129}{8}\)
2) \(x^3+\frac{1}{x^4}+\frac{129}{8}\)
3) \(x^4+\frac{1}{x^3}+\frac{129}{8}\)
4) \(x^3+\frac{1}{x^4}-\frac{129}{8}\)
Solution:
1) \(x^4+\frac{1}{x^3}-\frac{129}{8}\)
\(\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4} \Rightarrow f(x)=\int\left(4 x^3-\frac{3}{x^4}\right) d x=\not A \cdot \frac{x^4}{\not A}-\not z\left(\frac{-1}{\not \partial x^3}\right)=x^4+\frac{1}{x^3}+c\) …….(1)
Given, f(x) = 0 ⇒ 0 = 16 + \(\frac{1}{8}+c \Rightarrow c=-\left(\frac{129}{8}\right)(1) \Rightarrow f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}\)

Question 4.
\(\int \frac{10 x^9+10^x \log _e 10}{x^{10}+10^x}\)dx =
1) 10^x – 10¹⁰ + C
2) 10^x + x¹⁰ + C
3) (10^x – x¹⁰)^-1 + C
4) log(10^x + x¹⁰) + C
Solution:
4) log(10^x + x¹⁰) + C
\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log|f(x)| + c
I = \(\int \frac{10 x^9+10^x \log _e^{10}}{x^{10}+10^x} d x\) = log(x¹⁰ + x^x) + c

Question 5.
\(\int \frac{x^2}{1+x^3}\) dx =
1) \(\frac{1}{3} \log \left|1+x^3\right|+c\)
2) \(\frac{2}{3} \log \left|1+x^3\right|+c\)
3) log|1 + x³| + c
4) tan^-1(x^3/2 + c
Solution:
1) \(\frac{1}{3} \log \left|1+x^3\right|+c\)
Put 1 + x³ = t ⇒ 0 + 3x²dx = dt ⇒ x² dx = \(\frac{1}{3}\)dt
I = \(\int \frac{x^2}{1+x^3} d x=\frac{1}{3} \int \frac{1}{t} d t=\frac{1}{3} \log |t|+c=\frac{1}{3} \log \left|1+x^3\right|+c\)

Question 6.
\(\int \frac{d x}{\sin ^2 x \cos ^2 x}\) =
1) tan x + cot x + C
2) tan x – cot x + C
3)tan x cot x + C
4) tan x – cot 2x + C
Solution:
2) tan x – cot x + C
I = \(\int \frac{1}{\sin ^2 x \cos ^2 x} d x=\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x} d x\)
= \(\int \frac{\sin ^2 x}{\sin ^2 x \cos ^2 x} d x+\int \frac{\cos ^2 x}{\sin ^2 x \cos ^2 x} d x\) = ∫sec² dx + ∫cosec² dx = tan x – cot x + c

Question 7.
\(\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x}\)dx =
1) tanx + cot x + C
2) tan x + cosecx + C
3) -tan x + cot x + C
4) tan x + sec x + C
Solution:
1) tanx + cot x + C
I = \(\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x=\int \frac{1}{\cos ^2 x} d x-\int \frac{1}{\sin ^2 x} d x\)
= ∫sec² x dx – ∫cosec²x dx = tan x + cot x + c

Question 8.
\(\int \frac{\cos x+x \sin x}{x(x+\cos x)}\)dx = log|f(x)| + c then f(x) =
1) x(x + cos x)
2) \(\frac{x+\cos x}{x}\)
3) \(\frac{x}{x+\cos x}\)
4) \(\frac{1}{x(x+\cos x)}\)
Solution:
3) \(\frac{x}{x+\cos x}\)

Question 9.
\(\int \frac{e^x(1+x)}{\cos ^2\left(e^x x\right)}\)dx =
1) -cot(e^xx) + C
2) tan(xe^x) + C
3) tan(e^x) + C
4) cot(e^x) + C
Solution:
2) tan(xe^x) + C
Put, x.e^x = t ⇒ (xe^x + e^x)dx = dt ⇒ e^x(x + 1)dx = dt
I = \(\int \frac{e^x(1+x)}{\cos ^2\left(e^x x\right)} d x \Rightarrow I=\int \frac{d x}{\cos ^2 t}\) = ∫sec² dt = tan t + c = tan(xe^x) + c

Question 10.
\(\int \frac{d x}{x^2+2 x+2}\) =
1) x tan^-1(x + 1) + C
2) tan^-1 (x + 1) + C
3) (x + 1)tan^-1x + C
4) tan^-1x + C
Solution:
2) tan^-1 (x + 1) + C
I = \(\int \frac{\mathrm{dx}}{\mathrm{x}^2+2 \mathrm{x}+2} \mathrm{dx}=\int \frac{1}{\mathrm{x}^2+2 \mathrm{x}+1+1} \mathrm{dx}=\int \frac{1}{(\mathrm{x}+1)^2+1^2} \mathrm{dx}\) = tan^-1 (x + 1) + C

Question 11.
\(\int \frac{d x}{\sqrt{9 x-4 x^2}}\) =
1) \(\frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C\)
2) \(\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+C\)
3) \(\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C\)
4) \(\frac{1}{2} \sin ^{-1}\left(\frac{9 x-8}{9}\right)+C\)
Solution:
2) \(\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+C\)
9x – 4x² = \(-4\left[x^2-\frac{9}{4} x\right]=-4\left[x^2-2 \cdot x \cdot \frac{9}{8}+\left(\frac{9}{8}\right)^2-\left(\frac{9}{8}\right)^2\right]=4\left[\left(\frac{9}{8}\right)^2-\left(x-\frac{9}{8}\right)^2\right]\)

Question 12.
\(\int \frac{x d x}{(x-1)(x-2)}\) =
1) \(\log \left|\frac{(x-1)^2}{x-2}\right|+C\)
2) \(\log \left|\frac{(x-2)^2}{x-1}\right|+C\)
3) \(\log \left|\left(\frac{x-1}{x-2}\right)^2\right|+C\)
4) log|(x – 1)(x -2)| + C
Solution:
2) \(\log \left|\frac{(x-2)^2}{x-1}\right|+C\)
Using partial fractions \(\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\) ⇒ x = A(x – 2) + B(x – 1)
ar x = 1 we get A = -1; at x = 2 we get B = 2
I = \(\int \frac{x}{(x-1)(x-2)} d x=\int \frac{-1}{x-1} d x+\int \frac{2}{x-2} d x\) = -log|x – 1| + 2log|x – 2|
= -log|x – 1| + log|(x – 2)|² = \(\log \left|\frac{(x-2)^2}{(x-1)}\right|+c\)

Question 13.
\(\int \frac{d x}{x\left(x^2+1\right)}\) =
1) \(\log |x|-\frac{1}{2} \log \left|x^2+1\right|+C\)
2) \(\log |x|+\frac{1}{2} \log \left|x^2+1\right|+C\)
3) \(-\log |x|+\frac{1}{2} \log \left|x^2+1\right|+C\)
4) \(\frac{1}{2} \log |x|+\log \left|x^2+1\right|+C\)
Solution:
1) \(\log |x|-\frac{1}{2} \log \left|x^2+1\right|+C\)
\(\frac{1}{x\left(x^2+1\right)}=\frac{A}{x}+\frac{B x+C}{x^2+1}\) we get A = 1; B = -1; C = 0
I = \(\int \frac{1}{x\left(x^2+1\right)} d x=\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right) d x=\int \frac{1}{x} d x-\frac{1}{2} \int \frac{2 x}{x^2+1} d x=\log |x|-\frac{1}{2} \log \left|x^2+1\right|+c\)

Question 14.
\(\int \frac{x^2}{1-x^4} d x\) =
1) \(\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|-\frac{1}{2} \tan ^{-1} x+C\)
2) \(\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|+\frac{1}{2} \tan ^{-1} x+C\)
3) \(\frac{1}{4} \log \left|\frac{1+x^2}{1-x^2}\right|-\frac{1}{2} \tan ^{-1} x^2+C\)
4) \(\frac{1}{4} \log \left|\frac{1-x^2}{1+x^2}\right|-\frac{1}{2} \tan ^{-1} x^2+C\)
Solution:
1) \(\frac{1}{4} \log \left|\frac{1+x}{1-x}\right|-\frac{1}{2} \tan ^{-1} x+C\)

Question 15.
∫x²e^x dx =
1) \(\frac{1}{3}\)e^x3 + C
2) \(\frac{1}{3}\)e^x2 + C
3) \(\frac{1}{2}\)e^x3 + C
4) \(\frac{1}{2}\)e^x2 + C
Solution:
1) \(\frac{1}{3}\)e^x3 + C
Put, x³ = t ⇒ 3x²dx = dt ⇒ x²dx = \(\frac{1}{3}\)dt
I = ∫x²e^x3 dx = \(\frac{1}{3}\)∫e^tdt = \(\frac{1}{3}\). e^t + c = \(\frac{1}{3}\) e^x3 + c

Question 16.
∫ x sec² x dx =
1) xtanx – log|sec x| + C
2) xtanx – log|cos x| + C
3) xtanx – log|cosec x| + C
4) xtanx – log|sin x| + C
Solution:
1) xtanx – log|sec x| + C
Integration by parts we have
I = x(tan x) – ∫tan x dx = ∫ x sec²x dx = x(tan x) – log|sec| + c

Question 17.
∫e^x sec x( + tan x) dx =
1) e^xcos x + C
2) e^x sec x + C
3) e^x sin x + C
4) e^x tan x + C
Solution:
2) e^x sec x + C
∫ e^x(f(x) + f'(x)) dx = e^xf(x) + c
I = ∫e^xsecx(1 + tan x) dx = ∫e^x[sec x + sec x tan x]dx = e^x sec x + c

Question 18.
\(\int e^x\left(\frac{1+x \log x}{x}\right) d x\) =
1) xe^{log x} + C
2) e^x log x +C
3) e^x log x² + C
4) None
Solution:
2) e^x log x +C
I = ∫e^x[f(x) + f'(x)]dx = e^xf(x) + c
I = \(\int e^x\left(\frac{1+x \log x}{x}\right) d x=\int e^x\left(\frac{1}{x}+\log x\right) d x=\int e^x\left(\log x+\frac{1}{x}\right) d x\) = e^x(log x) + c

Question 19.
\(\int \sqrt{1+x^2} d x\) =
1) \(\frac{x}{2} \sqrt{1+x^2}+\frac{1}{2} \log \left|\left(x+\sqrt{1+x^2}\right)\right|+C\)
2) \(\frac{2}{3}\left(1+x^2\right)^{\frac{3}{2}}+C\)
3) \(\frac{2}{3} x\left(1+x^2\right)^{\frac{3}{2}}+C\)
4) \(\frac{x^2}{2} \sqrt{1+x^2}+\frac{1}{2} x^2 \log \left|x+\sqrt{1+x^2}\right|+C\)
Solution:
1) \(\frac{x}{2} \sqrt{1+x^2}+\frac{1}{2} \log \left|\left(x+\sqrt{1+x^2}\right)\right|+C\)
\(\int \sqrt{a^2+x^2} d x=\frac{x}{2} \sqrt{a^2+x^2}+\frac{a^2}{2} \log \left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1}\right|+c\)
I = \(\int \sqrt{1+\mathrm{x}^2} \mathrm{dx}=\frac{\mathrm{x}}{2} \sqrt{1+\mathrm{x}^2}+\frac{1}{2} \log \left|\mathrm{x}+\sqrt{\mathrm{x}^2+1}\right|+\mathrm{c}\)

Question 20.
\(\int \sqrt{x^2-8 x+7} d x\) =
1) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}+9 \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
2) \(\frac{1}{2}(x+4) \sqrt{x^2-8 x+7}+9 \log \left|x+4+\sqrt{x^2-8 x+7}\right|+C\)
3) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
4) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)
Solution:
4) \(\frac{1}{2}(x-4) \sqrt{x^2-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^2-8 x+7}\right|+C\)

Question 21.
\(\int \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}\) =
1) tan^-1(e^x) + C
2) tan^-1(e^-x) + C
3) log(e^x – e^-x) + C
4) log(e^x + e^-x) + C
Solution:
1) tan^-1(e^x) + C
I = \(\int \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}=\int \frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}}+\frac{1}{\mathrm{e}^{\mathrm{x}}}}=\int \frac{\mathrm{e}^{\mathrm{x}}}{\left(\mathrm{e}^{\mathrm{x}}\right)^2+1^2} \mathrm{dx}\). Put e^x = t ⇒ e^x dx = dt
I = \(\int \frac{d t}{t^2+1}\) = tan^-1(t) + c = tan^-1(e^x) + c

Question 22.
\(\int \frac{\cos 2 x}{(\sin x+\cos x)^2} d x\) =
1) \(\frac{-1}{\sin x+\cos x}+C\)
2) log|sin x – cos x| + C
3) log|sin x – cos x| + C
4) \(\frac{1}{(\sin x+\cos x)^2}\)
Solution:
2) log|sin x – cos x| + C
I = \(\int \frac{\cos 2 x}{(\sin x+\cos x)^2}=d x=\int \frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2} d x\)
= \(\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)(\cos x+\sin x)} d x\) = log|cos x + sin x| + c

Definite Intervals

Question 1.
\(\int_1^{\sqrt{3}} \frac{d x}{1+x^2}\) =
1) \(\frac{\pi}{3}\)
2) \(\frac{2 \pi}{3}\)
3) \(\frac{\pi}{6}\)
4) \(\frac{\pi}{12}\)
Solution:
4) \(\frac{\pi}{12}\)
Textual given key is 1.
I = \(\int \frac{1}{\left(1+x^2\right)} d x=\left(\tan ^{-1}(x)\right)_1^{\sqrt{3}}\) = tan^-1(\(\sqrt{3}\)) – tan^-1(1) = 60 – 45 = 15 = \(\frac{\pi}{12}\)

Question 2.
\(\int_0^{\frac{2}{3}} \frac{d x}{4+9 x^2}\) =
1) \(\frac{\pi}{6}\)
2) \(\frac{\pi}{12}\)
3) \(\frac{\pi}{24}\)
4) \(\frac{\pi}{4}\)
Solution:
3) \(\frac{\pi}{24}\)
Textual given key is 4.

Question 3.
The value of the integral \(\int_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^2} d x\) is
1) 6
2) 0
3) 3
4) 4
Solution:
1) 6

Question 4.
If f(x) = \(\int_0^x t\) sin t dt, then f'(x) is
1) cos x + x sin x
2) x sin x
3) x cos x
4) sinx + x cosx
Solution:
2) x sin x
f(x) = \(\int_0^x t \sin t d x\) Diff. w.r.t we get f'(x) = x sin x

Question 5.
The value \(\int_{-\pi / 2}^{\pi / 2}\left(x^3+x \cos x+\tan ^5 x+1\right) d x\) is
1) 0
2) 2
3) π
4) 1
Solution:
3) π
I = \(\int_{-\pi / 2}^{\pi / 2}\left(x^3+x \cos x+\tan ^5 x+1\right)=\int_{-\pi / 2}^{\pi / 2} 1 d x=(x)_{-\pi / 2}^{\pi / 2}=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\frac{\pi}{2}+\frac{\pi}{2}=\pi\)

Question 6.
The value of \(\int_0^\pi 2 \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x\) is
1) 2
2) 3/4
3) 0
4) -2
Solution:
3) 0
\(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)
I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x \quad \ldots \ldots \ldots .(1) \quad I=\int_0^{\pi / 2} \log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] d x\) …..(2)
I + I = \(\int_0^{\pi / 2}\left[\log \left(\frac{4+3 \sin x}{4+3 \cos x}\right)+\log \left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right] d x \Rightarrow 2 I=\int_0^{\pi / 2} \log (1) d x \Rightarrow 2 I=0 \Rightarrow I=0\)

Question 7.
If f(a + b – x) = f(x), then \(\int_a^b f(x) d x\) =
1) \(\frac{(a+b)}{2} \int_a^b f(b-x) d x\)
2) \(\frac{(a+b)}{2} \int_a^b f(b+x) d x\)
3) \(\frac{b-a}{2} \int_a^b f(x) d x\)
4) \(\frac{a+b}{2} \int_a^b f(x) d x\)
Solution:
4) \(\frac{a+b}{2} \int_a^b f(x) d x\)
\(\int_a^b x f(x) d x=\int_a^b(a+b-x) f(a+b-x) d x=\int_a^b[(a+b)-x] f(x) d x=\int_a^b(a+b) f(x) d x-\int_a^b x f(x) d x\)
\(\int_a^b x f(x) d x=(a+b) \int_a^b f(x) d x-\int_a^b x f(x) d x\) ⇒ \(2 \int_a^b x f(x) d x=(a+b) \int_a^b f(x) d x\)
⇒ \(\int_a^b x f(x) d x=\left(\frac{a+b}{2}\right) \int_a^b f(x) d x\)

Question 8.
\(\int_0^{\pi / 2} \frac{3 \sin x+5 \cos x}{\sin x+\cos x} d x\) =
1) 2π
2) π
3) 4π
4) 8π
Solution:
1) 2π
\(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)

Question 9.
\(\int_0^4|2-x| d x\) =
1) 12
2) 4
3) 8
4) 2
Solution:
2) 4
I = \(\int_0^4|2-x| d x\) |2 – x| = 2x if 2 – x ≥ 0; 2 ≥ x; x ≤ 2
= \(\int_0^2|2-x| d x+\int_2^4|2-x| d x=\int_0^2(2-x) d x+\int_2^4-(2-x) d x=\left(2 x-\frac{x^2}{2}\right)_0^2-\left(2 x-\frac{x^2}{2}\right)_2^4\)
= (4 – 2) – 0 – [(8 – 8) – (4 – 2)] = 2 – [0 – 2] = 2 + 2 = 4

Question 10.
\(\int_{-2}^2\left(4-x^2\right)^{\frac{3}{2}} d x\) =
1) 2π
2) 4π
3) 6π
4) 8π
Solution:
3) 6π

Practice AP Inter 2nd Year Maths Study Material Chapter 13 Probability MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Probability MCQ

Question 1.
If P(A) = \(\frac{1}{2}\), P(B) = (), then P(A|B) is
1) 0
2) \(\frac{1}{2}\)
3) not exist
4) 1
Solution:
3) not exist
P(A/B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{0}\) which is not defined

Question 2.
If A and B are events such that P(A|B) = P(B|A), then
1) A ⊂ B but A ≠ B
2) A = B
3) A ∩ B = Φ
4) P(A) = P(B)
Solution:
Given P(A/B) = P(B/A)
⇒ \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~B} \cap \mathrm{~A})}{\mathrm{P}(\mathrm{~A})} \Rightarrow \frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})}\) [∵ A ∩ B = B ∩ A]
⇒ \(\frac{1}{P(B)}=\frac{1}{P(A)}\) ⇒ P(A) = P(B)

Question 3.
The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
1) 0
2) \(\frac{1}{3}\)
3) \(\frac{1}{12}\)
4) \(\frac{1}{36}\)
Solution:
4) \(\frac{1}{36}\)
When two dice are rolled, the number of outcomes n(S) = 6² = 36.
The only even prime number is 2.
Let E be the event of getting an even prime number on each die. ∴ E = (2, 2) ⇒ P(E) = \(\frac{1}{36}\)

Question 4.
Two events A and B will be independent, if
1) A and B are mutually exclusive
2) P(A’B’) = [1 – P(A)] [1 – P(B)]
3) P(A) = P(B)
4) P(A) + P(B) = 1
Solution:
2) P(A’B’) = [1 – P(A)] [1 – P(B)]
A and B are independent ⇒ A’ and B’ are independent
⇒ P(A’ ∩ B’) = [1 – P(A)] [1 – P(B)] are independent

Question 5.
If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct?
1) P(A|B) = \(\frac{\mathrm{P}(\mathrm{~B})}{\mathrm{P}(\mathrm{~A})}\)
2) P(A|B) < P(A)
3) P(A|B) ≥ P(A)
4) P(A) = P(B)
Solution:
3) P(A|B) ≥ P(A)
If A ⊂ B, then A ∩ B ⇒ P(A ∩ B) = P(A). Also, P(A) < P(B)
P(A|B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A})}{\mathrm{P}(\mathrm{~B})}\) …..(1)
Since P(B) ≤ 1 ⇒ \(\frac{1}{\mathrm{P}(\mathrm{~B})} \geq 1 \Rightarrow \frac{\mathrm{P}(\mathrm{~A})}{\mathrm{P}(\mathrm{~B})} \geq \mathrm{P}(\mathrm{~A})\)
From (1), we have P(A|B) ≥ P(A)

Question 6.
If A and B are two events such that P(A) ≠ 0 and P(B | A) = I, then
1) A ⊂ B
2) B ⊂ A
3) B = Φ
4) A = Φ
Solution:
1) A ⊂ B
Given P(A) ≠ 0 and P(B|A) = 1,
∴ \(P(B \mid A)=\frac{P(B \cap A)}{P(A)} \Rightarrow 1=\frac{P(B \cap A)}{P(A)}\) ⇒ P(A) = P(B ∩ A) ⇒ A = A ∩ B ⇒ A⊂ B

Question 7.
If P(A|B) > P(A), then which of the following is correct :
1) P(B|A) < P(B)
2) P(A ∩ B) < P(A) . P(B) 3) P(B|A) > P(B)
4) P(B|A) = P(B)
Solution:
3) P(B|A) > P(B)
Given that
Given, P(A|B) > P(A) ⇒ \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}>\mathrm{P}(\mathrm{~A})\)
⇒ P(A ∩ B) > P(A) × P(B) ⇒ \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})}\) ⇒ P(B) ⇒ P(B|A) > P(B)

Question 8.
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
1) P(B|A) = 1
2) P(A|B) = 1
3) P(B|A) = 0
4)P(A|B) = 0
Solution:
2) P(A|B) = 1
Given that P(A) + P(B) – P(A and B) =P(A),
⇒ P(A) + P(B) – P(A ∩ B) = P(A) ⇒ P(B) – P(A ∩ B) = 0 ⇒ P(A ∩ B) = P(B)
P(A|B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~B})}{\mathrm{P}(\mathrm{~B})}\) = 1

Question 9.
A bag contains 7 red and 3 white balls. Three balls are drawn one after other without replacement. Then the probability that the first two are red and third one is white is
1) \(\frac{7}{40}\)
2) \(\frac{33}{40}\)
3) \(\frac{23}{40}\)
4) \(\frac{17}{40}\)
Solution:
1) \(\frac{7}{40}\)
7R + 3W = Total 10 balls
P(E) = P(Red and Red and White) = \(\left(\frac{7}{10}\right) \times \frac{6}{9} \times \frac{3}{8}=\frac{7}{40}\) (∵ Drawn ball is not replaced)

Question 10.
A book consists «f 20 pages. If two pages arc drawn (opened) at random, then the probability that both numbers are prime numbers is
1) \(\frac{17}{95}\)
2) \(\frac{16}{95}\)
3) \(\frac{2}{15}\)
4) \(\frac{14}{95}\)
Solution:
4) \(\frac{14}{95}\)
Total no. of pages = 20
Primes up to 20 are 2, 3, 5, 7, 11, 13, 17, 19 & the no. of these primes = 8
∴ P(E) = \(\frac{{ }^8 \mathrm{C}_2}{{ }^{20} \mathrm{C}_2}=\frac{8 \times 7}{20 \times 19}=\frac{2 \times 7}{5 \times 19}=\frac{14}{95}\)

Question 11.
A fair coin is tossed 3 times, then the probability of getting one head and two tails is
1) \(\frac{1}{8}\)
2) \(\frac{1}{4}\)
3) \(\frac{3}{8}\)
4) \(\frac{1}{2}\)
Solution:
3) \(\frac{3}{8}\)
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
E = {HTT, THT, TTH} ⇒ n(E) = 3; n(S) = 8 P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{3}{8}\)

Question 12.
A person appears lor an interview for two posts A and B. The selection of the posts are independent. If P( A) = \(\frac{1}{5}\), P(B) = \(\frac{1}{8}\) then P(A ∪ B) is
1) \(\frac{7}{10}\)
2) \(\frac{3}{10}\)
3) \(\frac{9}{10}\)
4) \(\frac{1}{10}\)
Solution:
2) \(\frac{3}{10}\)
Given A, B are independent event ⇒ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{1}{5}+\frac{1}{8}\) – [P(A).P(B)] = \(\frac{13}{40}-\left(\frac{1}{5} \times \frac{1}{8}\right)=\frac{13}{40}-\frac{1}{40}=\frac{12}{40}=\frac{3}{10}\)

Question 13.
Three events A, B, C are mutually exclussive and exhaustive and P(A) = 0.4, then P(B) + P(C) =
1) 0.4
2) 0.5
3) 0.6
4) 0
Solution:
3) 0.6
A, B, C are mutually exclusive and exhaustive ⇒ A ∪ B ∪ C = S ……..(1)
Given P(A) = 0.4 ……..(2)
(1) ⇒ P(A ∪ B ∪ C) = P(S) ⇒ P(A) + P(B) + P(C) = 1 ⇒ 0.4 + P(B) + P(C) = 1
⇒ P(B) + P(C) = 1 – 0.4 = 0.6

Question 14.
If P(A ∪ B) = 0.65 and P(A ∩ B) = 0.15, then P(\(\vec{A}\)) + P(\(\vec{B}\)) =3 J
1) 0.8
2) 0.6
3) 1.2
4) 1.4
Solution:
3) 1.2
\(\mathrm{P}(\overline{\mathrm{~A}})+\mathrm{P}(\overline{\mathrm{~B}})\) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)] = 2 – [P(A ∪ B) + P(A ∩ B)]
= 2 – [0.65 + 0.15] = 2 – [0.80] = 1.20 = 1.2

Question 15.
When two dice are rolled, the probability of getting unequal numbers on the faces is
1) \(\frac{1}{6}\)
2) \(\frac{35}{36}\)
3) \(\frac{5}{6}\)
4) \(\frac{1}{3}\)
Solution:
3) \(\frac{5}{6}\)
Two dice are rolled n(S) = 36
Equal number faces = {(1, 1) (2, 2) (3, 3) (4, 4)(5, 5)(6, 6)} ⇒ n(S) = 6
⇒ no.of unequal faces = 36 – 6 = 30 = n(E)
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{30}{36}=\frac{5}{6}\)

Question 16.
A fair coin whose faces are marked with I and 2 is thrown for four times. Then the probability of throwing a total of atleast 5 is
1) \(\frac{1}{16}\)
2) \(\frac{5}{16}\)
3) \(\frac{15}{16}\)
4) \(\frac{3}{16}\)
Solution:
3) \(\frac{15}{16}\)
Coin with faces 1. (say H); 2. (say T)
thrown 4 – times
Getting total at least 5 ⇒ total ≥ 5 ⇒ x ≥ 5
Now, P(x ≥ 5) = 1 – P(x < 5) = 1 – P (getting a total 4 faces 4 – times)
= 1 – P(every time a face 1) = 1 – \(\left[\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right]=1-\frac{1}{16}=\frac{16-1}{16}=\frac{15}{16}\)

Question 17.
If A and B are independent events of an experiment then which of the following statements is true
1) P(A ∪ B) = P(A) + P(B) – P(A) . P(B)
2) P(A|B) = P(A) and P(B|A) = P(B)
3) P(A ∩ B) = P(A) . P(B)
4) All the above
Solution:
4) All the above
By definition P(A ∩ B) = P(A).P(B)

Question 18.
If A and B are two events of a random experiment of throwing a die given by “A” : throwing an odd face and B : throwing a composite face.
Then which of the following statements is correct. ?
1)A and Bare equally likely
2) A and B are mutually exclusive
3) A and B are mutually exhaustive
4) A and B are linearly independent
Solution:
2) A and B are mutually exclusive
When a die is thrown
A : odd face (1, 3, 5); B : composite face (4, 6). Then P(A) =\(\frac{3}{6}=\frac{1}{2}\) and P(B) = \(\frac{2}{6}=\frac{1}{3}\)
1) A, B are likely (✗)
2) A, B are mutually exclusive(✓)
A ∩ B = Φ (or) P(A ∩ B) = 0
3) A, B mutually exhaustive (✗) ∵ A ∪ B ≠ S
4) P(A ∩ B) ≠ P(A).P(B) ⇒ NOT independent (✗)

Question 19.
If A, B and C are independent events of a random experiment such that P(A) = p, P(B) = q, P(C) = r, where p, a, r ∈ (0, 1). Then the probability of the event A only occurs is
1) p . q . r
2) p(1 – q)(1 – r)
3) p. q(1 – r)
4) (1 – p) (1 – q) (1 – r)
Solution:
2) p(1 – q)(1 – r)
A, B, C are independent
P(A only occurs) = \(\mathrm{P}(\mathrm{~A} \cap \overline{\mathrm{~B}} \cap \overline{\mathrm{C}})=\mathrm{P}(\mathrm{~A}) \cdot \mathrm{P}(\overline{\mathrm{~B}}) \cdot \mathrm{P}(\overline{\mathrm{C}})\) (∵ A, B, C are independent)
= P(A).[1 – P(B)][1 – P(C)] = P[1 – q][1 – r]

Question 20.
A fair die is rolled. Consider the events A = {I, 3, 5} and B = {2, 3}, then P(A|B) is
1) \(\frac{1}{2}\)
2) \(\frac{1}{3}\)
3) \(\frac{2}{3}\)
4) \(\frac{5}{6}\)
Solution:
1) \(\frac{1}{2}\)
S = {1, 2, 3, 4, 8, 6}; A = {1, 3, 5}, B = {2, 3} ⇒ P(B) =\(\frac{2}{6}=\frac{1}{3}\)
∴ (A ∩ B) = {3} ⇒ P(A ∩ B) = \(\frac{1}{6}\)
∴ P(A|B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{1 / 6}{1 / 3}=\frac{1}{6} \times \frac{3}{1}=\frac{1}{2}\)

Practice AP Inter 2nd Year Maths Study Material Chapter 12 Linear Programming MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Linear Programming MCQ

Question 1.
Region represented by x > 0, y ≥ 0 is
1) First quadrant
2) Second quadrant
3) Third quadrant
4) Fourth quadrant
Solution:
1) First quadrant
Region x ≥ 0, y ≥ 0 (non-negative) ⇒ Both x and y positive → first quadrant.

Question 2.
If the objective function Z = ax + by has both a maximum and a minimum value on the region R then R is
1) Bounded
2) Unbounded
3) Concave polygon
4) Infeasible
Solution:
1) Bounded
Both maximum and minimum exist only if region is closed and bounded.

Question 3.
The linear inequalities or equations or restrictions on the variables of a linear programming problem are called
1) constriants
2) Decision variables
3) objective function
4) Linear relations
Solution:
1) constriants
Linear restrictions in LPP are called constraints.

Question 4.
The optimal value of the objective function is attained at the points
1) On X-axis
2) On Y-axis
3) Which are at the corner points of the feasible region
4) Which are at the points of intersection of the inequation with Y-axis
Solution:
3) Which are at the corner points of the feasible region
In LPP, optimum value occurs at vertices of feasible region.

Question 5.
In a linear programming problem the objective function and constraints must be
1) Non-linear
2) Linear
3) Exponential
4) Logarithmic
Solution:
2) Linear
Objective function & constraints must be linear in LPP

Question 6.
The maximum value of Z = 3x + 4y subject to constraints x + y ≤ 4, x ≥ 0, y ≥ 0 is
1) 12
2) 14
3) 16
4) 10
Solution:
3) 16
Corner check points: (0, 0), (4, 0), (0, 4) then Z = 3x + 4y values: 0, 12, 16.
Maximum value is 16.

Question 7.
Maximize Z = 3x + 5y, subject to constriants x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
1) 20 at (1, 0)
2) 30 at (0, 6)
3) 37 at (4, 5)
4) 33 at (6, 3)
Solution:
3) 37 at (4, 5)
Corner check points of feasible region. Z = 3x + 5y
Best point: (4, 5) ⇒ Z = 12 + 25 = 37

Question 8.
The point which does not lie in the half plane 2x- + 3y – 12 < 0 is
1) (2, 1)
2) (1, 2)
3) (-2, 3)
4) (2, 3)
Solution:
4) (2, 3)
Point NOT in 2x + 3y – 12 < 0
Substitute each point → LHS < 0
Check (2, 3): 4 + 9 – 12 = 1 (NOT < 0)

Question 9.
Which of the following is not a component of linear programming problem?
1) Objective function
2) Constriant
3) Decision variable
4) Differential equation
Solution:
4) Differential equation
Not a component of LPP. LPP uses linear equations, not calculus.

Question 10.
The position of points 0(0, 0) and P(2, -3) in the region of graph of inequation 2x – 3y < 5 will be
1) O inside and P outside
2) O and P both inside
3) O and P both outside
4) O outside and P inside
Solution:
1) O inside and P outside
Check points in 2x – 3y < 5
0(0, 0): 0 < 5 → inside P(2, -3): 4 + 9 = 13 > 5 → outside

Practice AP Inter 2nd Year Maths Study Material Chapter 11 Three Dimensional Geometry MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Three Dimensional Geometry MCQ

Question 1.
If α, β, γ are the angles made by the line with positive direction of the coordinate axes, then sin²α + sin²β + sin² γ =
1) 1
2) 2
3) 3
4) \(\frac{3}{2}\)
Solution:
2) 2
α, β, γ are angle made by the line with +ve direction of coordinate axes
1 = cosα, m = cosβ, n = cosγ are dc’s of the line ⇒ l² + m² + n = 1
⇒ cos² α + cos² β + cos² γ = 1 ⇒ (1 – sin² α) + (1 – sin² β) + (1 – sin² γ) = 1
⇒ 3 – 1 = sin² α + sin² β + sin² γ ⇒ sin² a + sin² p + sin² γ = 2

Question 2.
The direction cosines of the median of the triangle formed by A(1, -3, 2) B(3, 1, 2) and C(-1, 3, -3) which passing through the vertex C is
1) \(\left(\frac{3}{5 \sqrt{2}}, \frac{4}{5 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
2) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{1}{5 \sqrt{2}}\right)\)
3) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{\sqrt{2}}\right)\)
4) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right)\)
Solution:
3) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{\sqrt{2}}\right)\)
Mid point of AB = F = \(\left(\frac{1+3}{2}, \frac{-3+1}{2}, \frac{2+2}{2}\right)\) = (2, -1, 2), C = (-1, 3, -3)
d.r’s of Median CF = (a, b, c) = (2 + 1, -1 – 3, 2 + 3) = (3, -4, 5) ⇒ \(\sqrt{9+16+25}=\sqrt{50}=5 \sqrt{2}\)
d.c’s = \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{5 \sqrt{2}}\right)\)

Question 3.
If the line joining the points A(2, 3, 4) and B(3, -2, 2) is parallel to the line joining C(1, -2, z) and D(-1, y, -1), then y + z =
1) 13
2) 3
3) -3
4) -13
Solution:
2) 3
Given AB || CD ⇒ \(\left(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\right)=\frac{-2}{1}=\frac{y+2}{-5}=\frac{-1-z}{-2}\)
⇒ \(\frac{-2}{1}=\frac{y+2}{-5}=\frac{1+z}{2} \Rightarrow-2=\frac{y+2}{-5} \text { and }-2=\frac{1+z}{2}\) ⇒ -4 = 1 + z ⇒ -5 = z
⇒ 10 = y + 2 ⇒ 8 = y ⇒ y + z ⇒ 8 + (-5) = 3

Question 4.
If the two lines \(\overline{\mathbf{r}}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\lambda(\mathbf{P} \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \text { and } \overline{\mathbf{r}}=(4 \hat{\mathbf{i}}-\mathbf{P} \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(3 \mathbf{P} \hat{\mathbf{i}}+5 \mathrm{P} \hat{\mathbf{j}}+3 \hat{\mathbf{k}})\) are perpendicular then p =
1) 2
2) 3
3) 6
4) 2 or 3
Solution:
4) 2 or 3
Dr’s of line (1) are (p, -3, 6); Dr’s of line (2) are (3p, 5p, 3)
Given lines are perpendicular
⇒ a₁a₂ + b₁b₂ + c₁c₂ = 0 ⇒ 3p(p) + 5p(-3) + 18 = 0 ⇒ 3p² – 15p + 18 = 0
⇒ p² – 5p + 6 = 0 ⇒ (p – 2)(p – 3) = 0 ⇒ p = 2 (or) p = 3

Question 5.
It the two lines \(\frac{x+1}{2 k}=\frac{y-3}{3}=\frac{z-4}{-7}\) and \(\frac{x-1}{1}=\frac{y+1}{-3 k}=\frac{z+2}{2}\) are perpendicular, then k =
1) 2
2) 1
3) -2
4) 14/11
Solution:
3) -2
Given lines are perpendicular ⇒ 2k(1) + 3(-3k) + (-7)(2) = 0 ⇒ 2k – 9k – 14 = 0
⇒ -7k = 14 ⇒ k = -2

Question 6.
The angle between the lines \(\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z+1}{1}\) and \(\frac{x+2}{1}=\frac{y+2}{1}=\frac{z-3}{2}\) is
1) \(\frac{\pi}{3}\)
2) \(\frac{\pi}{6}\)
3) \(\cos ^{-1}\left(\frac{5}{6}\right)\)
4) \(\cos ^{-1}\left(\frac{3}{4}\right)\)
Solution:
1) \(\frac{\pi}{3}\)
Dr’s of the lines are (a₁, b₁, c₁) = (2, -1, 1); (a₂, b₂, c₂) = (1, 1, 2)
∴ cos θ = \(\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}=\frac{|2-1+2|}{\sqrt{4+1+1} \sqrt{1+1+4}}=\frac{3}{\sqrt{6} \cdot \sqrt{6}}=\frac{3}{6}=\frac{1}{2}=\cos 60^{\circ} \Rightarrow \theta=\frac{\pi}{3}\)

Question 7.
If θ is the acute angle between the lines \(\overline{\mathbf{r}}=(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})\) and \(\vec{r}=(2 \hat{i}-3 \hat{j}-4 \hat{k})+\mu(4 \hat{i}+3 \hat{j}+12 \hat{k})\) then cosθ
1) \(\frac{34}{39}\)
2) \(\frac{22}{39}\)
3) \(\frac{26}{39}\)
4) \(\frac{14}{39}\)
Solution:
4) \(\frac{14}{39}\)
Dr’s of the lines are (a₁, b₁, c₁) = (1, 2, -2); (a₂, b₂, c₂) = (4, 3, 12)
cos θ = \(\frac{|(4+6-24)|}{\sqrt{1+4+4} \sqrt{16+9+144}}=\frac{14}{3 \sqrt{169}}=\frac{14}{3(13)}=\frac{14}{39}\)

Question 8.
Equation of the line passing through (2, 1, -4) and parallel to the line joining the points (1, 0, -1) and (3, 2, 2) is
1) \(\frac{x+1}{2}=\frac{y+1}{2}=\frac{z-4}{3}\)
2) \(\frac{x-2}{2}=\frac{y-1}{2}=\frac{z+4}{3}\)
3) \(\frac{x-2}{-2}=\frac{y-1}{-2}=\frac{z+4}{3}\)
4) \(\frac{x+2}{-2}=\frac{y+1}{-2}=\frac{z-4}{3}\)
Solution:
2) \(\frac{x-2}{2}=\frac{y-1}{2}=\frac{z+4}{3}\)
Dr’s of line joning points (1, 0, -1) and (3, 2, 2) are (3 – 1, 2 – 0, 2 + 1) = (2, 2, 3)
required line || to given line ⇒ Dr’s of the line = (a, b, c) = (2, 2, 3)
Also (x₁, y₁, z₁) = (2, 1, -4) is a point on the line
∴ Equation of required line = \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \Rightarrow \frac{x-2}{2}=\frac{y-1}{2}=\frac{z+4}{3}\)

Question 9.
Euation of the line passing through the point (1, 2, 3)and parallel to the z axis is
1) \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{0}\)
2) \(\frac{x-1}{0}=\frac{y-2}{1}=\frac{z-3}{1}\)
3) \(\frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1}\)
4) \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\)
Solution:
3) \(\frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1}\)
Dr’s of z-axis (a, b, c) = (0, 0, 1) . Aslo point on the line is (x₁, y₁, z₁) = (1, 2, 3)
∴ Equation of required line \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1}\)

Question 10.
The direction cossines of the line which is perpendicular to the lines \(\overline{\mathbf{r}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\lambda(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \(\stackrel{\rightharpoonup}{\mathbf{r}}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\boldsymbol{\mu}(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) is
1) \(\left(\frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)
2) \(\left(\frac{-3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{7}{\sqrt{59}}\right)\)
3) \(\left(\frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}}, \frac{7}{\sqrt{59}}\right)\)
4) \(\left(\frac{3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)
Solution:
4) \(\left(\frac{3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)
Given lines \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\mathrm{t} \overline{\mathrm{~b}} \Rightarrow \overline{\mathrm{~b}}=2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}}\) ………(1) and \(\overline{\mathrm{r}}=\overline{\mathrm{c}}+\mathrm{s} \overline{\mathrm{~d}} \Rightarrow \overline{\mathrm{~d}}=3 \overline{\mathrm{i}}-5 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) …………..(2)
Dr’s of the line which is perpendicular to both (1) and (2) and parallel to vector \(\overline{\mathbf{b}} \times \overline{\mathbf{d}}\)
Now \(\overline{\mathrm{b}} \times \overline{\mathrm{d}}=\left|\begin{array}{ccc}
\mathrm{i} & \mathrm{j} & \mathrm{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|=\overline{\mathrm{i}}(3)-\overline{\mathrm{j}}(1)+\overline{\mathrm{k}}(-7)\)
Dr’s of the line = (a, b, c) = (3, -1, -7) = \(\sqrt{3^2+(-1)^2+(-7)^2}=\sqrt{59}\)
∴ d.c’s = \(\left(\frac{3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)

Practice AP Inter 2nd Year Maths Study Material Chapter 10 Vector Algebra MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Vector Algebra MCQ

Question 1.
In triangle ABC (Fig), which of the following is not true:

1) \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
2) \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}\)
3) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}+\overrightarrow{\mathbf{A C}}=\overrightarrow{\mathbf{0}}\)
4) \(\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
Solution:
3) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}+\overrightarrow{\mathbf{A C}}=\overrightarrow{\mathbf{0}}\)
By Triangle Law of Addition of Vectors we have
\(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}} \text { (or) } \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=-\overrightarrow{\mathrm{CA}} \Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)

Question 2.
If \(\vec{a} \text { and } \vec{b}\) are two collinear vectors, then which of the following is correct
1) \(\overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{a}}\) for some scalar λ ≠ 0.
2) \(\vec{a}= \pm \vec{b}\)
3) the respective components of \(\vec{a} \text { and } \vec{b}\) are not proportional
4) both the vectors \(\vec{a} \text { and } \vec{b}\) have same direction, but different magnitudes.
Solution:
1) \(\overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{a}}\) for some scalar λ ≠ 0.
If \(\vec{a} \text { and } \vec{b}\) are two collinear vectors, then \(\vec{b}\) = λ\(\vec{a}\).
The other options (2) & (4) are only true for particular values of λ

Question 3.
If \(\overrightarrow{\mathbf{a}}\) is a nonzero vector of magnitude ‘a’ and λ. a nonzero scalar, then \(\lambda \overrightarrow{\mathbf{a}}\) is unit vector if
1) λ = 1
2) λ = – 1
3) a = |λ|
4) a = 1/| λ|
Solution:
4) a = 1/| λ|
\(|\lambda \bar{a}|=1 \Rightarrow|\lambda \| \vec{a}|=1 \Rightarrow|\vec{a}|=\frac{1}{|\lambda|} \Rightarrow a=\frac{1}{|\lambda|}\)

Question 4.
Let the vectors \(\vec{a} \text { and } \vec{b}\) be such that \(|\overrightarrow{\mathrm{a}}|=3,|\overrightarrow{\mathrm{~b}}|=\frac{\sqrt{2}}{3}\), then \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\) is a unit vector, if the angle between \(\vec{a} \text { and } \vec{b}\) is
1) π/6
2) π/4
3) π/3
4) π/2
Solution:
2) π/4
Given that |\(\vec{a}\)| = 3, |\(\vec{b}\)| = \(\frac{\sqrt{2}}{3}\) and \(\vec{a} \text { and } \vec{b}\) is a unit vector. ⇒ \(|\vec{a} \times \vec{b}|=1 \Rightarrow|\vec{a} \| \vec{b}| \sin \theta=1\)
⇒ \(3\left(\frac{\sqrt{2}}{3}\right) \sin \theta=1 \Rightarrow \sqrt{2} \sin \theta=1 \Rightarrow \sin \theta=\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4} \Rightarrow \theta=\frac{\pi}{4}\)

Question 5.
Area of a rectangle having vertices A, B, C and D with position vectors \(-\hat{\mathbf{i}}+\frac{1}{2} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \hat{\mathbf{i}}+\frac{1}{2} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \hat{\mathbf{i}}-\frac{1}{2} \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \text { and }-\hat{\mathbf{i}}-\frac{1}{2} \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\), respectively is
1) 1/2
2) 1
3) 2
4) 4
Solution:
3) 2
Given ABCD is a rectangle

Area of rectangle ABCD = Length × Breadth = (AB) × (AD) = 2(1) = 2 sq. units

Question 6.
If θ is the angle between two vectors \(\vec{a} \text { and } \vec{b}\), then \(\vec{a} \text.\vec{b}\) > 0 only when
1) 0 < θ < \(\frac{\pi}{2}\)
2) 0 ≤ θ ≤ \(\frac{\pi}{2}\)
3) 0 < θ < π
4) 0 ≤ θ ≤ π
Solution:
2) 0 ≤ θ ≤ \(\frac{\pi}{2}\)
We have \(\vec{a} \cdot \vec{b} \geq 0 \Rightarrow|\vec{a} \| \vec{b}| \cos \theta \geq 0 \Rightarrow \cos \theta \geq 0\) [∵ \(|\overrightarrow{\mathrm{a}}| \geq 0 \text { and }|\overrightarrow{\mathrm{b}}| \geq 0\)]
⇒ 0 ≤ θ ≤ \(\frac{\pi}{2}\) Hence \(\vec{a}\).\(\vec{b}\) ≥ 0 of 0 ≤ θ ≤ \(\frac{\pi}{2}\)

Question 7.
Let \(\vec{a} \text { and } \vec{b}\) be two unit vectors and θ is the angle between them. Then \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}\) is a unit vector if
1) θ = \(\frac{\pi}{4}\)
2) θ = \(\frac{\pi}{3}\)
3) θ = \(\frac{\pi}{2}\)
4) θ = \(\frac{2\pi}{3}\)
Solution:
4) θ = \(\frac{2\pi}{3}\)
We have \(\vec{a} \text { and } \vec{b}\) two unit vectors and θ is the angle between them. Then, |\(\vec{a}\)|=|\(\vec{b}\)|= 1
Now \(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\) is a unit vector if \(|\vec{a}+\vec{b}|=1 \Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=1 \Rightarrow \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=1\)
⇒ \(|\vec{a}|^2+2 \vec{a} \vec{b}+|\vec{b}|^2=1 \Rightarrow 1^2+2|\vec{a}| \vec{b} \cos \theta+1^2=1\)
⇒ 1 + 2(1)(1) cosθ + 1 = 1 ⇒ cos θ = \(-\frac{1}{2} \Rightarrow \theta=\frac{2 \pi}{3}\)

Question 8.
The value of \(\hat{\mathbf{i}} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})+\hat{\mathbf{j}} \cdot(\hat{\mathbf{i}} \times \hat{\mathbf{k}})+\hat{\mathbf{k}} \cdot(\hat{\mathbf{i}} \times \hat{\mathbf{j}})\) is
1) 0
2) -1
3) 1
4) 3
Solution:
3) 1
\(\hat{\mathrm{i}} \cdot \hat{\mathrm{j}} \times \hat{\mathrm{k}})+\hat{\mathrm{j}} \cdot(\hat{\mathrm{i}} \times \hat{\mathrm{k}})+\hat{\mathrm{k}} .(\hat{\mathrm{i}} \times \hat{\mathrm{j}})=\hat{\mathrm{i}} . \hat{\mathrm{i}}+\hat{\mathrm{j}} .(-\hat{\mathrm{j}})+\hat{\mathrm{k}} . \hat{\mathrm{k}}=1-1+1=1\)

Question 9.
If θ is the angle between any two vectors \(\vec{a} \text { and } \vec{b}\), then \(|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|\) when θ is equal to 10.
1) 0
2) \(\frac{\pi}{4}\)
3) \(\frac{\pi}{2}\)
4) π
Solution:
2) \(\frac{\pi}{4}\)
\(|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \Rightarrow|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta \Rightarrow \cos \theta=\sin \theta \Rightarrow \tan \theta=1 \Rightarrow \theta=\frac{\pi}{4}\)

Question 10.
The value of the dot product of \(\vec{a}-\vec{b} \text { and } \vec{a}+\vec{b}/latex] is
1) a² – b²
2) [latex](\vec{a} \times \vec{b})\)
3) \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\)
4) \(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}\)
Solution:
1) a² – b²
\((\bar{a}-\bar{b}) \cdot(\bar{a}+\bar{b})=\bar{a} \cdot \bar{a}+\bar{a}-\bar{b}-\bar{b} \cdot \bar{a}-\bar{b} \cdot \bar{b}=|\bar{a}|^2-|\bar{b}|^2=a^2-b^2 \text { where }|\bar{a}|=a ;|\bar{b}|=b\)

Question 11.
The position vector of the point (1, 2, 0)is
1) \(\overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}}\)
2) \(\overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{2j}}+\overrightarrow{\mathrm{k}}\)
3) \(\vec{i}+2 \vec{j}\)
4) \(2 \vec{j}+\vec{k}\)
Solution:
3) \(\vec{i}+2 \vec{j}\)
PV of P = (1, 2, 0) is \(\overline{\mathrm{OP}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+0 \overline{\mathrm{k}}\)

Question 12.
If \(|(\vec{a} \times \vec{b})|=4 \text { and }|\vec{a} \cdot \vec{b}|=2\) then \(\left.\overrightarrow{\mathbf{a}}\right|^2|\overrightarrow{\mathbf{b}}|^2\) is equal to
1) 4
2) 2
3) 20
4) 2
Solution:
3) 20
Relation between \(\vec{a} \text { and } \vec{b}\) and \(\bar{a} \cdot \bar{b} \text { is }|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=(\bar{a})^2(\bar{b})^2\)
⇒ (4)² + (2)² = \((\bar{a})^2(\bar{b})^2 \Rightarrow(\bar{a})^2 \cdot(\bar{b})^2\) = 16 + 4 = 20

Question 13.
The points with position vectors \(10 \bar{i}+3 \bar{j}, 12 i-5 \vec{j} \text { and } a \dot{i}+11 j\) are collinear, if a is
1) 2
2) -8
3) 4
4) 8
Solution:
4) 8

Question 14.
The vector cos α cosβ \(\vec{i}\) + cosα sinβ \(\vec{j}\) + sinα \(\vec{k}\) is
1) null vector
2) unit vector
3) constant vector
4) vector with magnitude > 1
Solution:
2) unit vector
Consider \(|(\cos \alpha \cdot \cos \beta) \overline{\mathrm{i}}+(\cos \alpha \cdot \sin \beta) \overline{\mathrm{j}}+(\sin \alpha) \overline{\mathrm{k}}|\)
= \(\sqrt{\cos ^2 \alpha \cos ^2 \beta+\cos ^2 \alpha \sin ^2 \beta+\sin ^2 \alpha}=\sqrt{\cos ^2 \alpha\left(\cos ^2 \beta+\sin ^2 \beta\right)+\sin ^2 \alpha}\)
= \(\sqrt{\cos ^2 \alpha+\sin ^2 \alpha}=\sqrt{1}=1\). Hence a Unit vector.

Question 15.
If \(\vec{a}\), \(\vec{b}\), \(\vec{b}\) are mutually perpendicular unit vectors, then the value of |\(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)| is
1) 1
2) \(\sqrt{2}\)
3) \(\sqrt{3}\)
4) 2
Solution:
3) \(\sqrt{3}\)
Given \(|\bar{a}|=|\bar{b}|=|\bar{c}|=1 \text { and } \bar{a} \cdot \bar{b}=\bar{b}-\bar{c}=\bar{c} \cdot \bar{a}=0\)
⇒ \(|\bar{a}+\bar{b}+\bar{c}|^2=(\bar{a})^2+(\bar{b})^2+(\bar{c})^2+2(\bar{a} \cdot \bar{b}+\bar{b}-\bar{c}+\bar{c}-\bar{a})\) = 1 + 1 + 1 + 0 = 3
⇒ \(|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=\sqrt{3}\)

Question 16.
If \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\), |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 5, |\(\vec{c}\)| = 7 then the angle between \(\vec{a} \text { and } \vec{b}\) is
1) \(\frac{\pi}{6}\)
2) \(\frac{2\pi}{3}\)
3) \(\frac{5\pi}{3}\)
4) \(\frac{\pi}{3}\)
Solution:
4) \(\frac{\pi}{3}\)
\(\bar{a}+\bar{b}+\bar{c}=0 \Rightarrow \bar{a}+\bar{b}=-\bar{c} \quad \Rightarrow|\bar{a}+\bar{b}|=\bar{c}\). Squaring on both sides, we get
⇒ \((\bar{a})^2+(\bar{b})^2+2 \bar{a} \cdot \bar{b}=(\bar{c})^2 \Rightarrow 9+25+2 \bar{a} \cdot \bar{b}=49 \Rightarrow 2 \bar{a} \cdot \bar{b}=15 \Rightarrow 2(\bar{a})(\bar{b}) \cos (\bar{a} \bar{b})=15\)
⇒ 2(3)(5) cos θ = 15 ⇒ 2 cos θ = 1 ⇒ cos θ = \(\frac{\pi}{2}\) = cos 60°
∴ θ = 60° = π/3

Question 17.
If \(\vec{a} \text { and } \vec{b}\) are two unit vectors inclined atan angle θ then the Value of |\(\vec{a}\) – \(\vec{b}\)| is
1) 2sin\(\frac{\theta}{2}\)
2) 2sinθ
3) 2cos\(\frac{\theta}{2}\)
4) 2cosθ
Solution:
1) 2sin\(\frac{\theta}{2}\)
Given \(\bar{a}=|\bar{b}|=1\langle\bar{a}, \bar{b}\rangle\) = θ
consider \(|\bar{a}-\bar{b}|^2=(\bar{a})^2+(\bar{b})^2-2 \bar{a}-\bar{b}=1+1-2(\bar{a})(\bar{b}) \cos \theta\) = 2 – 2 cosθ
= 2(1 – cosθ) = \(2 \sin ^2 \theta / 2 \Rightarrow|\bar{a}-\bar{b}|=\sqrt{4 \sin ^2(\theta / 2)}=2 \sin (\theta / 2)\)

Question 18.
If |\(\vec{a}\)|= 3 and -1 ≤ k ≤ 2 then | k\(\vec{a}\) |lies in the internal
1) [0, 6]
2) [-3, 6]
3) [3, 6]
4) [1, 2]
Solution:
1) [0, 6]
|k\(\vec{a}\)| ⇒ |k||\(\vec{a}\)| ⇒ 3|k|
-1 ≤ k ≤ 2
0 ≤ |k| ≤ 2
0 × 3 ≤ 3 |k| ≤ 2 × 3
0 ≤ 3|k| ≤ 6 ⇒ |k\(\vec{a}\)| ∈ [0, 6]

Practice AP Inter 2nd Year Maths Study Material Chapter 9 Differential Equations MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Differential Equations MCQ

I. Select the correct option from the given choices.

Question 1.
The degree of the differential equation \(\left(\frac{d^2 y}{d x^2}\right)^3+\left(\frac{d y}{d x}\right)^2+\sin \left(\frac{d y}{d x}\right)+1=0\) is
1) 3
2) 2
3) 1
4) not defined
Solution:
4) not defined
Given D.E is not a polynomial equation in its derivatives. Its degree is not defined.

Question 2.
The order of the differential equation \(2 x^2 \frac{d^2 y}{d x^2}-3 \frac{d y}{d x}+y=0\) is
1) 2
2) 1
3) 0
4) not defined
Solution:
1) 2
Highest order derivative present in the given D.E is \(\frac{d^2 y}{d x^2}\). Its order is two.

Question 3.
The number of arbitrary constants in the general solution of a differential equation of fourth order is
1) 0
2) 2
3) 3
4) 4
Solution:
4) 4
Number of constants in the GS= Order
Number of constants in the general solution of D.E of order n is equal to its order.
The number of constants in fourth order differential equation is 4.

Question 4.
The number of arbitrary constants in the particular solution of a differential equation of third order is
1) 3
2) 2
3) 1
4) 0
Solution:
4) 0
In a particular solution of a differential equation, there are no arbitrary constants.

Question 5.
The general solution of the differential equation \(\frac{d y}{d x}=e^{x+y}\) is
1) e^x + e^-y = C
2) e^x + e^y = C
3) e^-x + e^y = C
4) e^-x + e^-x = C
Solution:
1) e^x + e^-y = C
Given D.E is \(\frac{d y}{d x}\) = e^x+y = e^x.e^y ⇒ \(\frac{d y}{e^y}\) = e^x dx ⇒ e^-y dy = e^x dx
x ∫e^-y dy = ∫e^x dx ⇒ -e^-y = e^x + k ⇒ e^x + e^-y = -k ⇒ e^x + e^-y = C

Question 6.
A homogeneous differential equation of the from \(\frac{d x}{d y}=h\left(\frac{x}{y}\right)\) can be solved by making the substitution.
1) y = vx
2) v = yx
3) x = vy
4) x = v
Solution:
3) x = vy
For solving homogeneous equation of form \(\frac{d x}{d y}=h\left(\frac{x}{y}\right)\), we need to make substitution as x = vy
Thus, the correct option is C.

Question 7.
Which of the following is a homogeneous differential equation?
1) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
2) (xy) dx – (x³ + y³) dy = 0
3) (x³ + 2y²) dx + 2xy dy = 0
4) y² dx + (x² – xy – y²) dy = 0
Solution:
4) y² dx + (x² – xy – y²) dy = 0
F(x,y) is homogeneous function of degree n, if F (λx, λy) = λF\(x, y)
Consider D.E in (D) y²dx + (x² – xy² – y²)dy = 0 ⇒ \(\frac{d y}{d x}=\frac{y^2}{y^2+x y^2-x^2}\) F(x, y) = \(\frac{y^2}{y^2+x y^2-x^2}\)
F(λx, λy) = \(\frac{(\lambda y)^2}{(\lambda y)^2+(\lambda x)(\lambda y)^2-(\lambda x)^2}=\frac{\lambda^2 y^2}{\lambda^2\left(y^2+x y^2-x^2\right)}=\lambda^2\left(\frac{y^2}{y^2+x y^2-x^2}\right)\) = λ°F(x, y)
Differential equation given in D is a homogeneous equation

Question 8.
The Integrating Factor of the differential equation \(\frac{d y}{d x}-y=2 x^2\) is
1) e^-x
2) e^-y
3) \(\frac{1}{\mathrm{x}}\)
4) x
Solution:
3) \(\frac{1}{\mathrm{x}}\)
Given D.E is \(x \frac{d y}{d x}-y=2 x^2 \Rightarrow \frac{d y}{d x}-\frac{y}{x}=2 x\) This is in the \(\frac{d y}{d x}+P y=Q\) form
where, P = \(-\frac{1}{x}\) and Q = 2x ∴ IF = \(e^{-\int \frac{1}{x} d x}=e^{-\log x}=e^{\log \left(x^{-1}\right)}=x^{-1}=\frac{1}{x}\)

Question 9.
The Integrating Factor of the D.E (1 – y²)\(\frac{d x}{d y}\) + yx = ay, (-1 < y < 1) is
1) \(\frac{1}{y^2-1}\)
2) \(\frac{1}{\sqrt{y^2-1}}\)
3) \(\frac{1}{1-y^2}\)
4) \(\frac{1}{\sqrt{1-y^2}}\)
Solution:
4) \(\frac{1}{\sqrt{1-y^2}}\)
Given D.E is (1 – y²)\(\frac{d x}{d y}\) + yx = ay ⇒ \(\frac{d x}{d y}+\frac{y x}{1-y^2}=\frac{a y}{1-y^2}\) This is in the \(\frac{d y}{d x}+P y=Q\) form

Question 10.
The general solution of the differential equation \(\frac{y d x-x d y}{y}=0\) is
1) xy = C
2) x = Cy²
3) y = Cx
4) y = Cx²
Solution:
3) y = Cx
Given D.E. is \(\frac{y d x-x d y}{y}=0 \Rightarrow \frac{y d x-x d y}{x y}=0 \Rightarrow \frac{1}{x} d x-\frac{1}{y} d y=0\)
⇒ log |x| = log |y| = log k ⇒ \(\log \left|\frac{x}{y}\right|=\log k \Rightarrow \frac{x}{y}=k \Rightarrow y=\frac{1}{k} x \Rightarrow y=C x\) (where, C = \(\frac{1}{k}\))

Question 11.
The general solution of a D.E of the type \(\frac{d x}{d y}+P_1 x=Q_1\) (P₁, Q₁ are functions of y) is
1) \(y e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)
2) \(y . e^{\int P_1 d x}=\int\left(Q_1 e^{\int P_1 d x}\right) d x+C .\)
3) \(x e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)
4) \(x e^{\int P_1 d x}=\int\left(\mathbf{Q}_1 e^{\int \mathbf{P}_1 d \mathrm{x}}\right) \mathrm{dx}+\mathbf{C}\)
Solution:
3) \(x e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)
IF for \(\frac{d x}{d y}+P_1 x=Q_1^{\prime} \text { is } e^{\int P_1 d y} \Rightarrow x(\text { I.F. })=\left(\int Q_1 \times \text { IF }\right) d y+C \Rightarrow x . e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C\)

Question 12.
The general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0 is
1) x e^y + x² = C
2) x e^y + y² = C
3) y e^x + x² = C
4) y e^y + x² = C
Solution:
3) y e^x + x² = C
Given D.E is e^x dy + (ye^x + 2x)dx = 0 ⇒ e^x\(\frac{d y}{d x}\) + ye^x + 2x = 0 ⇒ \(\frac{d y}{d x}\) + y = \(\frac{2 x}{e^x}\) = 0
⇒ \(\frac{d y}{d x}\) + y = 2xe^-x = 0 ⇒ \(\frac{d y}{d x}\) + y = -2xe^-x
This is a Linear D.E form \(\frac{d y}{d x}\) + Py = Q where, P = I and Q = -2xe^-x
Now, IF = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \mathrm{dx}}=\mathrm{e}^{\mathrm{x}} \Rightarrow \overline{\mathrm{y}}(\mathrm{IF})=\int(\mathrm{Q} \times \mathrm{IF}) \mathrm{dx}+\mathrm{C}\)
∴ ye^x = \(\int\left(-2 x e^{-x} \cdot e^x\right) d x+C \Rightarrow y e^x=-\int 2 x d x+C\) ⇒ ye^x = -x² + C ⇒ ye^x + x² = C

Question 13.
General solution of the differential equation \(\log \left(\frac{d y}{d x}\right)\) = 2x + y is
1) \(e^{-y}=\frac{1}{2} e^{2 x}+C\)
2) \(\frac{1}{e^y}+\frac{1}{2} e^{2 x}=C\)
3) \(-e^{-y}=\frac{1}{2} e^{2 x}+C\)
4) \(e^y=\frac{1}{2} e^{2 x}+C\)
Solution:
3) \(-e^{-y}=\frac{1}{2} e^{2 x}+C\)
\(\log _{\mathrm{e}}\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]\) = 2x + y \(\frac{d y}{d x}\) = e^2x+y ⇒ \(\frac{d y}{d x}\) = e^2x.e^y \(\frac{1}{e^y}\)dy = e^2x dx
Integrating \(\int e^{-y} d y=\int e^{2 x} d x \Rightarrow-e^{-y}=\frac{e^{2 x}}{2}+c\)

Question 14.
General solution of differential equation \(\frac{d y}{d x}=\frac{y}{x}\) is
1) log y = Cx
2) y = Cx
3) xy = C
4) y = C log x
Solution:
2) y = Cx
\(\frac{d y}{d x}=\frac{y}{x} \Rightarrow \frac{d y}{y}=\frac{d x}{x} \Rightarrow \int \frac{1}{y} d y=\int \frac{1}{x} d x\) ⇒ log |y| = log |x| + log |c| ⇒ log |y| = log |cx| ⇒ y = cx.

Question 15.
The degree of the differential equation \(\left(1+\frac{d y}{d x}\right)^3=\left(\frac{d y}{d x}\right)^2\) is
1) 1
2) 2
3) 3
4) 4
Solution:
3) 3
order = 1; degree = 3

Question 16.
The degree of the differential equation \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\) is
1) 1
2) 2
3) 4
4) not defined
Solution:
4) not defined
The given equation is not a polynomial equation in \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\).
Here, its degree is not defined. Hence, degree not defined

Question 17.
Order of differential equation corresponding to family of curves y = Ae^2x + Be^-2x is
1) 2
2) 1
3) 3
4) 4
Solution:
1) 2
y = Ae^2x + Be^-2x arbitary constants = 2
∴ Order of D.E is ‘2’

Question 18.
The general solution of differential equation \(\frac{d y}{d x}=e^{x-y}\) is
1) e^y = e^x + C
2) e^x + e^y = C
3) e^x+y = C
4) e^x-y = C
Solution:
1) e^y = e^x + C
\(\frac{d y}{d x}=e^x \cdot e^{-y} \Rightarrow \frac{1}{e^{-y}} d y=e^x d x \Rightarrow e^y d y=e^x d x \Rightarrow \int e^y d y=\int e^x d x \Rightarrow e^y=e^x+c\)

Question 19.
The order and degree of the differential equation \(\frac{d y}{d x}=\left(\frac{d^2 y}{d x^2}+2\right)^{1 / 2}+\frac{d^2 y}{d x^2}+5\) are respectively
1) 2, 1
2) 2, 4
3) 2, 2
4) 2, 3
Solution:
3) 2, 2
Transposing the terms properly and squaring on both sides we get \(\left[\left(\frac{d y}{d x}\right)-\left(\frac{d^2 y}{d x^2}\right)-5\right]^2=\frac{d^2 y}{d x^2}+2\)
∴ order = 2 ; degree = 2

Question 20.
The differential equation for which ax + by = 1 is general solution (a, b are arbitrary constants) is
1) \(\frac{d y}{d x}=x+C\)
2) \(y \frac{d^2 y}{d x^2}+x=1\)
3) \(\frac{d^2 y}{d x^2}=0\)
4) \(\frac{d^3 y}{d x^3}=0\)
Solution:
3) \(\frac{d^2 y}{d x^2}=0\)
Given ax + by = 1 ⇒ a(1) + b\(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\) = 0 Again diff w.r.t ‘x’, 0 + b\(\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right)=0 \Rightarrow \frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{dx}^2}=0\)

Practice AP Inter 2nd Year Maths Study Material Chapter 8 Application of Integrals MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Application of Integrals MCQ

I. Select the correct option from the given choices.

Question 1.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
1) π
2) \(\frac{\pi}{2}\)
3) \(\frac{\pi}{3}\)
4) \(\frac{\pi}{4}\)
Solution:
1) π

Area(OAB) = \(\int_0^2 \mathrm{ydx}=\int_0^2 \sqrt{4-\mathrm{x}^2} \mathrm{dx}=\left[\frac{\mathrm{x}}{2} \sqrt{4-\mathrm{x}^2}+\frac{4}{2} \sin ^{-1} \frac{\mathrm{x}}{2}\right]_0^2=2\left(\frac{\pi}{2}\right)\) = π sq. units

Question 2.
Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is
1) 2
2) \(\frac{9}{4}\)
3) \(\frac{9}{3}\)
4) \(\frac{9}{2}\)
Solution:
2) \(\frac{9}{4}\)

Area (OAM) = \(\int_0^3 x d y=\int_0^3 \frac{y^2}{4} d y=\frac{1}{4}\left[\frac{y^3}{3}\right]_0^3=\frac{1}{12}(27)=\frac{9}{4} \text { sq.units }\)

Question 3.
Area bounded by the curve y = x³, the x-axis and the ordinates x = – 2 and x = 1 is
1) -9
2) \(\frac{-15}{4}\)
3) \(\frac{15}{4}\)
4) \(\frac{17}{4}\)
Solution:
4) \(\frac{17}{4}\)
Required area = \(-\int_{-2}^0 y d x+\int_0^1 y d x\)
= \(-\int_{-2}^0 x^3 d x+\int_0^1 x^3 d x=-\left[\frac{x^4}{4}\right]_{-2}^0+\left[\frac{x^4}{4}\right]_0^1=-\left[0-\frac{(-2)^4}{4}\right]+\left[\frac{1}{4}-0\right]=\left(4+\frac{1}{4}\right)=\frac{17}{4} \text { sq.units }\)

Question 4.
The area bounded by the curve y = x |x| , x-axis and the ordinates x = – 1 and x = 1 is given by [Hint: y = x² if x > 0 and y = -x² if x < 0|
1) 0
2) \(\frac{1}{3}\)
3) \(\frac{2}{3}\)
4) \(\frac{4}{3}\)
Solution:
3) \(\frac{2}{3}\)
Required area = \(\int_{-1}^1 y d x=\int_{-1}^1 x|x| d x=-\int_{-1}^0 x^2 d x+\int_0^1 x^2 d x\)
= \(\left[\frac{x^3}{3}\right]_{-1}^0+\left[\frac{x^3}{3}\right]_0^1=-\left(-\frac{1}{3}\right)+\frac{1}{3}=\frac{2}{3} \text { sq.units }\)

Question 5.
Area under the curve y = \(\sqrt{a^2-x^2}\) included between the lines x = 0 and x = a is
1) \(\frac{\pi \mathrm{a}^2}{2}\)
2) \(\frac{\pi \mathrm{a}^2}{4}\)
3) \(\frac{\pi \mathrm{a}}{2}\)
4) \(\frac{\pi \mathrm{a}}{4}\)
Solution:
1) \(\frac{\pi \mathrm{a}^2}{2}\)
Area \(\int_0^a \sqrt{a^2-x^2} d x\) = Area of the circle x² + y² = a² in 1st quadrant = \(\frac{1}{4}\)(πa²)

Question 6.
The area bounded by y = sin2x the x – axis and the lines x = \(\frac{\pi}{2}\) and x = \(\frac{3\pi}{4}\) is
1) 1sq units
2) 2sq. units
3) 4sq. units
4) \(\frac{3}{2}\)sq. units
Solution:
1) 1sq units
y = sin2x ⇒ y > 0 if x < 2x < π; i.e., 0 < x <\(\frac{\pi}{2}\) and y < 0 if π < 2x < 2π; i.e., \(\frac{\pi}{2}\) < x < π
A = \(\int_{\pi / 4}^{3 \pi / 4} \sin (2 x) d x=\int_{\pi / 4}^{\pi / 2} \sin (2 x) d x-\int_{\pi / 2}^{3 \pi / 4} \sin (2 x) d x=-\left[\frac{\cos (2 x)}{2}\right]_{\pi / 4}^{\pi / 2}-\left[-\frac{\cos (2 x)}{2}\right]_{\pi / 2}^{3 \pi / 4}\)
= \(-\frac{1}{2}\left[\cos \pi-\cos \frac{\pi}{2}\right]+\frac{1}{2}\left[\cos \left(\frac{3 \pi}{2}\right)-\cos \pi\right]=-\frac{1}{2}[-1-0]+\frac{1}{2}[0-(-11)]=\frac{1}{2}+\frac{1}{2}(1)=1 \text { sq.units }\)

Question 7.
The area bounded by the curve y = x² – 4, and the lines y = 0 and y = 5 is
1) \(\frac{38}{3}\)
2) \(\frac{76}{3}\)
3) \(\frac{16}{3}\)
4) \(\frac{8}{3}\)
Solution:
2) \(\frac{76}{3}\)
Given y = x² – 4 ⇒ x² = y + 4 ⇒ x = \(\sqrt{y+4}\)
Required Area A = \(2\left[\int_0^5 \mathrm{xdx}\right]=2\left[\int_0^5 \sqrt{\mathrm{y}+4} \mathrm{dy}\right]=2\left[\frac{2}{3}(\mathrm{y}+4) \sqrt{\mathrm{y}+4}\right]_0^5\)
= \(\frac{4}{3}[9 \sqrt{9}-(4 \sqrt{4})]=\frac{4}{3}[27-8]=\frac{4 \times 19}{3}=\frac{76}{3} \text { sq.units }\)

Question 8.
The area of the region bounded by parabola y² = 8x and latus rectum is
1) \(\frac{4}{3}\)
2) \(\frac{16}{3}\)
3) \(\frac{32}{3}\)
4) \(\frac{8}{3}\)
Solution:
3) \(\frac{32}{3}\)
y² = 8x y = \(\sqrt{8 x}=2 \sqrt{2 x}\)
Area = \(2 \int_0^2(y) d x=2\left[\int_0^2 2 \times 2 \sqrt{x} d x\right]=2 \times 2 \sqrt{2}\left(\frac{2}{3} x \sqrt{x}\right)_0^2=\frac{8 \sqrt{2}}{3}(2 \sqrt{2}-0)=\frac{16 \times 2}{3}=\frac{32}{3} \text { sq.units }\)

Question 9.
The area bounded by the curve y = 2x – x² and the line y = -x is
1) \(\frac{7}{2}\)
2) 7
3) \(\frac{9}{2}\)
4) 9
Solution:
3) \(\frac{9}{2}\)
Given y = 2x – x² ………..(1) (Upper curve) y = -x …….(2) (Lower curve)
Solving (1) and (2)
-x = 2x – x² ⇒ x² – x – 2x = 0 ⇒ x² – 3x = 0 ⇒ x(x – 3) = 0 ⇒ x = 0 x = 3
Area = \(\int_0^3\left(2 x-x^2\right)-(-x) d x=\int_0^3\left(3 x-x^2\right) d x=\left(3 \frac{x^2}{2}-\frac{x^3}{3}\right)_0^3\)
= \(\frac{3}{2} \times 9-\frac{27}{3}-(0)=\frac{27}{2}-\frac{27}{3}=27\left(\frac{1}{6}\right)=\frac{9}{2} \text { sq.units }\)

Question 10.
The area enclosed between the graph of y = x³ and the lines x = 0, y = 1, y = 8 is
1) 7
2) 14
3) \(\frac{45}{4}\)
4) \(\frac{54}{4}\)
Solution:
3) \(\frac{45}{4}\)
y = x³
x = 0 (y-axis), y = 1 , y =8
A = \(\int_1^8(x) d x=\int_1^8 y^{\frac{1}{3}} d x=\left(\frac{y^{\frac{1}{3}}+1}{\frac{1}{3}+1}\right)_1^8=\frac{3}{4}\left(y^{\frac{4}{3}}\right)_1^8=\frac{3}{4}\left[2^4-1\right]=\frac{3 \times 15}{4}=\frac{45}{4} \text { sq.units }\)

Question 11.
The area of the region bounded by the curve y² = x, the Y-axis and between y = 2 and y = 12.
1) \(\frac{52}{2}\)
2) \(\frac{54}{3}\)
3) \(\frac{56}{3}\)
4) \(\frac{58}{3}\)
Solution:
3) \(\frac{56}{3}\)
y² = x; y-axis(x = 0), y = 2, y = 4
Area = \(\int_2^4(x) d x=\int_2^4 y^2 d x=\left[\frac{y^3}{3}\right]_2^4=\frac{1}{3}[64-8]=\frac{1}{3}[56]=\frac{56}{3} \text { sq.units }\)

Question 12.
Area of the region bounded by the curve y = cos x between x – 0 and x = π and the X-axis is
1) 1
2) 2
3) 3
4) 4
Solution:
2) 2
y = cos x, x = 0 (y-axis), x = π, x-axis (y = 0)
Required Area = \(2 \int_0^{\pi / 2}(y) d x=2 \int_0^{\pi / 2} \cos x d x=2[\sin x]_0^{\pi / 2}=2\left[\sin 90^{\circ}-\sin 0^{\circ}\right]=2[1-0]=2\)

Question 13.
Area of the region bounded by the curve x = 2y + 3, the Y-axis and between y = -1 and y = 1 is
1) 6
2) 4
3) 8
4) 3/2
Solution:
1) 6
Given x = 2y + 3
Area A = \(\int_{-1}^1(x) d y=\int_{-1}^1(2 y+3) d y=\left(\frac{2 y^2}{2}+3 y\right)_{-1}^1\) = 1 + 3 – [1 – 3] = 4 – (-2) = 6 sq. units

Question 14.
The area bounded by the curve y = x³, X-axis and two ordinates x = 1 and x = 2 is
1) \(\frac{15}{2}\)
2) \(\frac{15}{4}\)
3) \(\frac{17}{2}\)
4) \(\frac{17}{4}\)
Solution:
2) \(\frac{15}{4}\)
y = x³ x – axis (y = 0) x = 1, x = 2
Area = \(\int_1^2(y) d x=\int_1^2 x^3 d x=\left(\frac{x^4}{4}\right)_1^2=\frac{16}{4}-\frac{1}{4}=\frac{15}{4} \text { sq.units }\)

Question 15.
The area bounded by the curves y² = 4x and y = x is equal to
1) \(\frac{1}{3}\)
2) \(\frac{8}{3}\)
3) \(\frac{35}{6}\)
4) \(\frac{7}{3}\)
Solution:
2) \(\frac{8}{3}\)
y² = 4x ⇒ y = 2\(\sqrt{x}\) …(1) (Upper curve) y = x ……(2) (Lower curve)
Solving (1) and (2) y² = 4y y(y – 4) = 0 y = 0; y = 4
Area = \(\int_1^4(2 \sqrt{x}-x) d x=2 \int_1^4 \sqrt{x} d x=\int_1^4 x d x=2 \frac{2}{3}(x \sqrt{x})_0^4-\left(\frac{x^4}{2}\right)_0^4=\frac{4}{3}[4 \sqrt{4}]-\frac{1}{2}\)
= \(\frac{32}{3}-\frac{16}{2}=\frac{32}{3}-8=\frac{32-24}{3}=\frac{8}{3} \text { sq.units }\)

Practice AP Inter 2nd Year Maths Study Material Chapter 6 Application of Derivatives MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Application of Derivatives MCQ

I. Select the correct option from the given choices.

Question 1.
The rate of change of the area of a circle with respect to its radius r at r = 6 cm
1) 10π
2) 12π
3) 8π
4) 11π
Solution:
2) 12π
Area of a circle A = πr²; Diff w.r.t ‘r’
Rate of change of Area = \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = ²(2r); Now, = \(\frac{\mathrm{dA}}{(\mathrm{dr})}\) = ²(2)(6) = 12² at r = 6

Question 2.
The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is
1)116
2) 96
3) 90
4) 126
Solution:
4) 126
Revenue = R(x) = 3x² + 36x + 5; Marginal Revenue = \(\frac{\mathrm{dR}}{\mathrm{dx}}\) = 3(2x) + 36
\(\begin{aligned}
&\frac{\mathrm{dR}}{\mathrm{dx}}\\
&\text { at } x=15
\end{aligned}\) = 6(15) + 36 = 90 + 36 = 126

Question 3.
Which of the following functions are decreasing on (0, \(\frac{\pi}{2}\))?
1) cos x
2) cos2x
3) cos3x
4) tanx
Solution:
1) cos x
Let f(x) = cosx. For decreasing interval f'(x) < 0 ⇒ -sin x < 0 ⇒ sin x > 0 ∀ x ∈ (0, \(\frac{\pi}{2}\))

Question 4.
On which of the following intervals is the function f given by f (x) = x¹⁰⁰ + sin x – 1 is decreasing ?
1) (0, 1)
2) (\(\frac{\pi}{2}\), π)
3) (0, \(\frac{\pi}{2}\))
4) (-π, \(\frac{\pi}{2}\))
Solution:
4) (-π, \(\frac{\pi}{2}\))
Give f(x) = x¹⁰⁰ + sinx – 1 ⇒ f’ (x) = 100x⁹⁹ + cos x. For decreasing interval f'(x) < 0
check option
1) In (0, 1) = (0, radian) = (0,57°) f'(x) = 100x⁹⁹ + cos x > 0 (+ve)
2) In (\(\frac{\pi}{2}\), π), f'(x) = 100x⁹⁹ + cosx – a large+ve value + ve (∵ -1 ≥ cos + ve)
3) In (0, \(\frac{\pi}{2}\)), f'(x) = +ve + +ve (+ve);
4) In (-π, \(\frac{-\pi}{2}\)), f'(x) = -ve- = -ve < 0

Question 5.
In which interval y = x² e^-x is increases
1) (-∞, ∞)
2) (-2, 0)
3) (2, ∞)
4) (0, 2)
Solution:
4) (0, 2)
f(x) = x².e^-x ⇒ f'(x) = x²(-e^=x) + e^-x(2x)
For increasing interval f'(x) > 0 ⇒ e^-x(2x – x²) > 0
⇒ 2x – x² > 0 ∵ e^-x > 0 ∀x ∈ R ⇒ x² – 2x < 0 ⇒ x(x – 2) < 0 ⇒ x ∈ (0, 2)

Question 6.
On the curve x² = 2y which is nearest to the pojnt (0, 5) is
1) (\(2 \sqrt{2}\), 4)
2) (\(2 \sqrt{2}\), 0)
3) (0, 0)
4) (2, 2)
Solution:
1) (\(2 \sqrt{2}\), 4)
Let P(t, \(\frac{t^2}{2}\)) is a point on x² = 2y and A = (0, 5)
consider PA² = (t – 0)² + (\(\frac{t^2}{2}\) – 5)² …..(1) ⇒ PA² = f(x) = t² + (\(\frac{t^2}{2}\) – 5)²
For maxima (or) minimum f'(x) = 0 ⇒ 2t + 2(\(\frac{t^2}{2}\) – 5)\(\left[\frac{2 \mathrm{t}}{2}\right]\) = 0 ⇒ 2t + (t² – 10)t = 0
⇒ 2t + t³ – 10t = 0 ⇒ t³ – 8f = 0 ⇒ f(t² – 8) = 0 ⇒ t = 0 (or) t = \(\sqrt{8}=2 \sqrt{2}\)
From (1) at t = 0 ⇒ PA² = 0 + (-5)² = 25; at t = \(\sqrt{8}\) ⇒ PA² = 8 + (-1)² = 9 minimum
∴ at t = \(\sqrt{8}\) ⇒ P = (\(\sqrt{8}\), 4) = (2\(\sqrt{2}\), 4) is nearest

Question 7.
For all real values of x, the minimum value of \(\frac{1-x+x^2}{1+x+x^2}\) is
1) 0
2) 1
3) 3
4) 1/3
Solution:
4) 1/3
f(x) = \(\frac{1-x+x^2}{1+x+x^2} \Rightarrow f^{\prime}(x)=\frac{\left(1+x+x^2\right)(-1+2 x)-\left(1-x+x^2\right)(1+2 x)}{\left(1+x+x^2\right)^2}=\frac{2\left(x^2-1\right)}{\left(1+x+x^2\right)^2}\)
∴ f'(x) = 0 ⇒ 2(x² – 1) = 0 ⇒ x² = 1 ⇒ x = ±1
By second derivative test, f is the minimum at x = 1 and f(1) = \(\frac{1-1+1}{1+1+1}=\frac{1}{3}\)

Question 8.
The maximum value of |x(x – 1) + 1|^{\(\frac{1}{3}\)}, 0 ≤ x ≤ 1 is
1) \(\left(\frac{1}{3}\right)^{\frac{1}{3}}\)
2) \(\frac{1}{2}\)
3) 1
4) 0
Solution:
3) 1
y = f(x) = \([x(x-1)+1]^{\frac{1}{3}}=\left(x^2-x+1\right)^{\frac{1}{3}}=\left(\left(x-\frac{1}{2}\right)+\frac{3}{4}\right)^{\frac{1}{3}}\)
Since, extreme values (maximum (or) minimum) occurs at critical points (or) at the end of the interval. Solving, f'(x) = 0 we get x = \(\frac{1}{2}\) (critical calue)
∴ f_max = Max of {(f(0), f(1), f\(\left(\frac{1}{2}\right)\)} = Max of {1, 1, \(\left(\frac{3}{4}\right), \frac{1}{3}\)} ⇒ f_max = 1

Question 9.
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
1) 1 m/h
2) 0.1 m/h
3) 1.1 m/h
4) 0.5 m/h
Solution:
1) 1 m/h
Given r = 10 = radius, \(\frac{d v}{d t}\) = 314, h = depth, \(\frac{d h}{d t}\) = ?
Volume = V = πr²h ⇒ V = π(100)h ⇒ V = (3.14)100h ⇒ V = (314)h
Diff w.r.t ‘f’ \(\frac{d v}{d t}\) = (314)\(\frac{d h}{d t}\) ⇒ (314) = (314)\(\frac{d h}{d t}\) ⇒ \(\frac{d h}{d t}\) = 1 ∴ \(\frac{d h}{d t}\) = 1 m/h

Question 10.
The function f(x) = x³ + 3x is increasing in interval
1) (-∞, 0)
2) (0, ∞)
3) R
4) (0, 1)
Solution:
3) R
f(x) = x³ + 3x
For increasing interval f'(x) > 0 ⇒ 3x² + 3 > 0 ∀x ∈ R

Question 11.
The interval in which the function f(x) = 2x³ + 9x² + 12x – 1 is decreasing
1) (-1, ∞)
2) (-2, -1)
3) (-0, -2)
4) (-1, 1)
Solution:
2) (-2, -1)
f(x) = 2x³ + 9x² + 12x – 1
For decreasing interval f'(x) < 0 ⇒ 2(3x²) + 9(2x) + 12 < 0
⇒ x² + 3x + 2 < 0 ⇒ (x + 1)(x + 2) < 0 x ∈ (-2, -1)

Question 12.
At which point the function f(x) = |x – 3| attains minimum value
1) x = 1
2) x < 3 3) x = 3 4) x > 3
Solution:
3) x = 3

y = f(x) = |x – 3| graph
clearly f(x) is maximum at x = 3

Question 13.
Minimum value of the function f(x) = |x – 2| + |x – 5| is
1) 1
2) 2
3) 3
4) 4
Solution:
3) 3
f(x) = |x – 2| + |x – 5| = |x – a| + |x – b|
Range of f(x) is [|a – b|, ∞) ⇒ f_Minimum = |a – b|
f_min = |2 – 5| = |3| = 3

Question 14.
The maximum value of is \(\frac{\log x}{x}\) is 0 < x < ∞ is
1) ∞
2) e
3) 1
4) e^-1
Solution:
4) e^-1
f(x) = \(\frac{\log x}{x} \Rightarrow f^{\prime}(x)=\frac{x\left(\frac{1}{x}\right)-\log x(1)}{x^2}=\frac{1-\log x}{x^2}\)
For maxima (or) Minima f'(x) = 0 ⇒ 1 – log x = 0 ⇒ log_e x = 1 ⇒ x = e
f_max at x = e = \(\frac{\log _{\mathrm{e}}}{\mathrm{e}}=\frac{1}{\mathrm{e}}=\mathrm{e}^{-1}\)

Question 15.
The minimum value of (x – α) (x – β) is
1) 0
2) αβ
3) \(\frac{1}{4}(\alpha-\beta)^2\)
4) \(\frac{-1}{4}(\alpha-\beta)^2\)
Solution:
4) \(\frac{-1}{4}(\alpha-\beta)^2\)
f(x) = (x – α)(x – β) = x²(α + β)x + αβ = ax² + bx + c
⇒ A = 1 (+ve)
⇒ f_min = \(\frac{4 a c-b^2}{4 a}=\frac{4(1)(\alpha \beta)-(\alpha+\beta)^2}{4}=\frac{-\left[(\alpha+\beta)^2+4 \alpha \beta\right]}{4}=-\left(\frac{(\alpha-\beta)^2}{4}\right) .\)