Plus Two

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Kerala Plus Two English Previous Year Question Paper March 2019

Board	SCERT
Class	Plus Two
Subject	English
Category	Plus Two Previous Year Question Papers

Time: 2 1/2 Hours
Cool off time: 15 Minutes
Maximum: 80 Score

General Instructions to Candidates:

• There is a ‘cool off-time’ of 15 minutes in addition to the writing time of 2^1/2 hours.
• You are not allowed to write your answers or to discuss anything with others during the cool off time’.
• Read questions carefully before answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Electronic devices except non-programmable calculators are not allowed in the Examina­tion Hall.

(Question Nos. 1 – 6): Answer the questions as directed. (14)

Read the excerpt from ‘When, a Sapling is Planted’ and answer the questions that follow: ‘Later, they became aware of the widespread destruction of the ecosystems, especially through deforestation, Climatic instability and contamination of the soil and waters – all contributed to excruciating poverty and subsequent riots.’

Question 1.
Who does the speaker refer to as ‘they’? (1)
Answer:
The women of Kenya/African women/People of Kenya/Women

Question 2.
What was the main reason for the destruction of the ecosystem? (1)
Answer:
Deforestation/cutting down of trees/destruction of forests

Question 3.
Which word from the options given, can best replace the word ‘excruciating’? (1)
a) Entertaining
b) Progressing
c) Agonising
Answer:
c) Agonising

Question 4.
Here is a diary entry by Sudha Murty.
Fill in the blanks with appropriate similes or adjectives from the options given in brackets. (3)
Village days were memorable. Days were full of fun and frolic and we were all as ……. (a) …….. (dull/cheerful/sharp) as a lark. How I miss those days! In Mumbai, everyone is as busy as a ……… (b) ………. (snail/sloth/bee) and women too, are no different. Girls who were regarded to be as …….. (c) ……… (sour/chill/sweet) as a rose have now transformed themselves into leaders, office and torchbearers.
Answer:
a) cheerful; b) bee; c) sweet

Question 5.
Imagine that you are invited for an interview for the post of the chef at Taj Group of Hotels. Complete the interview suitably. (4)
Interviewer: ……. a ………?
You: I have specialized in Chinese and Continental dishes.
Interviewer: ……. (b) …….?
You: Yes, of course. I am comfortable with all types of delicacies.
Interviewer: ……. (c) …….?
You: I expect a comfortable work atmosphere, fantastic team work, and opportunities to develop.
Interviewer: Where do you see yourselves ten years from now?
You: Ten years from now, I ……. (d) …….
Answer:
a) What have you specialized in?/What are your areas of specialization?
b) Are you comfortable with delicacies?/Do you know how to make delicacies?
c) What are your expectations from this Organization?/ What do you expect from us?
d) I would start a hotel of my own./I may become the Chief Chef here. I would be working in a prestigious hotel abroad.

Question 6.
The passage given below contains a few errors. (4)
Edit the passage appropriately. The main conflict in the story ‘Amigo Brothers’ is that off ambition. Both friends want to be champions. But only one of them can be the champ. So they decide in fight out it. In the end, they realize that they value their bonding more than their ambition.
Answer:
off – of; wants – want; fight out it – fight it out; there bonding – their bonding

(Question Nos. 7 – 12): Answer any 5 questions in not less than 60 words. Each carries 4 scores. (5 × 4 = 20)

Question 7.
Your class is conducting a debate on the topic ‘PUNISHMENTS ARE ESSENTIAL FOR STUDENT DISCIPLINE’.
Write four arguments either favouring or opposing the topic.
Answer:
Punishments are essential for student discipline.
Arguments for:

1. A proverb says, “Spare the rod, spoil the child”.
2. Students can be easily corrected with appropriate punishments.
3. If punishments are not given, students will tend to ignore their studies.
4. Punishments will force the students to be punctual, respectful, obedient and hard-working.

Question 8.
Elaborate the idea conveyed in the lines given below:
‘I am the pillars of the house;
The keystone of the arch am I
Take me away, and roof and wall
Would fall to ruin me utterly.’
Answer:
The main idea in these lines is the importance of a woman in the house. She may be a mother or wife. She says she is the pillars that keep the house in place. She is the keystone of the arch and if she is removed from the house, it will completely collapse. The poem stresses the importance of women for the house and its proper functioning.

Question 9.
The activities of the literary club of your school will be inaugurated by the cine artist and state award winner, Sri. Indrans. The one-act play, ‘Post Early for Christmas’ will be staged on the day. Prepare a notice inviting all for the programme.
Answer:

GOVERNMENT HIGHER SECONDARY SCHOOL,
IRINJALAKUDA
LITERARY CLUB

Date 1 July 2019

NOTICE

The activities of our Literary Club will be inaugurated by the famous Cine Artiste and State Award Winner Sri Indrans. The inaugural meeting will take place at 10.00 a.m. in the Hall of the School. After the inaugural meeting, the one-act play titled “Post Early for Christmas” by R.H. Wood will be staged by our Literary Club Members.

“Post Early for Christmas” is a hilarious comedy in which you find different kinds of people coming to the Post Office and the havoc created by a parcel that was ticking. Some people think it is a bomb. There are many frantic activities going on. See what happens on stage!

All of you are invited!

Eva Saifu
Secretary

Question 10.
You have come to know that your close friend is having a severe toothache. Frame four sets of suggestions you would make before him/her.
You may use expressions like:

I think you should ……/ If I were you, I ……/ Would you like ……/ Perhaps you could ……/ I’d like to suggest ……
Answer:
I think you should immediately keep some salty, warm water in your mouth as it will reduce your pain. If I were you, I would then go to the dentist. Would you like to have a warm drink or take some pain killers? Perhaps you could take a Panadol to suppress your pain for the time being till you meet the doctor.

Question 11.
Prepare a profile of Shaheen Mistri using the hints given below:
Birth: 16th March 1971
Place of birth: Mumbai
Academic Qualifications: B.A. degree (Sociology), M.A. in Sociology
Almamater: the University of Mumbai, University of Manchester
Occupation: CEO, Teach for India
Famous as: Indian social activist and educator
Founder: Akansha Foundation, Teach for India
Awards won: Ashoka Fellow (2001), Global Leader for tomorrow (2002), Asia Society 21 Leader (2006)
Authored: Redrawing India – The Teach for India Story (2014)
Answer:

PROFILE OF SHAHEEN MISTRI

Shaheen Mistri was born in Mumbai on 16 March 1971. After her +2, she joined the University of Mumbai and got her B.A. in Sociology. Later she went to England and passed her M.A. in Sociology with a First Class, from the University of Manchester. Currently, she is the CEO of the Organization called “Teach India” which is doing a laudable job in educating slum children.

She is famous as a social activist and educator. She is the Founder of two famous Organizations named “Akanksha” and “Teach India”. She has received numerous awards which include Ashoka Fellow (2001), Global Leader for Tomorrow (2002), and Asia Society 21 Leader (2006). She has authored a well-known Book – “Redrawing India – The Teach for India Story” which was published in 2014.

Question 12.
Wangari Maathai says, ‘Women are often the first to become aware of environmental damage ………..’
Do you agree with her opinion? Substantiate your answer.
Answer:
I quite agree with the opinion of Wangari Maathai that “Women are often the first to become aware of environmental damage”. Wangari Maathai is a Kenyan. She is an environmentalist and political activist. She got the Nobel Prize in 2004 for her contribution to sustainable development, democracy, and peace. She made the remark quoted above in her Nobel Prize Acceptance speech.

Being a Kenyan she was especially speaking of the women of Africa. She spoke from her childhood experiences and observations of nature in rural Kenya. As she was growing up, she noticed that forests were cleared and were replaced by commercial plantations. This destroys the local biodiversity and the ability of the forests to conserve water.

Because of large scale deforestation, the African women lacked firewood, clean drinking water, balanced diets, shelter, and income. In Africa, women are the primary caretakers. They till the land and feed their families. As a result, they are the first to notice the environmental damage as resources become scarce, making it difficult for them to maintain their families.

(Question Nos. 13 – 19): Answer any 5 questions in about 80 words. Each carries 6 scores. (5 × 6 = 30)

Question 13.
Read the following lines from the poem ‘Rice’.
‘I’ll reach home in good time, at last
just as my mother drains the well-cooked rice.’
This is better money – what good times!’
How is the expression ‘good times’ used in the above two contexts? Substantiate your answer.
Answer:
The expression “good time/s” is used with two different meanings in the poem. I’ll reach home in good time means I will reach home soon or promptly or without much delay. The son, who is in North India, studying 15. for his doctorate, is tired of eating chappatis all the time and now he is dreaming of eating the well-cooked rice which his mother has drained. So he wants to reach home as quickly as he can. He hopes he will reach home in good time, that is very soon, to eat the rice. But in the second use ‘what good times’ the phrase good times refers to prosperity, the time when one has plenty of money. People of his village have abandoned rice cultivation and have started cultivating cash crops for good times and better money.

Question 14.
You have dreams of setting up a business of your own after your studies. You are greatly inspired by Irfan Alam’s views and ideas. Draft a letter of inquiry to Sammaan foundation asking for clarification of the doubts regarding the financial investment of the organization, mode of operation, the security of the members, etc.
Answer:
Jerry John
16/IV, Azad Road
Irinjalakuda North P.O.
Kerala
PIN 680125
3 July 2019

Mr. Irfan Alam
CEO, SammaaN
Kalina
Mumbai-21

Dear Sir,
Sub : Information about SammaaN

First of all let me congratulate you for founding such a useful organization like SammaaN which is helping thousands of auto-rickshaw workers to live better lives without getting into any kind of debt traps.

Being inspired by your model, I also want to embark upon some such entrepreneurship after my +2 studies. The financial condition of my family does not permit me to go for higher education as my father is sick and my mother does not go for any outside work.

I want to get some matters clarified by you as I also want to bean entrepreneur like you. What was your initial investment in the Organization? Did the banks or any government agency advance you some funds to start the operations?

I also would want to know the modes of your operation. Do you give financial help to the members who join you? What are the security measures you have in place if a member runs into a loss or is disabled due to some disease or accident?

Please find some time to send a reply to my letter. You can contact me on the phone. My phone number is 282 228450. My email address is jerry@amail.com.
Thank you in advance,

Yours sincerely,
Jerry John.

Question 15.
After the competition at Tompkin’s Square Park, the Amigo Brothers are interviewed.
Reporter: Has the bout affected your friendship?
Felix: No, we both never take fighting into our hearts.
Reporter: So then, who among you must have won the fight?
Antonio: We both are always winners. The question itself is unimportant.
How would you report the above conversation?
Answer:
The reporter asked the Amigo brothers if/whether the bout had affected their friendship. Felix replied in the negative saying that they both never took fighting into their hearts. Then the reporter asked them who among them must have won the fight. Antonio replied saying that they both were always winners and that the question itself was unimportant.

Question 16.
As the secretary of the Anti-Narcotics Club in your school, you decide to write an article on the dangers of drug abuse to make aware of this growing menace among schoolmates. Prepare the article to be published in your school magazine.
Answer:

DANGERS OF DRUG ABUSE

The problem of drug abuse has become a serious menace threatening the future and even the life of the youth, especially the school and college-going youth. This age group is more vulnerable to drug abuse because most of them are teenagers. Teenage is a time of stress and strain. To overcome their stress and strain, many youngsters abuse drugs. These drugs give them temporary pleasure, relieving them of their tensions for some time.

There are many reasons for drug abuse among youth. They get a lot of money from their parents. Since most families have only one child or two, the parents pamper them. Many parents do not find time to spend with their children. Another reason for drug abuse is peer influence. A third reason is the blind imitation of the so-called glamour boys and girls in films and sports who use drugs. A 4th reason is the easy availability of drugs. Another reason is the moral laxity of the times. Law-enforcing agencies do not do their work sincerely. There is corruption everywhere and one can do anything if one has money.

Sensual drugs play havoc with the body and mind of their users. The drug addict experiences sensory deprivation. He has a general feeling of physical discomfort and there are personality changes in him. The addict feels depressed. His mental disturbance can be like paranoia. The addict knows he has a problem. But he does not know its source. He looks at external objects with suspicion. Anything outside scares him and he withdraws further and further into himself.

Drugs also affect the body adversely. Dirty needles and solutions can cause liver diseases, venereal diseases and infection of the kidney and brain. Sniffing cocaine and amphetamines can damage the tissue of the nose. Marijuana and tobacco smoking can cause lung diseases. Babies of women who are addicted to drugs are likely to be born with a lot of problems. A drug addict can easily get pneumonia, tuberculosis and have problems of malnutrition and weight loss. An overdose of any drug can cause respiratory or cardiac failure and death.

The drug problems can be solved only through the concerted efforts of the parents, teachers, community leaders and the law-enforcing agencies. Awareness programmes should be conducted about the dangerous effects of drug abuse. The community leaders should help the youth to channel their energies in the right directions. The police must ensure that sensual drugs are not easily available.

Drug abuse is a demon that should be exterminated from our midst with all the might we can muster.

Question 17.
Ahoregallu is essential in any journey. It is more so when people around us are too busy in their own worlds. You feel that it is necessary to post counsellors in schools so that students can reveal their fears and joys to them. Write a letter to the Minister of Education requesting him to take necessary steps in this regard.
Answer:

15/XV Azad Road
Irinjalakuda, North P.O.
10, July 2019

Prof. C. Ravindranath
Education Minister of Kerala
The Secretariat
Trivandrum – 001

Honourable Education Minister,

Sub.: Posting Counsellors in schools

Every day we watch on the TV and read in the newspapers about cases of child abuse. It is strange that even parents, guardians, teachers and religious . leaders abuse children who are entrusted to their care. These people are supposed to work for the welfare of the children and lead them into the right paths to succeed in life and make their valuable contributions to humanity at large. But unfortunately the fact is that many children are abused in many ways, even sexually, by the very people who are supposed to protect them.

The main reason for child abuse is the lack of ‘horegallus’. A horegallu is a stone bench where the weary traveller sits and talks to a sympathetic listener. Thus a horegallu is a patient and sympathetic listener. In the journey of life we all need horegallus. Children need them all the more. A child is afraid to -talk to others about the abuse he/she is facing from one of his relatives, teachers, neighbours or religious leaders. So he/she silently suffers the abuse which can even change his/her personality and attitude towards life and other people.

If there are counsellors in schools, they can listen to the problems of the children patiently and suggest appropriate solutions. By telling their problems to a sympathetic counsellor, the abused children feel relieved. The counsellors can initiate actions against the offenders and-th*is free the abused children from further abuses.

I humbly request your honour to post Counsellors to all schools so that the children can have horegallus to whom they can reveal their hearts.

Yours sincerely,
Mehboob Saithu

Question 18.
Robert Baldwin in ‘The Hour of Truth’ stands as an epitome of honesty throughout the play. His decisions are never influenced by any financial offers. Prepare a character sketch of Robert Baldwin.
Answer:
Robert Baldwin is the hero of the play “The Hour of Truth” by Percival Wilde. Baldwin lives in a trim cottage with his wife Martha and their son John and daughter Evie. He is working as the governor of a bank owned by John Gresham. His salary is very low, 60 dollars a week. His son John earns only 30 dollars a week. With this limited income they live reasonably happily.

Suddenly there is a problem in their lives. John Gresham has misappropriated bank money and he is in jail. The bank is closed and Baldwin will have no job. Only Baldwin is the witness for the misappropriation. If he gives his true testimony, John Gresham will definitely go to jail for a long period. Baldwin is honest and he has taught his family the importance of honesty.

When Baldwin returns home after the arrest of John Gresham, his wife and children rush to him to know the latest news. Initially his wife and the children tell him that he should say the truth during the trial even if it means jail term for Gresham. But when they come to know that Gresham has offered him a bribe of 100,000 dollars just to say ‘I don’t remember’ when some incriminating questions are asked in the court, they change their stance. Theytrytofind out all kinds of loopholes to make Baldwin accept the bribe. Martha, John and Evie do their best to make Baldwin change his mind. But he asserts that he wants to go to his grave clean.

He does love Gresham. He even named his son after him. Even now he remembers how he worked with him for so long. They were boys together. They worked side by side. All this is true, but he is not willing to tell a lie to save his friend and employer even when he is offered a colossal sum of 100,000 dollars.

Baldwin’s honesty is repaid abundantly. Even Gresham is proud of him and he recommends him to Mr. Marshall, the President of the Third International. At the end of the play we see Marshall coming to the house of Baldwin and offering him a job at the Third International. We see Baldwin crying in the end. He must have been crying for two reasons – his friend Gresham has confessed his guilt and he will surely be punished. Secondly he is shedding tears of joy in gratitude to God who has amply repaid his honesty. Honesty, Baldwin proves, beyond an iota of doubt, is the best policy.

Question 19.
The story of Amigo Brothers is always heart, warming. You want to share your appreciation of the story with your friend in Bengaluru. Draft an e-mail conveying the essence of the story so as to inspire your friend to read it at the earliest.
Answer:
ashalatha@gmail.com
Hi Asha, I got your mail. Good to know that you are planning to visit the Taj Mahal during the Onam Vacation.

I am writing this email with a special purpose. I read a story titled “Amigo Brothers” written by Piri Thomas. The story impressed me greatly. It is about friendship. As you are my best friend, I want you to read this story. I’m sure you will enjoy it.

Antonio Cruz and Felix Vargas were both 17. They were so together in friendship that they felt like brothers. They had known each other from childhood. They grew up in the same building on the Lower East Side of Manhattan. They both had a dream – becoming a light-weight champion of the world.

Whenever they got a chance they exercised. They would run everyday morning. They had a collection of fight magazines. They also kept the torn tickets of all the boxing matches they had gone to see. They also had some clippings of their own.

After a series of elimination bouts, they were told that they were to meet each other in the division finals which would be 2 weeks away. The winner would represent the Boys’ Club in the Golden Gloves Championship Tournament.

They were in a big fix. They both wanted to win. But neither of them wanted to defeat the other. They decided to go into the ring as if they had never met. They have to fight it out.

The fight was at Tompkins Square Park. The Park was full of people as the fight was well publicised. Antonio and Felix enter the ring. The crowd explodes into a roar. The referee whistles to start the fight. The fight is on. Punches fly back and forth but none is floored. Each punch is applauded thunderously by the crowd.

It is now the 3rd round. Each contestant wants to win. They punch each other very hard. Nobody falls to the canvas. The final bell rings. But the fight continues. The bell rings again and again. But there is no stopping to the fighting. People start getting worried. It looks as if they are witnessing a do or die fight and not a mere contest. There is utter silence. Then the referee and the trainers separate the contestants.

Now the contestants embrace. They have forgotten they were fighting like bulls up to a moment ago. The announcer makes an announcement. He is trying to name the winner. But Antonio and Felix had left like good old friends, arm in arm, declaring their bond of friendship.

In my mind both are winners, aren’t they Asha?
I am coming to Bengaluru next week. We will talk more about the story then! Bye for now!

Love,
Johny

(Question Nos. 20 – 22): Answer any 2 questions, each in about 140 words. Each carries 8 scores. (2 × 8 = 16)

Question 20.
Given below is a poster displayed at the Assembly Hall in connection with the International Women’s Day celebrations conducted by the Souhrida Club of your school. As the convenor of the programme, you are asked to give a detailed talk on ‘EMPOWERING WOMEN – CHALLENGES AHEAD’.Prepare the likely speech taking hints from the poster.

Answer:
Respected Principal and teachers and my dear friends,

Today we are celebrating International Women’s Day and I am asked to talk about “Empowering Women – Challenges Ahead”. It is a great pity that Indians speak very highly of women, but they do not show much respect to them in their actions. They will say woman is the mother, Devi guru and the light of the home. They even name their daughters with names of goddesses and saintly persons. Even in our own school, we have so many Lakshmis and Saraswatis, Aishas and Khadeejas and Marys and Theresas. But do they receive the respect they deserve from the society? Even in the buses and roads they are troubled and teased. We hear of so many crimes against women every day. Rapes and murder of women have ceased to shock people as they have become everyday affairs.

I say we are doing only lip-service to women. Even in Kerala which has more women than men, women are not given their rights. Look at the number of women candidates in the Lok Sabha elections for 2019. We have three coalitions in Kerala, UDF, LDF and NDA. Each of these groups is fielding 20 candidates for the 20 Lok Sabha seats. What percentage of women candidates do they field? Just 10%. No political party is free from this fault.

Women don’t get equal pay with men in non¬governmental services. A man gets Rs. 800 a day but a woman is given only Rs. 400 although she is doing similar or even harder work than the man. Women are refused opportunities for employment because employers feel that married women will take maternity leave.

I strongly feel that the Parliament and State Assemblies should have equal number of men and women representatives. Men have no business keeping women under a patriarchal system. Let the women also feel they are persons with the dignity and freedom that men enjoy in our society.

We have to change our mindset. Man and woman are created as equals and they should be treated equally by the society in all walks of life.

As I urge all the women here to fight for their rights,
I wish them success in their endeavors.

Jai Hind!

Question 21.
Read the poem given below. Compare and contrast it with ‘Mending Wall’ and attempt a critical appreciation.

NEIGHBOUR

Iain Crichton Smith

Build me a bridge over the stream
to my neighbour’s house
where he is standing in dungarees*
in the fresh morning.

O ring of snowdrops
spread wherever you want
and you also blackbird
sing across the fences.

My neighbour, if the rain fails on you,
let it fall on me also
from the same black cloud
that does not recognize gates.

*dungarees: a garment consisting of trousers held up by straps over the shoulders.
Answer:
“Mending Wall” by Robert Frost is a delightful poem. He once said: “A poem begins in delight and ends in wisdom.” He starts the poem in a delightful way saying that there is something that does not like a wall. That something makes the ground under the wa|l swell which results in cracks in the wall. Gradually the stones that make the wall fall to either side.

The fallen stones have lost their shapes. It is not easy to keep them back in their place. The gaps are so big that even two people can walk abreast through them. The poet strongly feels there is no need for a wall between him and his neighbour because he grows apple trees and the neighbour grows pine. Apple trees won’t go and eat the pine and pines won’t come to eat the apples. A wall was fine if they had cows as they could get mixed up without a wall. When the poet says there is no need for a wall between them, the neighbour tells him “Good fences make good neighbours.” Thus the poem ends in wisdom.

In “Mending Wall”, the poet has used many poetic devices such a metaphor, simile, personification and repetition. The language is simple and the imagery is exquisite. It gives a fine message, a priceless truth. “Neighbour” by lain Crichton Smith is a beautiful poem of just 12 lines. It gives a fine message. We see two neighbours here separated by a stream, ring of snowdrops, fences and gates. The poet can see his neighbour standing in his house wearing his trousers with shoulder straps. He wants a bridge to be built across the stream to connect the two neighbours. The blackbird sings across the fences and the rain falls on both the neighbour and himself. The rain which comes from the black cloud does not differentiate his neighbour from him. Nature is treating them equally and so the poet thinks that the neighbours should be linked by a bridge.

The poem is highly melodious. The imagery is superb as we can visualize the stream, the ring of snowdrops, the blackbird singing, and the rain falling. We can see the neighbour standing in his house in his trousers held by shoulder straps. The message is loud and clear – There should be love between neighbours.

Between the two poems, I prefer “Mending Wall”, as it is more dramatic and action packed.

Question 22.
In the light of reading excerpts from the lives of Irfan Alam and Shaheen Mistri, you arrive at the conclusion that ‘Entrepreneurs are not made, they are born.’ Our nation benefits from such enterprises too. Write an essay on the topic highlighting the merits of promoting entrepreneurship among the youth.
(HINTS : youth made responsible – dignity of labour – self-confident – self-reliance – hard work and commitment – job satisfaction – financial stability – creating employment opportunities – initiatives like kudumbasree, mango cabs, ubereats, flipkart, etc.)
Answer:

ENTREPRENEURS ARE NOT MADE, THEY ARE BORN

There is a wrong tendency in our society which makes the youth to look for government jobs. After getting their degrees, they go on writing PSC tests in the
hope of getting clerical jobs in government services. This is really bad for our country because the government can’t provide jobs for all the people. Here comes the importance of entrepreneurship.

Entrepreneurship makes us responsible and the success of the enterprise depends on our planning, vision and hard work. It also adds to the concept of dignity of labour’. An entrepreneur is prepared to do any kind of work that is needed of him for the success of his enterprise. In the Government service employees are graded as Class I, II, III and IV. Class IV employees are usually sweepers, cleaners and peons. Class III consists of clerks. Class II is of the Supervisory staff and Class I is that of High Officers. But in a private enterprise, there is no such division.

Entrepreneurship enhances self-confidence, self reliance and hard work. There is a lot of job satisfaction when we see our enterprises are succeeding. Entrepreneurs give jobs to others instead of being job hunters. It brings financial stability to ourselves and also the people whom we employ.

There are so many private enterprises which are very popular today. We have the Kudumbasree, Mango Cabs, Uber Eats and Flipkart. They are all successful enterprises and they originated in the minds of ordinary people. With some thinking, willingness to work hard, determination and self-confidence we too can embark upon some project that will prove successful.

Rome was not built in a day. Let us remember Dhirubhai Ambani, the father of the richest man in India today, Mukesh Ambani. Dhirubhai Ambani worked as a clerk in Yemen and returned to India and started a textile trading company in 1958. His initial capital was a mere Rs. 15,000. Let’s remember, “Where there is a will there is a way.”

HSE Kerala Board Syllabus Plus Two Zoology Previous Year Model Question Papers and Answers Pdf HSSLive Free Download in both English medium and Malayalam medium are part of SCERT Kerala Plus Two Previous Year Question Papers and Answers. Here HSSLive.Guru have given Higher Secondary Kerala Plus Two Zoology Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus.

Board	SCERT, Kerala Board
Textbook	NCERT Based
Class	Plus Two
Subject	Zoology
Papers	Previous Papers, Model Papers, Sample Papers
Category	Kerala Plus Two

Kerala Plus Two Zoology Previous Year Question Papers and Answers

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We hope the given HSE Kerala Board Syllabus Plus Two Zoology Previous Year Model Question Papers and Answers Pdf HSSLive Free Download in both English medium and Malayalam medium will help you. If you have any query regarding HSS Live Kerala Plus Two Zoology Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

HSSLive Plus Two

HSE Kerala Board Syllabus Plus Two Botany Previous Year Model Question Papers and Answers Pdf HSSLive Free Download in both English medium and Malayalam medium are part of SCERT Kerala Plus Two Previous Year Question Papers and Answers. Here HSSLive.Guru have given Higher Secondary Kerala Plus Two Botany Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus.

Board	SCERT, Kerala Board
Textbook	NCERT Based
Class	Plus Two
Subject	Botany
Papers	Previous Papers, Model Papers, Sample Papers
Category	Kerala Plus Two

Kerala Plus Two Botany Previous Year Question Papers and Answers

• Kerala Plus Two Botany Question Paper March 2024
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HSE Kerala Board Syllabus Plus Two Accountancy with Computerised Accounting Previous Year Model Question Papers and Answers Pdf HSSLive Free Download in both English medium and Malayalam medium are part of SCERT Kerala Plus Two Previous Year Question Papers and Answers. Here HSSLive.Guru have given Higher Secondary Kerala Plus Two Accountancy with Computerised Accounting Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus.

Board	SCERT, Kerala Board
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Class	Plus Two
Subject	Accountancy with Computerised Accounting
Papers	Previous Papers, Model Papers, Sample Papers
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Kerala Plus Two Accountancy Computerised Accounting Previous Year Question Papers and Answers

Plus Two Computerised Accounting Previous Year Question Papers and Answers

• Kerala Plus Two Accountancy Question Paper March 2023
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• Kerala Plus Two Accountancy Question Paper March 2021
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• Kerala Plus Two Accountancy Board Model Paper 2023
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HSE Kerala Board Syllabus Plus Two Computer Science Previous Year Model Question Papers and Answers Pdf HSSLive Free Download in both English medium and Malayalam medium are part of SCERT Kerala Plus Two Previous Year Question Papers and Answers. Here HSSLive.Guru have given Higher Secondary Kerala Plus Two Computer Science Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus.

Board	SCERT, Kerala Board
Textbook	NCERT Based
Class	Plus Two
Subject	Computer Science
Papers	Previous Papers, Model Papers, Sample Papers
Category	Kerala Plus Two

Kerala Plus Two Computer Science Previous Year Question Papers and Answers

• Kerala Plus Two Computer Science Question Paper March 2024
• Kerala Plus Two Computer Science Question Paper March 2023
• Kerala Plus Two Computer Science Question Paper March 2022
• Kerala Plus Two Computer Science Question Paper March 2021
• Kerala Plus Two Computer Science Question Paper March 2020
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• Kerala Plus Two Computer Science Model Question Paper Set 3

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HSSLive Plus Two

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf March 2021 helps in understanding answer patterns.

Kerala Plus Two Maths Previous Year Question Paper March 2021

Time: 2 Hours
Total Score: 60 Marks

Answer the following questions from 1 to 29 up to a maximum score of 60.

PART – A

Questions from 1 to 10 carry 3 scores each. (10 × 3 = 30)

Question 1.
Find the values of x for which \(\left|\begin{array}{ll}
3 & x \\
x & 1
\end{array}\right|\) = \(\left|\begin{array}{ll}
3 & 2 \\
4 & 1
\end{array}\right|\) (3)
Answer:
Given \(\left|\begin{array}{ll}
3 & x \\
x & 1
\end{array}\right|\) = \(\left|\begin{array}{ll}
3 & 2 \\
4 & 1
\end{array}\right|\)
Expanding on both sides
3 – x² = 3 – 8 = – 5
x² = 8
x = ±√8 = ±2√2

Question 2.
Let A = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right|\)
(i) Find adj A (2)
(ii) FindA.adjA. (1)
Answer:

Question 3.
Find the value of k so that the function is continuous at x = 5(3)
Answer:
Given f(x) =
Since f(x) is continuous at x = 5,
LHL = RHL = f(5)
f(5) = k(5) + 1 = 5k + 1 _______ (1)
RHL = \(\lim _{x \rightarrow 5^{+}} f(x)\) = \(\lim _{x \rightarrow 5^{+}} 3 x-5\) = 10 ________ (2)
From (1) and (2),
5k + 1 = 10.
5k = 9
k = \(\frac{9}{5}\)

Question 4.
Verify Rolle’s theorem for the function
f(x) = x² + 2x – 8, x ∈ [-4, 2] (3)
Answer:
Given f(x) = x² + 2x – 8, x ∈ [-4, 2]
Since f(x) is a polynomial function, it is continuous in [-4, 2]
f (x) = 2x + 2
∴It is differentiable in (-4, 2)
f(a) = f(-4)= 16 – 8 – 8 = 0
f(b) = f(2) = 4 + 4 – 8 = 0
∴ f(a) = f(b)
f (c) = 0 ⇒ 2c + 2 = 0 ⇒ c = -1 ∈ (-4, 2)
Hence Rolle’s theorem verified.

Question 5.
Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm. (3)
Answer:
Let A = π r²
\(\frac{d A}{d r}\) = 2 πr
\(\left.\frac{d A}{d r}\right]_{r=5}\) = 2π × 5 = 10 π

Question 6.
Find the projection of vector î + 3ĵ + 7k̂ on the vector 7î – ĵ + 8k̂ (3)
Answer:
Let \(\vec{a}\) = î + 3ĵ + 7k̂
\(\vec{b}\) = 7î – ĵ + 8k̂
\(\vec{a}\) . \(\vec{b}\) = 7 – 3 + 56 = 60
|\(\vec{b}\)| = \(\sqrt{49+1+64}\) = \(\sqrt{144}\)
Projection of \(\vec{a}\) on \(\vec{b}\) = \(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}{|\overline{\mathrm{~b}}|}\) = \(\frac{60}{\sqrt{114}}\)

Question 7.
Find the equation of a plane passing through the point (1, 4, 6) and the normal to the plane is î – 2ĵ + k̂. (3)
Answer:
Given (x₁, y₁, z₁ ) = (1, 4, 6)
Direction ratios of normal,
<a, b, c> = <1, -2, 1>
Equation of plane is
a(x – x₁) + b(y – y₁) + (z – z₁) = 0
1(x – 1) – 2(y – 4) + 1(z – 6)= 0
x – 2y + z + 1 = 0

Question 8.
(i) Which of the following can be the domain of the function cos^-1 x?
(a) (0, π)
(b) (0, π)
(c) (-π, π)
(d) (\(\frac{-\pi}{2}\) \(\frac{\pi}{2}\)) (1)
(ii) Find the value of cos^-1 (-1/2) + 2 sin^-1(1/2). (2)
Answer:
(i) There is no correct option.
Answer is [-1, 1]

(ii) cos^-1(\(\frac{-1}{2}\)) + 2 sin^-1(\(\frac{1}{2}\))
= π – \(\frac{\pi}{3}\) + 2 × \(\frac{\pi}{6}\) = \(\frac{2 \pi}{3}\) + \(\frac{\pi}{3}\) = \(\frac{2 \pi}{3}\) = π

Question 9.
Find the are of a triangle with vertices (-2, -3), (3, 2) and (-1,-8). (3)
Answer:
Given vertices are (-2, -3, (3, 2), (-1, -8)
Area = \(\frac{1}{2}\)\(\left|\begin{array}{ccc}
-2 & -3 & 1 \\
3 & 2 & 1 \\
-1 & -8 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[|-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)|]
= \(\frac{1}{2}\)[|-20 + 12 – 22|] = \(\frac{1}{2}\)|-30|
= 15 sq. units

Question 10.
Find the general solution of the differential equation \(\frac{d y}{d x}\) – y = cos x (3)
Answer:
Given \(\frac{d y}{d x}\) – y = cos x
Which is a linear differential equation of the form \(\frac{d y}{d x}\) +Py = Q
Where P = -1, Q = cos x
Integrating factor (I. F) = \(e^{\int P d x}\) = \(e^{\int-1 d x}\) = e^-x
∴ Solution is Y (I. F) = \(\int Q(I F) \mathrm{d} x\)
ye^-x = \(\int \cos x \cdot e^{-x} d x\)
ye^-x = \(\cos x \cdot \frac{e^{-x}}{-1}-\int-\sin x \cdot \frac{e^{-x}}{-1} \mathrm{~d} x\)
ye^-x = \(\frac{e^{-x}}{2}[\sin x-\cos x]+C\)

PART – B

Questions from 11 to 22 carry 4 scores each. (12 × 4 = 48)

Question 11.
Consider the matrices
A = \(\left[\begin{array}{cc}
3 & 4 \\
-5 & -1
\end{array}\right]\) and 3A + B = \(\left[\begin{array}{cc}
2 & 8 \\
3 & -4
\end{array}\right]\)
(i) Find the matrix B. (2)
(ii) Find AB. (2)
Answer:

Question 12.
If A = \(\left[\begin{array}{c}
-2 \\
4 \\
5
\end{array}\right]\) and B = [1, 3, -6]
(i) What is the order of AB? (1)
(ii) Verify (AB)’ = B’A’ (3)
Answer:
(i) Order of A= 3 × 1
Order of B = 1 × 3
∴ Order of AB = 3 × 3

Question 13.
(i) If xy < 1, tan^-1 x + tan^-1 y = __________
(a) tan^-1 \(\frac{x-y}{1+x y}\)
(b) tan^-1 \(\frac{1-x y}{x+y}\)
(c) tan^-1 \(\frac{x+y}{1-x y}\)
(d) tan^-1 \(\frac{x-y}{1+x y}\) (1)
(ii) Prove that tan^-1\(\frac{2}{11}\) + tan^-1\(\frac{7}{24}\) = tan^-1\(\frac{1}{2}\) (3)
Answer:
(i) (c) tan^-1(\(\frac{x+y}{1-x y}\))

Question 14.
Find \(\frac{d y}{d x}\)
(i) x² + xy + y² = 100 (2)
(ii) y = sin^-1(\(\frac{2 x}{1+x^2}\)), -1 ≤ x ≤
Answer:
(i) Given x² + xy + y² = 100
Differentiating with respect to x,
2x + x\(\frac{d y}{d x}\) + y + 2y \(\frac{d y}{d x}\) = 0
(x + 2y)\(\frac{d y}{d x}\) = – 2x – y
\(\frac{d y}{d x}\) = \(\frac{-(2 x+y)}{x+2 y}\)

(ii) We have sin^-1(\(\frac{2 x}{1+x^2}\)) = 2 tan^-1 x
∴ y = 2 tan^-1 x
\(\frac{d y}{d x}\) = 2 × \(\frac{1}{1+x^2}\) = \(\frac{2}{1+x^2}\)

Question 15.
Find the intervals in which the function f given by f(x) = 2x² – 3x is
(i) increasing
(ii) decreasing (4)
Answer:
Given f(x) = 2x² – 3x
f'(x) = 4x – 3
f'(x) = 0 ⇒ 4x – 3 = 0
x = \(\frac{3}{4}\)

Intervals are (-∞, 3/4), and (3/4, ∞)
In (-∞, 3/4), f'(0) = 0 – 3 = -3 < 0 In (\(\frac{3}{4}\), ∞) f'(1) = 4 – 3 = 1 > 0
∴ f(x) is decreasing in (-∞, 3/4)
and increasing in (3/4, ∞)

Question 16.
(i) Find the order and degree of the differential equation (\(\frac{d s}{d t}\))⁴ + \(\frac{3 \mathrm{~d}^2 \mathrm{~s}}{\mathrm{dt}^2}\) = 0. (1)
(ii) Find the general solution of the differential equation \(\frac{d y}{d x}\) = (1 + x²)(1 + y²). (3)
Answer:
(i) Order = 2
Degree = 1

(ii) Given \(\frac{d y}{d x}\) = (1 + x² )(1 + y²)
⇒ \(\frac{d y}{1+y^2}\) =(1 + x²)dx
Which is variable seperable.
∴ Solution is \(\int \frac{d y}{1+y^2}=\int\left(1+x^2\right) d x\)
⇒ tan^-1 y = x + \(\frac{x^3}{3}\) + C
or y = tan¹ x + \(\frac{x^3}{3}\) + C

Question 17.
Find a unit vector both perpendicular to the vectors if
\(\vec{a}\) = 2î + ĵ + 3k̂ and \(\vec{b}\) = 3î + 5ĵ – 2k̂ (4)
Answer:
Given \(\vec{a}\) = 2î + ĵ + 3k̂, \(\vec{b}\) = 3î + 5ĵ – 2k̂ vector perpendicular to both \(\vec{a}\) & \(\vec{b}\) is

Question 18.
Find the shortest distance between the skew lines
\(\vec{r}\) = î + 2ĵ + k + λ(î – ĵ + k) and
\(\vec{r}\) = 2î – ĵ – k̂+ μ(2î + ĵ + 2k̂) (4)
Answer:
\(\vec{a}\)₁ = î + 2ĵ + k̂, \(\vec{b}\)₁ = i – ĵ + k̂
\(\vec{a}\)₂ = 2î – ĵ – k̂, \(\vec{b}\)₂ = 2i + ĵ + 2k̂
\(\vec{a}\)₂ – \(\vec{a}\)₁ = î – 3ĵ – 2k̂

Question 19.
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4. Find
(i) P(A∩B) (2)
(ii) P(A | B) (1)
(iii) P(A∪B) (1)
Answer:
(i) P(B/A) = \(\frac{P(A \cap B)}{P(A)}\)
∴ P(A∩B) = P(A). P(B/A)
= 0.8 × 0.4 = 0.32

(ii) P(A/B) = \(\frac{P(A \cap B)}{P(B)}\) = \(\frac{0.32}{0.5}\) = 0.64

(iii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.8 + 0.5 – 0.32 = 0.98

Question 20.
(i) Let R be the relation on a set A = {1, 2, 3}, defined by R = {(1, 1), (2, 2), (3, 3), (1, 3)}. Then the ordered pair to be added to R to make it a
smallest equivalence relation is ___________
(a) (2, 1)
(b) (3, 1)
(c) (1, 2)
(d) (1, 3) (1)
(ii) Determine whether the relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x} is reflexive, symmetric and transitive. (3)
(a) (2, 1)
(b) (3, 1)
(c) (1, 2)
(d) (1, 3)
Answer:
(i) (b)(3, 1)

(ii) Given A = {1, 2, 3, 4, 5, 6}
R = {(x, y): y is divisible by x}
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4),
(5, 5), (6, 6)}
Since (a, a) ∈ R for all a ∈ A, R is reflexive.
(1, 2) ∈ R, but (2, 1) ∉ R,
∴ R is not symmetric.
(a, b) ∈ R,(b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c, ∈ A
∴ R is transitive.

Question 21.
Find \(\frac{d y}{d x}\)
(i) x^x (2)
(ii) x = 2at² ; y = at⁴ (2)
Answer:
(i) Let y = x³
Taking ‘log’ on both sides
Log y = x. log x
Differentiating with respect to x,
\(\frac{1}{y}\)\(\frac{d y}{d x}\) = x.\(\frac{1}{x}\) + logx.1
\(\frac{d y}{d x}\) = y[1 + log x] = x^x [1 + log x]

(ii) Given x = 2at², y = at⁴
\(\frac{d x}{d t}\) = 4 at, \(\frac{d y}{d x}\) 4a t³
∴ \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\) = \(\frac{4 a t^3}{4 a t}\) = t²

Question 22.
Integrate:
\(\int \frac{x}{(x+1)(x+2)} d x\) (4)
Answer:
Let \(\frac{x}{(x+1)(x+2)}\) = \(\frac{\mathrm{A}}{x+1}\) + \(\frac{\mathrm{B}}{x+2}\)
x = A(x + 2) + B(x + 1)
Put x = -1, -1 = A+ 0 ⇒ A = -1
x = -2, -2 = 0 + B(-1) ⇒ B = 2
\(\frac{x}{(x+1)(x+2)}\) = \(\frac{-1}{x+1}\) + \(\frac{2}{x+2}\)
\(\int \frac{x}{(x+1)(x+2)} d x\) = \(-\int \frac{1}{x+1} d x\) + \(2 \int \frac{1}{x+2} d x\)
= -log| x + 1| + 2log|x + 2| + C

Part – C

Questions from 23 to 29 carry 6 scores each. (7 × 6 = 42)

Question 23.
(i) construct a 3 × 2 matrix A = [a_ij] whose elements are given by a_ij = 3î – ĵ (2)
(ii) Express \(\left[\begin{array}{cc}
2 & 3 \\
1 & -4
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix. (4)
Answer:
(i) Given a_ij = 3î – ĵ
a₁₁= 2 a₁₂ = 1
a₂₁ = 5 a₂₂ = 4
a₃₁ = 8 a₃₂ = 7
∴ A = \(\left[\begin{array}{ll}
2 & 1 \\
5 & 4 \\
8 & 7
\end{array}\right]\)

Hence A = P + Q
where P is symmetric and Q is skew symmetric.

Question 24.
Solve the following system of equations by matrix method
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4 (6)
Answer:
Let A = \(\left[\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(\left[\begin{array}{l}
8 \\
1 \\
4
\end{array}\right]\)
Here the system of equations can be written as AX = B.
|A| = \(\left|\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right|\)
= 3(2 – 3) + 2(4 + 4) + 3(- 6 – 4)
= – 3 + 16 – 30 = -17 ≠ 0
∴ System of eqns is consistent.

Question 25.
(i) Let f : {1, 3, 4} → {1, 2, 5} and g:{1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1))} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof. (3)
(ii) Consider f: R → R given by f(x) = 2x +1. Show that f is invertible. Find the inverse of f. (3)
Answer:
(i) Given f = {(1, 2), (3, 5), (4, 1)}, g = {(1, 3), (2, 3), (5, 1)}
∴ f(1) = 2, f(3) = 5, f(4) = 1, g(1) = 3, g(2) = 3, g(5) = 1
(gof)(1) = g(f(1)) = g(2) = 3
(gof) (3) = g(f(3)) = g(5) = 1
(gof) (4) = g(f(4))= g(1) = 3
∴ gof ={(1,3), (3,1), (4, 3)}

(ii) Given f(x) = 2x + 1
f(x₁) = f(x₂) ⇒ 2x₁ + 1 = 2x₂ + 1
2x₁ = 2x₂
x₁ = x₂
∴ is one one
Let y = 2x + 1 ⇒ 2x = y – 1
x = \(\frac{y-1}{2}\)
f(x) = f(\(\frac{y-1}{2}\)) = 2(\(\frac{y-1}{2}\)) + 1 = y – + 1 = y
∴ for every y, there exists x
such that f(x) = y
Hence f is onto
∴ f is bijective
Inverse of f = f^-1(x) = \(\frac{x-1}{2}\)

Question 26.
(i) Find the slope of the tangent to the curve
y = x³ – x at x = 2. (2)
(ii) Find the equation of tangent to the above curve. (2)
(iii) What is the maximum value of the function sin x + cos x ? (2)
Answer:
(i) y = x³
\(\frac{d y}{d x}\) = 3x²2 – 1
Slope of tangent = \(\left.\frac{d y}{d x}\right]_{x=2}\) = 3(4) – 1 = 11

(ii) Point is (2, 6)
Equation of tangent is
y – 6 = 11 (x – 2)
y – 6 = 11x – 22
11x – y – 16 = 0

(iii) f(x) = sin x + cos x, f'(x) = cos x – sin x
f”(x) = – sin – cos x = -(sin x + cos x)
f'(x) = 0 ⇒ cos x = sin x ⇒ x = \(\frac{\pi}{4}\)
f”(π/4) = -(\(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)) = -√2 < 0
∴ f is maximum, when x = π/4
Max. value = f(π/4) = \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = \(\frac{2}{\sqrt{2}}\) = √2

Question 27.
Integrate:
(i) \(\int \sin x \sin (\cos x) d x\) (3)
(ii) \(\int_0^1 \frac{\tan ^{-1} x}{1+x^2} \mathrm{~d} x\) (3)
Answer:
(i) ∫sin x sin(cos x)dx
|Put t = cos x
dt = -sin x dx
-dt = sinx dx|
= ∫sin t(-dt)
= -∫sin t dt = -(-cost) + C
= cos t + C = cos(cos x) + C

Question 28.
Solve the following problem graphically
Maximise: z = 3x + 2y
Subject to: x + 2y ≤ 10
3x + y ≤ 15,
x, y ≥ 0 (6)
Answer:
x + 2y = 10

x	0	10
y	5	0

3x + y = 15

x	0	5
y	15	0

∴ Z has maximum when x = 4, y = 3
Maximum value = 18

Question 29.
(1) Find the area of the region bounded by the curve y² = x and the line x = 1 & x = 4 and the x-axis. (3)
(ii) Find the area of the region bounded by two parabolas y = x² and y² = x. (3)
Answer:

(ii) Given y = x² (1)
f = x ________ (2)
Solving (1) and (2),

Reviewing Kerala Syllabus Plus Two Physics Previous Year Question Papers and Answers Pdf March 2023 helps in understanding answer patterns.

Kerala Plus Two Physics Previous Year Question Paper March 2023

Time : 2 1/2 Hours
Maximum : 80 scores

SECTION-A

Answer any 5 questions from 1 to 7. Each carries 1 score. (5 × 1 = 5)

Question 1.
State true of false:
Two field lines never intersect.
Answer:
True

Question 2.
The SI unit of resistance is …………
Answer:
ohm(Ω)

Question 3.
Current loop behaves as a ………… (Magnetic dipole/ Electric dipole)
Answer:
Magnetic dipole

Question 4.
An accelerating charge produces ………….. waves.
(a) electric
(b) magnetic
(c) electromagnetic
(d) none of these
Answer:
(c) electromagnetic

Question 5.
When the speed of light is independent of direction, the secondary waves are
(a) Spherical
(b) Cylindrical
(c) Plane
(d) Rectangular
Answer:
(a) Spherical

Question 6.
X-rays were discovered by in 1895.
(a) Roentgen
(b) J.J. Thompson
(c) William Crookes
(d) Rutherford
Answer:
(a) Roentgen

Question 7.
Atoms of same element differing in mass are called ………..
(a) Isotones
(b) Isobars
(c) Isotopes
(d) Isomers
Answer:
(c) Isotopes

SECTION-B

Answer any 5 questions from 8 to 14. Each carries 2 scores. (5 × 2 = 10)

Question 8.
Define magnetisation. Give its dimension.
Answer:
Manetisation is the net magnetic moment per unit volume.
Dimension of magnetisation is M = \(\frac{m}{V}\)

Question 9.
State laws of electromagnetic induction.
Answer:
Faraday’s law of electromagnetic induction states that the magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
Mathematically, the induced emf is given by
∈ = \(\frac{d \phi}{d t}\)
Lenz’s law states that the direction emf (or current) is such tfiat it opposes the change in magnetic flux which produces it
Mathematically the Lenz’s law can be written as
∈ = \(\frac{-d \phi}{d t}\)

Question 10.
Obtain the expression for the current flowing through a resistor when an a.c. voltage is applied to it.
Answer:

Consider a circuit containing a resistance ‘R’ connected to an alternating voltage.
Let the applied voltage be
V = V₀ sin ωt ………….(1)
According to Ohm’s law, the current at any instant can be written as
I = \(\frac{V_0 \sin \omega t}{R}\)

Where I₀ = V₀/R is the peak value of current
I = I₀ sin ωt

Question 11.
How Maxwell modified Ampere’s law?
Answer:
Ampere circuital theorem is
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} l}\) = μ₀ I_c
I_C is the conduction current. This equation is not applicable for ac circuits. Maxwell introduced a cor-rection term related to displacment current
I_d =

The modified Maxwell equation is
\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} l}\) = μ₀ (i_c + i_d)

\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} l}\) = μ₀ I_c + μ₀ ε_c(\(\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{~d} l}\))

Question 12.
What is total internal reflection?
Answer:
When light passes from denser medium to rarer medium and angle of incidence is greater than critical angle, light gets reflected back to denser me¬dium itself. The phenomenon is called total internal reflection.

Question 13.
Explain work function.
Answer:
The minimum energy required by an electron to escape from metal surface is called work function of metal.

Question 14.
Differentiate between nuclear fission and nuclear fusion.
Answer:
In nuclear fission, a heavier nuclei splits into lighter ones releasing hugeenprgy. In nuclear fusion, lighter nuclei combine to form heavier nuclei releasing energy.
Fission can be controlled. But fusion cannot be controlled.

SECTION-C

Answer any 6 questions from 15 to 21. Each carries 3 scores. (6 × 3 = 18)

Question 15.
Explain the basic properties of electric charge.
Answer:
Basic properties of electric charges are:
(a) Unlike charges attract and like charges repel.
(b) Charge is conserved
(c) Electric Charge is Quantized
(d) Additivity of Charges

Question 16.
(a) Derive the expression for the capacitance of a parallel plate capacitor.
Answer:

Expression for capacitance of a capacitor:
Potential difference between two plates
V = Ed
= \(\frac{\sigma}{\varepsilon_0}\) d (E = \(\frac{\sigma}{\varepsilon_0}\))
V = \(\frac{Q}{A \varepsilon_0}\) d ………..(1) (σ = \(\frac{Q}{A}\))

Capacitance C of the parallel plate capacitor,
C = \(\frac{Q}{V}\) …………..(2)
Sub. eq. (1) ineq. (2)
C = \(\frac{Q}{\frac{Q}{A \varepsilon_0} d}\)
C = \(\frac{A \varepsilon_0}{d}\)

(b) What happens to the capacitance if a medium of dielectric constant K is introduced between the plates?
Answer:
Capacitance increases.

Question 17.
(a) State Biot-Savart law.
Answer:
The magnetic field at any point due to an element of current carrying conductor is

1) Directly proportional to the strength of the current (I)
2) Directly proportional to the length of the element(dl)
3) Directly proportional to the sine of the angle (θ) between the element and the line joining the midpoint of the element to the point.
4) Inversely proportional to the square of the distance of the point from the element
I dl sinθ
dB ∝ \(\frac{I dl sinθ}{r^2}\)

dB =\(\frac{\mu_0}{4 \pi}\) \(\frac{I dl sinθ}{r^2}\)

(b) Obtain the expression for the magnetic field on the axis of a circular current loop.
വൃത്താകൃതിയിലുള്ള കറണ്ട് ലൂപ്പിന്റെ അക്ഷത്തിലൂടെ യുള്ള കാന്തിക മണ്ഡല തീവ്രതയുടെ സമവാക്യം രൂപീകരിക്കുക. (2)
Answer:

Consider a circular loop of radius ‘a’ and carrying current “I”. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A.
The magnetic field at ‘p’ due to this small element dl,
dB =\(\frac{\mu_0 \text { Idl } \sin 90}{4 \pi x^2}\)

dB = \(\frac{\mu_0 \mathrm{Idl}}{4 \pi \mathrm{x}^2}\) …………(1)

The dB can be resolved into dB cosϕ (along Py) and dB sinϕ (along Px).
Similarly consider a small element at B, which pro-duces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get.
dB sinϕ (along px) and dB cosϕ (along py1)
dB cosϕ components cancel each other, because
they are in opposite direction. So only dB sinϕ components are found at P, so total filed at P is
B = ∫ dB sinϕ
= \(\int \frac{\mu_0 \mathrm{Idl}}{4 \pi \mathrm{x}^2} \sin \phi\)
but from AOP we get, sin ϕ = a / x

Question 18.
Differentiate between paramagnetic, diamagnetic and ferromagnetic substances.
പാരാമാഗ്നറ്റിക്, ഡയാമാഗ്നറ്റിക്, ഫെറോമാഗ്നറ്റിക് വസ്തു ക്കളെ തരം തിരിക്കുന്നതെങ്ങനെ?
Answer:

Diamagnetism: Diamagnetic substances are those which have tendency to move from stronger to the weaker part of the external magnetic field.

Examples : Some diamagnetic materials are bismuth, copper, lead, silicon, nitrogen (at STP), water and sodium chloride.
Paramagnetism : Paramagnetic substances are those which get weakly magnetized in an external magnetic field. They get weakly attracted to a magnet.

Some paramagnetic materials are aluminium, sodium, calcium, oxygen (at STP) and copper chloride.

Ferromagnetism : Ferromagnetic substances are those which gets strongly magnetized in an external magnetic field. They get strongly attracted to a magnet.
Common examples of ferromagnetic substances are Iron, Cobalt Nickel, etc.

Question 19.
(a) State the principle of a.c. generator.
എ.സി. ജനറേറ്ററിന്റെ തത്വം പ്രസ്താവിക്കുക. (1)
Answer:
Electromagnetic Induction

(b) Obtain the expression for the emf generated by an a.c. generator.
എ.സി. ജനറേറ്റർ ഉത്പാദിപ്പിക്കുന്ന ഇ.എം. എഫിന്റെ സമവാക്യം രൂപീകരിക്കുക. (2)
Answer:

Take the area of coil as A and magnetic field pro¬duced by the magnet as B .Let the coil be rotating about an axis with an angular velocity ω.
Let θ be the angle made by the areal vector with the magnetic field B. The magnetic flux linked with the coil can be written as
ϕ =B. A
ϕ = BA cos θ
ϕ = BA coscot [since θ = ωt]
If there are N turns
ϕ = NBA cos ωt

∴ The induced e.m.f. in the coil,
ε = \(\frac{-\mathrm{d} \phi}{\mathrm{dt}}\)
ε = \(\frac{-\mathrm{d}}{\mathrm{dt}}(\mathrm{NBA} \cos \omega \mathrm{t})\)
ε = NBA cos ωt ω
Let ε = NBA ω,
then ε = ε₀ sinωt

Question 20.
Derive the expression for the refractive index of a prism with the help of a diagram.
ഒരു പ്രിസത്തിന്റെ റിഫ്രാക്ടീവ് ഇൻഡക്സിന്റെ സമവാക്യം ചിത്രത്തിന്റെ സഹായത്തോടെ രൂപീകരിക്കുക.
Answer:

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism.
A ray PQ incidents on the face AB at an angle i₁. QR is the refracted ray inside.the prism, which makes two angles r₁ and r₂ (inside the prism). RS is the emergent ray at angle i₂.
The angle between the emergent ray and incident ray is the deviation‘d’.
In the quadrilateral AQMR,
∠Q + ∠R = 180°
[since N,M and NM are normal]
ie, ∠A + ∠M = 180° —(1)
In the ∆QMR,
∴ r₁ + r₂ + ∠M = 180° —(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ∠A —(3)
From the ∆QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the opposite inte-rior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ – r₂ = A
.-. (i₁ + i₂) – A = d
(i₁ + i₂) = d + A …….(4)

It is found that for a particular angle of incidence, the deviation is found to be minimum value ‘D’.

At the minimum deviation positio
i₁ = i₂ = i, r₁ = r₂ = r and d = D
Hence eq (3) can be written as,
r + r = A
or r = \(\frac{A}{2}\) ……(5)
Similarly eq (4) can be written as,
i + i = A + D
i = \(\frac{A+D}{2}\) ………(6)
Let n be the refractive index of the prism,
then we can write,
n = \(\frac{sin i}{sin r}\) ……….(7)
Substituting eq (5) and eq (6) in eq (7),
n = \(\frac{\sin \frac{A+D}{2}}{\sin \frac{A}{2}}\)

Question 21.
Explain Rutherford’s, alpha particle scattering experiment.
റൂഥർഫോർഡിന്റെ ആൽഫാ പാർട്ടിക്കിൾ സ്കാറ്ററിംഗ് പരി ക്ഷണം വിവരിക്കുക.
Answer:
Rutherford’s scattering experiment

Experimental arrangement : a particles are incident on a gold foil (very small thickness) through a lead collimator. They are scattered at different angles. The scarred particles are counted by a particle detector.

Observations : Most of the alpha particles are scattered by small angles. Afew alpha particles are scattered at an ang|e greater than 90°.

Conclusions:
1) Major portion of thfe atom is empty space.
2) All the positive! charges of the atom are concentrated in a small portion of the atom.
3) The whole mass of the atom is concentrated in a small portion of the atom.

Rutherford’s model of atom:
1) The massive part of the atom (nucleus) is concentrated at the centre of the atom .
2) The nucleus contains all the positive charges of the atom.
3) The size of the nucleus is the order of 10^-15m.
4) Electrons move around the nucleus in circular orbits.
5) The electrostatic force of attraction (between proton and electron) provides centripetal force.

SECTION-D

Answer any 3 questions from 22 to 25. Each carries 4 scores.
22 മുതൽ 25 വരെയുള്ള ചോദ്യങ്ങളിൽ ഏതെങ്കിലും 3 എണ്ണത്തിന് ഉത്തരമെഴുതുക. 4 സ്കോർ വീതം. (3 × 4 = 12)

Question 22.
(a) What is an electric dipole?
വൈദ്യുത ഡൈപോൾ എന്നാലെന്ത്? (1)
Answer:
Electric dipole : A pair of equal and opposite charges separated by small distance is called electric dipole.

(b) Obtain the expression for the electric field intensity at a point on the axial line of an electric dipole.
വൈദ്യുത ഡൈപോളിന്റെ അക്ഷത്തിലുള്ള ബിന്ദുവിലെ വൈദ്യുത മണ്ഡല തീവ്രതയുടെ സമവാക്യം രൂപീകരി ക്കുക. (3)
Answer:
Electric field at a point on the axial line of an electric dipole : Consider an electric dipole of moment P = 2aq. Let ‘S’ be a point at a distance V from the centre of the dipole.

Electric field at ‘S’ due to point charge at ‘A’
E_A = \(\frac{1}{4 \pi \varepsilon_0}\) \(\frac{q}{(r+a)^2}\)
Directed as shown in figure.
Electric field at ‘S’ due to point charge at ‘B’
E_B = \(\frac{1}{4 \pi \varepsilon_0}\) \(\frac{q}{(r-a)^2}\)
Directed as shown in figure.
Therefore resultant electric field at ‘S’ And its magnitude
E = E_B + -E_A

The direction is along E_B
The field due to an electric dipole is directed from negative charge to positive charge along the axial line.

Question 23.
(a) Derive the expression for the torque on a . rectangular current loop in a uniform magnetic field with the help of a diagram. (2)
Answer:

Consider a rectangular coil PQRS of N turns which is suspended in a magnetic field, so that it can rotate (about yy1). Let ‘l’ be the length (PQ) and ‘b’ be the breadth (QR).
When a current l flows in the coil, each side produces a force. The forces on the QR and PS will not produce torque. But the forces on PQ and RS will produce a Torque.
Which can be written as
τ = Force × ⊥ distance ………..(1)
But, force = BI l ………(2) [since θ = 90°]
And from ∆QTR , we get
⊥ distance (QT) = b sin θ ………..(3)
Substituting the vales of eq (2) and eq (3) in eq(1) we get
τ = BIl b sin θ
= BIA sin θ [since lb = A (area)]
τ = IAB sin θ
τ = mB sin θ [since m = IA]
τ = m x B
If there are N turns in the coil, then

T = NIAB sin θ

(b) A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. What is the magnetic moment of this coil? (2)
Answer:
r = 10 cm = 10 × 10-2 m
I = 3.2A
N = 100
M = NIA .
m = 100 × 3.2 × 3.14 × (10 × 10^-2)²
= 10.046 Am²

Question 24.
(a) With a neat diagram, derive lens makers formula. (2)
Answer:
Consider a thin lens of refractive index n2 formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁.
Let an object ‘O’ is placed in the medium of refractive index n₁. Hence the incident ray OM is in the medium of refractive index n1 and the refracted ray MN is in the medium of refractive index n₂.

The spherical surface ABC (radius of curvature R₁) forms the image at I₁. Let ‘u’ be the object distance and ‘v₁’ be the image distance.
Then we can write,
This image I₁, will act as the virtual object for the surface ADC and forms the image at v.
Then we can write

(b) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (2)
Answer:
R₁ = 10 cm
R₂ = -15 cm
f = 12 cm
\(\frac{1}{f}\) = (n-1)[\(\frac{1}{R_1}\) – \(\frac{1}{R_2}\)]
\(\frac{1}{12}\) = (n-1)[\(\frac{1}{10}\) + \(\frac{1}{15}\)]
\(\frac{1}{12}\) = (n-1) [latex]\frac{15+10}{150}[/latex]
\(\frac{1}{12}\) = (n-1) × \(\frac{1}{6}\)
n – 1 = \(\frac{6}{12}\) = \(\frac{1}{2}\)
n = \(\frac{1}{2}\) + 1 = 1.5

Question 25.
(a) Give the classification of materials based on energy band diagram. (3)
Answer:
Classification on the basis of Energy bands
Conductors: Conduction band is partially filled and valence band is partially empty.

Or

Conduction band and valence band are overlaped so that E_g = 0 ev

Due to overlapping, electrons are partially filled in conduction band. These partially filled elec¬trons are responsible for current conduction.
Insulators

Conduction band is empty. Valence band may fully or partially filled. There is a wide energy gap between valence band and conduction band (E_g > 3ev).

Semiconductors : Conduction band may be empty or lightly filled. Valence band is fully filled. The energy gap is very small (< 3ev)

(b) Differentiate between intrinsic and extrinsic semiconductors. (1)
Answer:
A semiconductor in its pure form is called intrinsic semiconductor. The number of of free electrons is equal to number of holes. Dopped semi conductors (n type or p type) are called extrinsic semiconductors.

SECTION-E

Answer any 3 questions from 26 to 29. Each carries 5 scores. (3 × 5 = 15)

Question 26.
(a) Give the relation between electric field and potential. (1)
Answer:
E = \(\frac{-dv}{dx}\)

(b) Derive the expression for the potential due to an electric dipole. (2)
Answer:
Potential due to an electric dipole

Consider dipole of length ‘2a’. Let P be a point at distance r₁ from +q and r₂ from -q . Let r be the distance of P from the centre ‘O’ of the dipole. Let 0 be angle between dipole and line OP.

The potential due to +q , V₊ = \(\frac{1+q}{4 \pi \varepsilon_0 \quad r_1}\)

The potential due to -q , V_– = \(\frac{1-q}{4 \pi \varepsilon_0 \quad r_1}\)

Therefore total potential,
V = \(\frac{1+q}{4 \pi \varepsilon_0 \quad r_1}\) + \(\frac{1-q}{4 \pi \varepsilon_0 \quad r_1}\)
= \(\frac{1}{2}\) (\(\frac{1}{r_1}\) – \(\frac{1}{r_2}\))
= \(\frac{1}{2}\) (\(\frac{r_2-r_1}{r_1 r_2}\)) ……..(1)

From ∆ABC , we get (r₂ – r₁) = 2a cosθ
we can also take r₂= r₁ = r (since ‘2a’. is very small) Substituting these values in equation (1),we get
V = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0}\) (\(\frac{2 a \cos \theta}{r^2}\))
V = \(\frac{1, \mathrm{P} \cos \theta}{4 \pi \varepsilon_0 \quad \mathrm{r}^2}\) (Since P = q2a)
But \(\overrightarrow{\mathrm{P}}\) . \(\hat{\mathbf{r}}\) = P cosθ, where \(\hat{\mathbf{r}}\) = \(\frac{\vec{r}}{|r|}\)
V = \(\frac{1 \overrightarrow{\mathrm{P}} \cdot \hat{r}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

(c) Calculate the potential at a point due to a charge of 4×10^-7 located 9 cm away. (2)
Answer:
q = 4 × 10^-7 C
r = 9 cm = 9 × 10^-2 m
V = 9 × 10⁹ × \(\frac{4 \times 10^{-7}}{9 \times 10^{-2}}\) = 4 × 10⁴ V

Question 27.
(a) State Kirchhoff’s law. (2)
Answer:
First law (Junction rule): The total current en-tering the junction is equal to the total current leaving the junction.
Second law (loop rule) : In any closed circuit the algebraic sum of the product of the current and resistance in each branch of the circuit is equal to the net emf in,that branch.
OR
Total emf in a closed circuit is equal to sum of voltage drops.

(b) Obtain the balancing condition of Wheatstone’s bridge with the help of a diagram. (3)
Answer:
Four resistances P,Q,R and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂,I₃ and I₄ be the four currents passing through RR,Q and S respectively.

Working
The voltage across R.
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, V_AB = I₁P
The voltage across Q, V_BC = I₃Q ………(1)
The voltage across R, V_AD = I₂R
The voltage across S, V_DC = I₄S
The value of R is adjusted to get zero deflection in galvanometer. Under this condition,
I₁ = I₃ and I₂ = I₄ ………(2)
Using Kirchoffs second law in loopABDA and BCDB, we get
V_AB = V_AD …………(3)
and V_BC = V_DC ……….. (4)
Substituting the values from eq(1) in to (3) and (4), we get
I₁P = I₂R …………..(5)
and I₃Q = I₄S ………….(6)
Dividing Eq(5)byEq(6)
\(\frac{I_1 P}{I_3 Q}\) = \(\frac{\mathrm{I}_2 \mathrm{R}}{\mathrm{I}_4 \mathrm{~S}}\)
\(\frac{P}{Q}\) = \(\frac{R}{S}\) [since I₁ = I₃ and I₂ = I₄]
This is called Wheatstone condition.

Question 28.
(a) State the principle of a transformer. (1)
Answer:
Mutual Induction

(b) Explain the working of a transformer. (2)
Answer:

A transformer consists of two insulated coils wound over a core. The coil to which energy is given is called primary and that from which energy is taken is called secondary.

Working and mathematical expression : Let V₁, N₁ be the voltage and number of turns in the primary. Similarly let V₂, N₂ be voltage and number of turns in the secondary.

When AC is passed, a change in magnetic flux is produced in the primary. This magnetic flux passes through secondary coil.

If ϕ₁ and ϕ₂ are the magnetic flux of primary and secondary, we can write ϕ₁∝N₁ and ϕ₂∝N₂.
Dividing ϕ₁ and ϕ₂
\(\frac{\phi_1}{\phi_2}\) = \(\frac{N_1}{N_2}\)
[since ϕ is proportional to number of turns]
or ϕ = \(\frac{N_1}{N_2}\) ϕ₂
Taking differentiation on both sides we get
\(\frac{d \phi_1}{d t}\) = \(\frac{N_1}{N_2} \frac{d \phi_2}{d t}\); V₁ = \(\frac{N_1}{N_2}\) V₂
[since V₂ = \(\frac{-d \phi_2}{d t}\) and V₁ = \(\frac{-d \phi_1}{d t}\)]

i.e., \(\frac{V_1}{V_2}\) = \(\frac{N_1}{N_2}\)

(c) Differentiate between step up transformer and step down transformer. (2)
Answer:
Step up Transformer : If the output voltage is qreater than input voltaqe, the transformer, is called step up transformer. In a step up transformer N₂ > N₁ and V₂ > V₁.
Step down transformer: If the output voltage is less than the input voltage, then the transformer is called step down transformer. In a step down transformer N₂ < N₁ and V₂ < V₁

Question 29.
(a) Sate Huygens principle. (2)
Answer:
Huygen’s principle

1. Every point in a wavefront acts as a source of seconda ry wavelets.
2. The secondary wavelets travel with the same velocity as the original value.
3. The envelope of all these secondary wavelets gives a new wavefront.

(b) Explain the refraction of plane wave using Huygens principle. (3)
Answer:
AB is the incident wavefront and c, is the velocity of the wavefront in the first medium. CD is the refracted wavefront and c2 is the velocity of the wavefront in the second medium. AC is a plane separating the two media.

The time taken for the ray to travel from P to R is
t = \(\frac{PO}{C_1}\) + \(\frac{OR}{C_2}\)

O is an arbitrary point. Hence AO is a variable. But the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq.(2) should be zero.

AO (\(\frac{\sin i}{C_1}\) – \(\frac{\sin r}{C_2}\)) = 0
\(\frac{\sin i}{C_1}\) = \(\frac{\sin r}{C_2}\)
where 1 n2 is the refractive index of the second medium w.r.t. the first. This is the law of refraction.

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf March 2022 helps in understanding answer patterns.

Kerala Plus Two Maths Previous Year Question Paper March 2022

Time: 2 Hours
Total Score: 60 Marks

Part – I

A. Answer any 5 questions from 1 to 9. Each carries 1 score. (5 × 1 = 5)

Question 1.
Which of the following relations on A = {1, 2, 3} is an equivalence relation?
(a) {(1, 1), (2, 2), (3, 3)}
(b) {(1, 1), (2, 2), (3, 3), (1, 2)}
(c) {(1, 1), (3, 3), (1, 3), (3, 1)}
(d) None of these
Answer:
(a) {(1, 1), (2, 2), (3, 3)}

Question 2.
The value of sin^-1(sin(\(\frac{1}{2}\))) = …………………..
(a) \(\frac{1}{2}\)
(b) π – \(\frac{1}{2}\)
(c) –\(\frac{1}{2}\)
(d) \(\frac{\pi}{6}\)
Answer:
(a) \(\frac{1}{2}\)

Question 3.
If A is a 3 × 3 matrix, then |adj(A)| = ……………………..
(a) |A|
(b) |A|²
(c) |A|³
(d) 3|A|
Answer:
(b) |A|²

Question 4.
A fair, die is rolled. If the events are E = {1, 3, 5}, F = {2, 3}, then P(E|F) = …………
Answer:
P(E|F) = \(\frac{P(E \cap F)}{P(F)}\) = \(\frac{\frac{1}{6}}{\frac{2}{6}}\) = \(\frac{1}{2}\)

Question 5.
The area bounded by the curve y = 2x between x = 0, x = 2 and x-axis is ………………
Answer:
Required area = \(\int_0^2 y d x\)
= \(\int_0^2 2 x d x\) = \(\left[x^2\right]_0^2\) = 4 sq. units

Question 6.
Slope of the tangent to the curve y = x² + 1 at the point (2, 5) is ……………….
Answer:
y = x² + 1
\(\frac{d y}{d x}\) = 2x
\(\left.\frac{d y}{d x}\right]_{(2,5)}\) = 4
∴ Slope of tangent = 4

Question 7.
Write the vector from the point A(1, 3, 5) to B (4, 3, 2)
Answer:
\(\overrightarrow{A B}\) = 3î – 3k̂

Question 8.
Which of the following is a point on the plane 3x + 2y + 4z = 0?
(a) (1, 2, 1)
(b) (2, 3, 2)
(c) (2, 1, -2)
(d) (2, 1, 2)
Answer:
(c) (2, 1, -2)

Question 9.
Write the degree of the differential equation
2\(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\)) = 0
Answer:
Degree = 1

B. Answer all questions from 10 to 13. Each carries 1 score. (4 × 1 = 4)

Question 10.
The value of sin^-1(\(\frac{1}{\sqrt{2}}\)) = ………………….
Answer:
sin^-1(\(\frac{1}{\sqrt{2}}\)) = \(\frac{\pi}{4}\)

Question 11.
The vertices of a triangle are (0, 2), (0, 3), (4, 6), then area of the triangles ____________
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
Area = \(\frac{1}{2}\)\(\left|\begin{array}{lll}
0 & 2 & 1 \\
0 & 3 & 1 \\
4 & 6 & 1
\end{array}\right|\) = |\(\frac{1}{2}\)[4 × -1]|
= |\(\frac{-4}{2}\)| = |-2|
= 2 sq. units

Question 12.
Find the direction cosines of the vector 3î – 2ĵ + 5k̂
Answer:
l = \(\frac{3}{\sqrt{38}}\), m= \(\frac{-2}{\sqrt{38}}\), n = \(\frac{5}{\sqrt{38}}\)

Question 13.
Derivative of log (x³) is ………………….
Answer:
y = log (x³)
\(\frac{d y}{d x}\) = \(\frac{1}{\dot{x}^3}\) × 3x² = \(\frac{3}{x}\)

Part – II

A. Answer any 2 questions from 14 to 17. Each carries 2 scores. (2 × 2 = 4)

Question 14.
\(\left[\begin{array}{cc}
x+y & 2 \\
5+x & 8
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 2 \\
6 & 8
\end{array}\right]\), find x and y.
Answer:
5 + x = 6
∴ x = 1

x + y = 5
1 + y = 5
y = 4

Question 15.
The length x of a rectangle is increasing at the rate of 4 cm/s and the width y is decreasing at the rate of 5 cm/s. Find the rates of change of its area when x =10 cm and y = 5 cm.
Answer:
\(\frac{d x}{d t}\) = 4 cm/s. \(\frac{d y}{d t}\) = -5 cm/s
A = xy
\(\frac{d A}{d t}\) = x . \(\frac{d y}{d t}\) + y . \(\frac{d x}{d t}\)
= – 5x + 4y
\(\left.\frac{d A}{d t}\right]_{x=10, y=5}\) = -50 + 20 = -30
Area decreases at the rate of 30 cm2/s

Question 16.
Show that the function f(x) = x³ + 3x + 5 is strictly increasing on R.
Answer:
f(x) = x³ + 3x + 5
f'(x) = 3x² + 3
= 3(x² + 1) > 0 for all x ∈ R
∴ f(x) is increasing on R

Question 17.
Solve the differential equation \(\frac{d y}{d x}\) = \(\frac{2 x}{y^2}\)
Answer:
Given \(\frac{d y}{d x}\) = \(\frac{2 x}{y^2}\)
y² dy = 2x dx,
which is variable separable.
Solution is \(\int y^2 d y\) = \(\int 2 x d x\)
\(\frac{y^3}{3}\) = x² + C

B. Answer any 2 questions from 18 to 20. Each carries 2 scores. (2 × 2 = 4)

Question 18.
Find the value of λ if the vectors î – ĵ + k̂, 3î + ĵ + 2k̂ and î + λĵ – 3k̂ are coplanar.
Answer:
Let \(\bar{a}\) = î – ĵ + k̂, \(\bar{b}\) = 3î + ĵ + 2k̂, \(\bar{c}\) = î + λĵ – 3k̂
Since \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are coplanar,
[\(\bar{a}\), \(\bar{b}\), \(\bar{c}\)] = 0
\(\left|\begin{array}{ccc}
1 & -1 & 1 \\
3 & 1 & 2 \\
1 & \lambda & -3
\end{array}\right|\) = 0
1(- 3 – 2λ) + 1(-11) + 1(3λ – 1) = 0
– 3 – 2λ – 11 + 3λ – 1 = 0
– 15 + λ = 0
λ = 15

Question 19.
If y = x^{sin x} find \(\frac{d y}{d x}\)
Answer:
Let y = x^{sin x}
log y = sin x . log x
\(\frac{1}{y}\)\(\frac{d y}{d x}\) = sin x.\(\frac{1}{x}\) + log x cos x
\(\frac{d y}{d x}\) = y[\(\frac{\sin x}{x}\) + log x cos x]
= x^{sin x}[\(\frac{\sin x}{x}\) + log x cos x]

Question 20.
Find the integrating factor of the differential equation
x\(\frac{d y}{d x}\) – y = 2x²
Answer:
Given x \(\frac{d y}{d x}\) – y = 2x²

Part – III

A. Answer any 3 questions from 21 to 24. Each carries 3 scores. (3 × 3 = 9)

Question 21.
Express the matrix A = \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\) as the sum of a symmetric matrix and a skew symmetric matrix.
Answer:

Question 22.
R = {(x, y): x, y ∈ Z, (x – y) is an integer}. Show that R is an equivalence relation.
Answer:
Given R = {(x, y): x, y ∈ Z, x – y is an integer} 0 is an integer
⇒ x – x is an integer for all x ∈ Z
⇒ (x, x) ∈ R for all x ∈ Z

∴ R is reflexive
Let (x, y) ∈ R ⇒ x – y is an integer
⇒ -(x – y) is an integer
⇒ y – x is an integer
⇒ (y, x) ∈ R for all x, y ∈ Z

∴ R is symmetric
Let (x, y) ∈ R, (y, z) ∈ R
⇒ x – y is an integer and y – z is an integer
⇒ x – y + y – z is an integer
⇒ x – z is an integer
⇒ (x, z) ∈ R for all x, y, z ∈ R
∴ R is transitive
Hence R is an equivalence relation

Question 23.
Bag 1 contains 5 red and 3 black balls while another Bag 2 contains 3 red and 7 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag 2.
Answer:

Let E₁ : ball is drawn from bag I
E₂ : ball is drawn from bag II
A : Getting a red ball.

Question 24.
Consider the vector \(\vec{a}\) = 2î + ĵ + k̂ and \(\vec{b}\) = î + ĵ + k̂
(a) Find \(\vec{a}\) × \(\vec{b}\) (2)
(b) Find a unit vector perpendicular to \(\vec{a}\) and \(\vec{b}\) (1)
Answer:
(a) \(\vec{a}\) × \(\vec{b}\) = \(\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= î(0) – ĵ(1) + k̂(1)
= ĵ + k̂

(b) Unit vector perpendicular to \(\vec{a}\) and \(\vec{b}\)
= \(=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\) = \(\frac{-\hat{j}+\hat{k}}{\sqrt{2}}\)
= \(\frac{-1}{\sqrt{2}}\)ĵ + \(\frac{1}{\sqrt{2}}\)k̂

B. Answer any 2 questions from 25 to 27. Each carries 3 scores. (2 × 3 = 6)

Question 25.
Using elementary operations, find the inverse of the matrix A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
Answer:

Question 26.
If * is a binary operation of R defined by a * b = \(\frac{a b}{3}\),
(a) Find the identity element of*. (1)
(b) Find the inverse of 3. (2)
Answer:
(a) Let ‘e’ be the identity element.
∴ a * e = a
\(\frac{a e}{3}\) = a
e = 3

(b) Let ‘b’ be the inverse of 3
∴ 3 * b = e
\(\frac{3 b}{3}\) = 3
b = 3

Question 27.
Evaluate \(\int_0^2 x^2 d x\) as the limit of a sum.
Answer:
a = 0, b = 2, nh = b – a = 2
f(x) = x²
f(a) = f(0) = 0
f(a + h) = f(0 + h) = (0 + h)² = h²
f(a + 2h) = f(0 + 2 h) = (0 + 2 h)² = 4h²
f(a + 3h) = f(0 + 3h) = (0 + 3h)² = 9h²
f(a + (n – 1 )h = f(0 + (n – 1)h = (n – 1)²h²

Part – IV

A. Answer any 3 questions from .28 to 31. Each carries 4 scores. (3 × 4 = 12)

Question 28.
Show that 2 tan^-1(\(\frac{1}{2}\)) + tan^-1(\(\frac{1}{7}\)) = tan^-1(\(\frac{31}{17}\)).
Answer:

Question 29.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Answer:

Question 30.
(a) Discuss the continuity of the function (2)

(b) Verify Rolle’s theorem for the function
f(x) = 2x² – 12x + 1 in [2, 4] (2)
Answer:

LHL = 3(3) + 1 = 9 + 1 = 10
RHL = (3)2 +1 = 9 + 1 = 10
f(3) = 9 + 1 = 10
LHL = RHL = f(3)
∴ f(x) is continuous at x = 3
Hence f(x) is a continuous function.

(b) f(x) = 2x² – 12x + 1
Since f(x) is a polynomial function, it is continuous in [2, 4]
f'(x) = 4x – 12
∴ It is differentiable in (2, 4)
f(a) = f(2) = 8 – 24 + 1 = -15
f(b) = f(4) = 32 – 48 + 1 = -15
∴ f(a) = f(b)
∴ f'(c) = 0
⇒ 4c – 12 = 0
4c = 12
c = 3 ∈ (2, 4)
Hence Rolle’s theorem verified.

Question 31.
(a) Find the equation of the line passing through the pints (2, 1, 0) and (4, 4, 3)
(b) Find the equation of the plane which is perpendicular to the above line and passing through the point (1, 1, 2)
Answer:
(a) (x₁, y₁, z₁) = (2, 1, 0)
(x₂, y₂, z₂) = (4, 4, 3)
Equation of line is
\(\frac{x-x_1}{x_2-x_1}\) = \(\frac{y-y_1}{y_2-y_1}\) = \(\frac{z-z_1}{z_2-z_1}\) = λ
(ie) \(\frac{x-2}{2}\) = \(\frac{y-1}{3}\) \(\frac{z}{3}\) = λ

(b) < a, b, c > = < 2, 3, 3 >
(x₁, y₁, z₁,) = (1, 1, 2)
Equation of plan is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
2(x – 1) + 3(y – 1) + 3(z – 2) = 0
2x – 2 + 3y – 3 + 3z – 6 = 0
2x + 3y + 3z – 11 = 0

B. Answer any 1 question from 32 to 33. Carries 4 scores. (1 × 4 = 4)

Question 32.
Find the mean of the number obtained on a throw of an unbiased die. (4)
Answer:
Let X be the number obtained.
∴ X = 1, 2, 3, 4, 5, 6

Mean = Σx_ip_i = \(\frac{21}{6}\) = \(\frac{7}{2}\) = 3.5

Question 33.
Consider the planes 3x – 2y + z + 6 = 0 and 2x + y + 2z – 6 =0;
(a) Find the angle between the planes. (2)
(b) Find the equation of the plane passing through the line of intersection of above planes and through the point (0, 0, 0). (2)
Answer:
p₁ : 3x – 2y + z + 6 = 0
p₂ : 2x + y + 2z – 6 = 0

(b) Eqn. of plane is p₁ + λp₂ = 0
(3x – 2y + z + 6) + λ(2x + y + 2z – 6) = 0 ………….. (1)
Since it passes through (0,0,0)
6 + λ(-6) = 0
-6λ = -6
λ = 1
Sub in (1),
3x – 2y + z + 6 + 1(2x + y + 2z – 6) = 0 5x – y + 3z = 0

Part – V

Answer any 2 questions from 34 to 36. Each carries 6 scored. (2 × 6 = 12)

Question 34.
Solve the following system of equations by matrix method.
x – y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
Answer:
A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right]\)

∴ System of equations can be written as AX = B.
|A| = 1(8 – 6) + 1(0 + 9) + 2(0 – 6)
= 2 + 9 – 12
= -1 ≠ 0
∴ System is consistent
Unique solution is X = A^-1B
A₁₁ = 2, A₁₂ = -9, A₁₃ = -6
A₂₁ = 0, A₂₂, = -2, A₂₃ = -1
A₃₁ = -1, A₃₂, = 3, A₃₃ = -2

Question 35.
Find the following integrals:
(a) \(\int \frac{x}{(x+1)(x+2)} d x\) (3)
(b) \(\int_0^{\frac{\pi}{2}} \frac{\sin ^4 x}{\sin ^4 x+\cos ^4 x} d x\) (3)
Answer:

Question 36.
Solve the following linear programming problem graphically
Maximise Z = 3x + 2y
Subject to
x + 2y ≤ 10
3x + y ≤ 15
x, y ≥ 0
Answer:
x + 2y = 10

x	0	10
y	5	0

3x + y = 15

x	0	5
y	15	0

∴ z has maximum, when x = 5, y = 0
Maximum value = 15

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf March 2024 helps in understanding answer patterns.

Kerala Plus Two Maths Previous Year Question Paper March 2024

Time: 2 Hours
Total Score: 60 Marks

Answer any 6 questions from 1 to 8. Each carries 3 scores. (6 × 3 = 18)

Question 1.
Let R be a relation on a set A = {1, 2, 3, 4, 5, 6} defined as R = {(x, y : y = 2x – 1}.
(i) Write R in roster form and find its domain and range. (2)
(ii) Is R is an equivalence relation? Justify.(1)
Answer:
(i) R = {{x, y) : y = 2x – 1]
R = {(1, 1),(2, 3), (3, 5)}
Domain = {1,2, 3}
Range ={1,3,5}

(ii) R is not an equivalence relation
∵ (2, 2)∉ R and hence not reflexive.

Question 2.
A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\) show that
A² – 5A + 7I = O
(Where I is the identity matrix)
Answer:
A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\)
A² = A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2\end{array}\right]\)
\(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\)

∴ A² – 5A + 7I = O

Question 3.
(i) Check the continuity of the function
f (x) = 2x + 3 at x =1. (1)
(ii) Determine the value of k so that the function

is continuous at x = 5.
Answer:
(i) \(\lim _{x \rightarrow 1} f(x)[latex] = [latex]\lim _{x \rightarrow 1} 2 x+3[latex]
2 × 1 + 3 = 5
f(1) = 2 × 1 + 3 = 5
∴ [latex]\lim _{x \rightarrow 1} f(x)[latex] = f(1)
Hence, f is continuous at x = 1

(ii) Given f(x) is continuous at x = 5
∴ [latex]\lim _{x \rightarrow 5^{-}} f(x)\) = \(\lim _{x \rightarrow 5^{+}} f(x)\)
⇒ \(\lim _{x \rightarrow 5^{-}} k x+1\) = \(\lim _{x \rightarrow 5^{+}} 3 x-5\)
⇒ k × 5 + 1 = 3 × 5 – 5
⇒ 5k + 1 = 15 – 5 = 10
⇒ 5k = 9
k = \(\frac{9}{5}\)

Question 4.
(i) Find the principal value of sin^-1(\(\frac{1}{2}\)) (1)
(ii) Find the value of
tan^-1 [2 cos(2 sin^-1\(\frac{1}{2}\)) (2)
Answer:
(i) Let y = sin^-1(\(\frac{1}{2}\))
sin y = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
∴ y = \(\frac{\pi}{6}\)
Hence, the principal value of
sin^-1(\(\frac{1}{2}\)) = \(\frac{\pi}{6}\)

(ii) tan^-1[2 cos(2 sin^-1\(\frac{1}{2}\))]
= tan^-1[2 cos(2 × \(\frac{\pi}{6}\))]
= tan^-1[2 cos\(\frac{\pi}{3}\)]
= tan^-1 [2 × \(\frac{1}{2}\)]
= tan^-1(1)
= \(\frac{\pi}{4}\)

Question 5.
(i) Which of the following function is increasing in its domain:
(A) sin x
(B) cos x
(C) -2x
(D) log x (1)
(ii) Find the intervals in which the function f given by f(x) = x² – 4x + 6 is
(a) increasing
(b) decreasing. (2)
Answer:
(i) Option (D) log x

(ii) f(x) = x² – 4x + 6
f'(x) = 2x – 4
f'(x) = 0 ⇒ 2x – 4 = 0
⇒ 2x = 4
⇒ x = 2
∴ We have two intervals, (-∞, 2) and (2, ∞).
Consider (-∞, 2)
Put x = 0 ∈(-∞, 2) in f'(x)
f'(0) = -4 < 0
∴ f is strictly decreasing in (-∞, 2)
Consider (2, ∞),
Put x = 3 ∈ (2, ∞) in f'(x)
f'(3) = 2 × 3 – 4
= 2 > 0
∴ f is strictly increasing in (2, ∞)

Question 6.
(i) If θ is the angle between two non zero vectors \(\vec{a}\) and \(\vec{b}\) and |\(\vec{a}\) . \(\vec{b}\)| = |\(\vec{a}\) × \(\vec{a}\)| then θ = ___________ . (1)
(ii) Find the projection of the vector \(\vec{a}\) = î – ĵ on the vector \(\vec{b}\) = î + ĵ (2)
Answer:
(i) θ = \(\frac{\pi}{4}\)
∵ |\(\vec{a}\) . \(\vec{b}\)| = |\(\vec{a}\) × \(\vec{b}\)|
⇒ ab cos θ = ab sin θ
⇒ cos θ = sin θ
⇒ θ = \(\frac{\pi}{4}\)

(ii) \(\vec{a}\) = î – ĵ
\(\vec{b}\) = î + ĵ
Projection of \(\vec{a}\) on \(\vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
= \(\frac{(\hat{i}-\hat{\mathrm{j}}) \cdot(\hat{i}+\hat{\mathrm{j}})}{\sqrt{1^2+1^2}}\)
= \(\frac{0}{\sqrt{2}}\) = 0

Question 7.
Let A and B are independent events with P(A) = 0.3, P(B) = 0.4, find
(i) P(A∩B)
(ii) P(A∪B)
(iii) P(A/B)
Answer:
(i) A and B are independent events.
∴ P (A∩B) = P(A) × P(B)
= 0.3 × 0.4
= 0.12

(ii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.3 + 4 – 0.12
= 0.58

(iii) P(A / B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{P(\mathrm{~B})}\)
= \(\frac{0.12}{0.4}\) = 0.3

Question 8.
Evaluate \(\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x\)
Answer:
\(\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x\)
Put u = x⁵ + 1
du = 5x⁴dx
x = 1 ⇒ w = 2
x = -1 ⇒ u = 0

Answer any 6 questions from 9 to 16. Each carries 4 scores. (6 × 4 = 24)

Question 9.
(i) What is the minimum number of ordered
pairs to form a reflexive relation on a set of 4 elements? (1)
(ii) Let A = R – {3}, B= R – {1}
Consider the function f : A B
defined by f(x) = \(\frac{x – 2}{x – 3}\)
Check whether f is one-one and onto. (3)
Answer:
(i) 4 ordered pairs

(ii) f(x) = \(\frac{x – 2}{x – 3}\)
f : A → B,
A = R – {3},B = R – {1}
For x₁, x₂ ∈ A
f(x₁) = f(x₂) ⇒ \(\frac{x_1-2}{x_1-3}\) = \(\frac{x_2-2}{x_2-3}\)
⇒ (x₁ – 2) (x₂ – 3) = (x₁ – 3)(x₂ – 2)
⇒ x₁ x₂ – 3x₁ – 2x₂ + 6 = x₁ x₂ – 2x₁ – 3x₂ + 6
⇒ -3x₁ – 2x₁ = -2x₁ – 3x₁
⇒ 3x₂ – 2x₂ = 3x₁ – 2x₁
⇒ x₁ = x₂

Question 10.
(i) If A is a skew symmetric matrix then A’= _____________ (1)
(ii) Express the matrix
A = \(\left[\begin{array}{ccc}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right]\)
as the sum of a symmetric and skew symmetric matrix. (3)
Answer:
(i) A¹ = -A

which is the symmetric matrix.

which is the skew symmetric matrix.

Question 11.
A wire of length 28 m is cut into two pieces, one of the pieces is to be made into a square and other into a circle. What should be the length of the two pieces so that the combined area of square and circle is minimum?
Answer:
Let length of the wire made into square be
x and made into circle is 28 – x
ie, perimeter of square = x
4a = x
a = \(\frac{x}{4}\)
⇒ Area, A₁ = \(\frac{x^2}{16}\)
Circumference of circle = 28 – x
2πr = 28 – x
r = \(\frac{28-x}{2 \pi}\)
⇒ Area, A₂ = π × (\(\frac{28-x}{2 \pi}\))² = (\(\frac{28-x}{2 \pi}\))²
Combained area A = A₁ + A₂
= \(\frac{x^2}{16}\) + \(\frac{(28-x)^2}{4 \pi}\)
\(\frac{d A}{d x}\) = \(\frac{2 x}{16}\) + \(\frac{2(28-x) \times(-1)}{4 \pi}\)
= \(\frac{x}{8}\) – \(\frac{28-x}{2 \pi}\)
\(\frac{d A}{d x}\) = 0 ⇒ \(\frac{x}{8}\) – \(\frac{28-x}{2 \pi}\) = 0
⇒ \(\frac{x}{8}\) = \(\frac{28-x}{2 \pi}\)
⇒ 2π x = 224 – 8x
⇒ 2π x + 8x = 224,
⇒ 2π x + 8x = 224
⇒ x = \(\frac{224}{2 \pi+8}\) = \(\frac{112}{\pi+4}\)
\(\frac{d^2 A}{d x^2}\) = \(\frac{1}{8}\) + \(\frac{1}{2 \pi}\) > 0
∴ \(\frac{d^2 A}{d x^2}\) > 0 at x = \(\frac{112}{\pi+4}\)
ie., Area is minimum at x = \(\frac{112}{\pi+4}\)
∴ Length of two pieces are,
\(\frac{112}{\pi+4}\) for square
28 – \(\frac{112}{\pi+4}\) = \(\frac{28 \pi+112-112}{\pi+4}\)
= \(\frac{28 \pi}{\pi+4}\) for circle

Question 12.
(i) Write the order and degree of the differential equation
xy\(\frac{d^2 y}{d x^2}\))³ + (\(\frac{d y}{d x}\))² – sin (\(\frac{d y}{d x}\)) = 0 (1)
(ii) Find the integrating factor of the differential equation
x\(\frac{d y}{d x}\) + 2y = x² (2)
(iii) Solve the differential equation
x\(\frac{d y}{d x}\) + 2y = x² (1)
Answer:
(i) Order = 2
Degree = not defined

(ii) x\(\frac{d y}{d x}\) + 2y = x²
\(\frac{d y}{d x}\) + \(\frac{2 y}{x}\) = x
which is of the form \(\frac{d y}{d x}\) + py = Q
P = \(\frac{2}{x}\) Q = x
Integrating Factor = \(e^{\int p d x}\)
= \(e^{\int \frac{2}{x} d x}\)
= \(e^{2 \log x}\)
= \(e^{\log x^2}\)
= x²

(iii) y.IF = \(\int Q \cdot I F d x\)
y.x² = \(\int x \cdot x^2 d x\)
x²y = \(\int x^3 d x\)
x²y = \(\frac{x^4}{4}\) + c

Question 13.
If \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = î + 2ĵ + 3k̂
(i) Find \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\). (1)
(ii) Find \(\vec{a}\) unit vector perpendicular to both \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\) (3)
Answer:
(i) \(\vec{a}\) = î + ĵ + k̂ \(\vec{b}\) = î + 2ĵ + 3k̂
\(\vec{a}\) + \(\vec{b}\) = (î + ĵ + k̂) + (î + 2ĵ + 3k̂) = 2î + 3j + 4k̂
\(\vec{a}\) – \(\vec{b}\) = {î + ĵ + k̂) + (î + 2ĵ + 3k̂)
= – ĵ – 2k̂

(ii) Vector perpendicular to both
\(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\)
(\(\vec{a}\) + \(\vec{b}\)) × (\(\vec{a}\) + \(\vec{b}\))
= \(\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
0 & -1 & -2
\end{array}\right|\)
= î(- 6 + 4) – ĵ(- 4 – 0) + k̂(- 2 – 0)
= – 2î + 4j – 2k̂
|(\(\vec{a}\) + \(\vec{b}\)) × (\(\vec{a}\) + \(\vec{b}\))| = \(\sqrt{(-2)^2+4^2+(-2)^2}\)
= \(\sqrt{4+16+4}\)
= \(\sqrt{24}\)
∴ Unit vector perpendicular to both \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\)
= \(\frac{-2 \hat{i}+4 \hat{j}-2 \hat{k}}{\sqrt{24}}\)

Question 14.
Find the shortest distance between the lines
\(\vec{r}\) = (î + ĵ + λ(2î – ĵ + k̂) and
\(\vec{r}\) = (2î + ĵ – k̂) + µ(3î + 5ĵ + 2k̂)
Answer:
\(\vec{r}\) = (î + ĵ) + λ(2î – ĵ + k̂)
\(\vec{r}\) = (2î + ĵ – k̂) + μ(3î – 5ĵ + 2k̂)
\(\vec{a}\)₁ = î + ĵ \(\vec{a}\)₂ = 2î + ĵ – k̂
\(\vec{b}\)₁ = 2î – ĵ + k̂ \(\vec{b}\)₂ = 3î – 5ĵ + 2k̂
\(\vec{b}\)₁ × \(\vec{b}\)₂ = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|\)
= î(- 2 + 5) – ĵ(4 – 3) + k̂(- 10 + 3)
= 3î – ĵ – 7k̂
|\(\vec{b}\)₁ × \(\vec{b}\)₂| = \(\sqrt{3^2+(-1)^2+(-7)^2}\)
= \(\sqrt{9+1+49}\)
= \(\sqrt{59}\)
∴ Shortest distance = \(\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|\)
= \(\left|\frac{(\hat{i}-\hat{k}) \cdot(3 \hat{i}-\hat{j}-7 \hat{k})}{\sqrt{59}}\right|\)
= \(\frac{3+0+7}{\sqrt{59}}\) = \(\frac{10}{\sqrt{59}}\)

Question 15.
In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 45% of the bolt. Of their output 5%, 4% and 2% are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that is manufactured by machine B?
Answer:
Let E₁, E₂, E₃ be the events that bolts produced by machines A, B, C respectively.
P(E₁) = \(\frac{25}{100}\), P(E₂) = \(\frac{35}{100}\), P(E₃) = \(\frac{45}{100}\)
Let F be the event that bolt is defective.
P(F|E₁) = \(\frac{5}{100}\)
P(F|E₂) = \(\frac{4}{100}\)
P(F|E₃) = \(\frac{2}{100}\)

Question 16.
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}\) + \(\frac{y^2}{9}\) = 1 using integration.
Answer:

Answer any 3 questions from 17 to 20. Each carries 6 scores. (3 × 6 = 18)

Question 17.
Solve the following system of equations by matrix method:
2x – 3y + 5z = 11
3x + 2y – 4z = -5
x + y – 2z = -3
Answer:
A = \(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\) B = \(\left[\begin{array}{l}
11 \\
-5 \\
-3
\end{array}\right]\) X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
AX = B ⇒ x = A^-1B
A^-1 = \(\frac{{adj} \mathrm{A}}{|\mathrm{~A}|}\)
A₁₁ = 0 A₁₂ = 0 A₁₃ = 1
A₂₁ = -1 A₂₂ = -9 A₂₃ = -5
A₃₁ = 2 A₃₂ = 23 A₃₃ = 13

= 2[- 4 + 4] + 3(- 6 + 4) + 5 (3 – 2)
= 0 – 6 + 5
= -1

Question 18.
(i) sin x + cos y = xy find \(\frac{d y}{d x}\) (2)
(ii) x = a cos³ t ; y = a sin³ t find \(\frac{d y}{d x}\) (2)
(iii) If y = (sin^-1 x)² then show that
(1 – x²)\(\frac{d^2 y}{d x^2}\)) – x (\(\frac{d y}{d x}\)) = 2 (2)

Answer:
(i) sin x + cos y = xy
Differentiating with respect to x
cos x – sin y \(\frac{d y}{d x}\) = x\(\frac{d y}{d x}\) + y. 1
(x + sin y)\(\frac{d y}{d x}\) = cos x – y
\(\frac{d y}{d x}\) = \(\frac{\cos x-y}{x+\sin y}\)

(ii) x = a cos 3t y = a sin 3t
\(\frac{d x}{d t}\) = a × 3cos t × – sin t = -3a cos t sin t
\(\frac{d y}{d t}\) = a × 3sin t × cos t = 3a sin t cos t
∴ \(\frac{d y}{d x}\) = \(\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \sin t}\)
= \(\frac{-\sin t}{\cos t}\) = – tan t

(iii) y = (sin^-1 x)²
\(\frac{d y}{d x}\) = 2 sin ^-1 × \(\frac{1}{\sqrt{1-x^2}}\)
\(\sqrt{1-x^2}\) \(\frac{d y}{d x}\) = 2 sin^-1 x
Squaring,
(1 – x²) (\(\frac{d y}{d x}\))² = 4(sin^-1 x)²
(1 – x²) (\(\frac{d y}{d x}\))² = 4y
Differentiating
(1 – x²)2\(\frac{d y}{d x}\) \(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\))² × -2x = 4\(\frac{d y}{d x}\)
÷ by 2\(\frac{d y}{d x}\)
(1 – x²)\(\frac{d^2 y}{d x^2}\) – x\(\frac{d y}{d x}\) = 2

Question 19.
(i) Find \(\int \frac{x-1}{(x-2)(x-3)} d x\) (3)
(ii) Prove that \(\int_0^{-\frac{\pi}{4}} \log (1+\tan x) d x\) = \(\frac{\pi}{8}\)log 2 (3)
Answer:
(i) \(\int \frac{x-1}{(x-2)(x-3)} d x\)
\(\frac{x-1}{(x-2)(x-3)}\) = \(\frac{\mathrm{A}}{x-2}\) + \(\frac{\mathrm{B}}{x-3}\)
x – 1 = A(x – 3) + B(x – 2)
Put x = 3
3 – 1 = A(3 – 3) + B(3 – 2)
2 = B
Put x = 2
2 – 1 = A(2 – 3) + B(2 – 2)
1 = -A ⇒ A = -1
∴ \(\frac{x-1}{(x-2)(x-3)}\) = \(\frac{-1}{x-2}\) + \(\frac{2}{x-3}\)
Integrating on both sides,
\(\int \frac{x-1}{(x-2)(x-3)} d x\) = \(\int \frac{-1}{x-2} d x\) + \(\int \frac{2}{x-3} d x\)
= -log|x – 2| + 2log|x – 3| + C

Question 20.
Solve the following linear programming problem graphically:
Maximise Z = 60x + 15
Subject to the constraints
x + ≤ 50
3x + y ≤ 90
x ≥ 0,y ≥ 0
Answer:
z = 60x + 15y
x + y = 50

x	0	50
y	50	0

3x + y = 90

x	0	30
y	90	0

Corner points	z = 60x + 15y
A (0, 0) B (30, 0) C (20, 30) D (0, 50)	z = 0 z = 1800 z = 1650 z = 750

∴ Maximum value of z is 1800 at (30,0)

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf March 2023 helps in understanding answer patterns.

Kerala Plus Two Maths Previous Year Question Paper March 2023

Time: 2 Hours
Total Score: 60 Marks

Answer any 6 questions from 1 to 8. Each carries 3 scores. (6 × 3 = 18)

Question 1.
Let A = {1, 2, 3}
B = {2, 3, 4, 5}
and f : A → B defined by f(x) = {(x, y) : y = x + 1}
(i) Write f in roster from. (1)
(ii) Check whether y is one-one and onto. (2)
A = {1, 2, 3}
B = {2, 3, 4, 5}
Answer:
f(x) = {(1, 2), (2, 3),(3, 4)}
(ii) f is one-one but not onto

Question 2.
Find the matrices X and Y so that
X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\) (3)
Answer:
Adding these two equations,

Question 3.
Find the equation of line through the points A (1, 3) and B(0, 0) using determinants. (3)
Answer:
Let (x, y) be a point on line AB. Then A, B, P are collinear.
∴ \(\left|\begin{array}{lll}
1 & 3 & 1 \\
0 & 0 & 1 \\
x & y & 1
\end{array}\right|\) = 0
1(0 – y) – 3(0 – x) + 1(0 – 0) = 0
– y + 3x + 0 = 0
3x – y = 0

Question 4.
Find the value of ‘a’ and ‘b’ if

is a continuous function. (3)
Answer:
At x = 3
LHL = RHL = f(3)
3a + b = 10 ________ (1)
At x = 4
LHL = RHL = f(4)
4a + b = 20 ________ (2)
Solving (1) and (2)
a = 10 , b = -20

Question 5.
Find the local maxima or local minima of the function f(x) = sin x + cos x, 0< x < \(\frac{\pi}{2}\) if it exists.
f(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\) (3)
Answer:
f(x) = sin x + cos x
f'(x) = cos x – sin x
f”(x) = – sin x – cos x
f'(x) = 0 ⇒ x = \(\frac{\pi}{4}\)
f”π/4 = –\(\frac{1}{\sqrt{2}}\)–\(\frac{1}{\sqrt{2}}\) = –\(\frac{-2}{\sqrt{2}}\) = -√2 < 0
∴ f(x) has a maximum at x = \(\frac{\pi}{4}\)
Max value = f(\(\frac{\pi}{4}\)) = sin\(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)
= √2

Question 6.
Consider the vectors:
\(\vec{a}\) = î + 2ĵ + 3k̂
\(\vec{b}\) = 3î + 2ĵ + k̂
(i) Find \(\vec{a}\) . \(\vec{b}\) (1)
(ii) Find the angle between \(\vec{a}\) and \(\vec{b}\) (2)
Answer:
(i) \(\vec{a}\) . \(\vec{b}\) = 3 + 4 + 3 = 10

(ii) cos = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\) = \(\frac{10}{\sqrt{1+4+9} \sqrt{9+4+1}}\)
= \(\frac{10}{\sqrt{14} \cdot \sqrt{14}}\) = \(\frac{10}{14}\) = \(\frac{5}{7}\)
θ = cos-1(\(\frac{5}{7}\))

Question 7.
Find the vector and Cartesian equation of the line passing through (1, 2, 3) and parallel to the vector 3î + 2ĵ – 2k̂ (3)
Answer:
Vector form:-
\(\vec{r}\) =(î + 2ĵ + 3k̂) + λ(3î +2ĵ – 2k̂)
Cartesian form:-
\(\frac{x – 1}{3}\) = \(\frac{y – 2}{2}\) = \(\frac{z – 3}{-2}\)= λ

Question 8.
If A and B are two events such that P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\) and P(A’∪B’) = \(\frac{1}{4}\)
(i) FindP(A∩B) (1)
(ii) Check whether A and B are independent events. (2)
Answer:
Given P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\)
(i) P(A’∪B’) = \(\frac{1}{4}\)
(ie) P[(A∩B)’] = \(\frac{1}{4}\)
P(A∩B) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)

(ii) P(A).P(B) = \(\frac{1}{2}\) × \(\frac{7}{12}\) = \(\frac{7}{24}\) ≠ P(A∩B)
∴ A and B are not independant.

Answer any 6 questions from 9 to 16. Each carries 4 scores.

Question 9.
(i) Let R be a relation on the set Z, et of integers defined by
R = {(a, b): 2 divides (a – b)}
Choose the right answer:
(A) (2, 4) ∈ R
(B) (3, 8) ∈ R
(C)(7, 6) ∈ R
(D) (8, 7) ∈ R (1)
R = {(a, b): 2 divides (a – b)}
(ii) Check the above relation R is an equivalence relation. (3)
Answer:
(i) (A) (2, 4) ∈ R
(ii) R = {{a, b): 2 divides (a – b)}
2 divides 0,
(ie)2divides (a – a) ∀a ∈ Z
∴ (a, a) ∈ R ∀a ∈ Z
∴ R is reflexive.
Let (a, b) ∈ R
⇒ 2 divides (a – b)
⇒ 2 divides – (a – b)
⇒ 2 divides (b – a)
∴ (b, a) ∈ R
∴ R is symmetric
Let (a, b) ∈ R, (b, c) ∈ R
⇒ 2 divides (a – b), 2 divides (b – c)
⇒ 2 divides (a – b) + (b – c)
⇒ 2 divides (a – c)
⇒ (a, c) ∈ R
∴ R is transitive
Hence R is an equivalence relation.

Question 10.
(i) The principal value of sin-1(\(\frac{1}{2}\)) is ___________ (1)
(ii) Find tan-1[2 cos(2 sin-1\(\frac{1}{2}\))] (3)
Answer:
(i) \(\frac{\pi}{6}\)

(ii) tan^-1[2 cos(2 sin^-1\(\frac{1}{2}\))]
= tan^-1[2 cos(2\(\frac{\pi}{6}\))]
= tan^-1(2 cos \(\frac{\pi}{3}\))
= tan^-1(2 × \(\frac{1}{2}\))
= tan^-1(1) = \(\frac{\pi}{4}\)

Question 11.
(i) Which, among the following is not true?
(A) (A’)’ = A
(B) (A + B)’ = A’ + B’
(C) (AB)’ = A’.B’
(D) (kA)’ = k.A’ (1)
(ii) If A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), then verify that (A + A’) is symmetric and (A – A’) is skew -symmetric.
[A’ denotes the transposes of the matrix A] (3)
Answer:
(i) (C) (∵ (AB)’ = B’.A’)

(ii) A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), A’ = \(\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]\)
A + A’ = \(\left[\begin{array}{ll}
2 & 11 \\
11 & 14
\end{array}\right]\) it is symmetric.
A – A’ = \(\left[\begin{array}{ll}
0 & -1 \\
1 & 0\end{array}\right]\) it is skewsymmetric

Question 12.
Using integration find the area enclosed by the ellipse \(\frac{x^2}{9}\) + \(\frac{y^2}{4}\) = 1 (4)
Answer:
\(\frac{x^2}{9}\) + \(\frac{y^2}{4}\) = 1
\(\frac{y^2}{4}\) = 1 – \(\frac{x^2}{9}\)
= \(\frac{9-x^2}{9}\),
y² = \(\frac{4}{9}\) (9 – x²)
y = \(\frac{2}{3}\)\(\sqrt{9-x^2}\)

Question 13.
(i) The degree of the differential equation (1)
(\(\frac{d^2 y}{d x^2}\))³ + (\(\frac{d y}{d x}\))² + (\(\frac{d y}{d x}\)) + 1 = 0
(A) 1
(B) 2
(C) 3
(D) not defined
(ii) Solve the differential equation (3)
\(\frac{d y}{d x}\) = (1 + x²)(1 + y²).
Answer:
(i) (C) 3

(ii) Given \(\frac{d y}{d x}\) = (1 + x²) (1 + y² )
\(\frac{d y}{1+y^2}\) = (1 + x²)dx (which is variable separable)
∴ Solution is \(\int \frac{d y}{1+y^2}\) = \(\int\left(1+x^2\right) d x\)
tan^-1 y = x + \(\frac{x^3}{3}\) + C
y = tan^-1(x + \(\frac{x^3}{3}\) + C )

Question 14.
Consider the vectors:
\(\vec{a}\) = î – 7ĵ + 7k̂
\(\vec{b}\) = 3î – 2ĵ + 2k̂
(i) Find \(\vec{a}\) × \(\vec{b}\) (2)
(ii) Find the unit vector perpendicualr to both \(\vec{a}\) and \(\vec{b}\) (1)
(iii) Find the area of parallelogram whose adjacent sides are \(\vec{a}\) and \(\vec{a}\) (1)
Answer:

(iii) Area of parallelogram =| \(\vec{a}\) x \(\vec{b}\) |
= 19√2 sq. units

Question 15.
Find the shortest distance between the lines:
\(\vec{r}\) = (î + 2j + k̂) + λ(î + ĵ + k̂) and
\(\vec{r}\) = (2î – ĵ + 4k̂) + µ(2î + ĵ + k̂)
Answer:
Let \(\vec{a}\) = î + 2ĵ + k̂,
\(\vec{b}\)₁ = î + ĵ + k̂
\(\vec{a}\)₁ = 2î – ĵ + 4k̂,
\(\vec{a}\)₂ = 2î + ĵ + 2k̂
\(\vec{a}\)₂ – \(\vec{a}\)₁ = î – 3ĵ + 3k̂

Question 16.
Bag – I contains 3 red and 4 black balls, while Bag – II contains 5 red and 6 black balls. One of the bags is selected at random and a ball is drawn out of it. If the ball drawn is found to be red, find the probability that it was from Bag – II. (4)
Answer:

E₁ : bag II is selected
E₂ : bag II is selected
A : Getting a red ball

Answer any 3 questions from 17 to 20. Each carries 6 scores. (3 × 6 = 18)

Question 17.
Solve the following system of equations using matrix method: (6)
x + y + z
2x + y + z = 4
2x – y + z = 2
Answer:,
A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 1 & 1 \\
2 & -1 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(=\left[\begin{array}{l}
3 \\
4 \\
2
\end{array}\right]\)
∴ System of equations can be written as
AX = B
|A| = 1(1 + 1) – 1(2 – 2) + 1(-4)
= 2 – 4= -2 ≠ 0
∴ System is consistent.
Unique solution is X = A^-1B

Question 18.
(i) If y = x^x find \(\frac{d y}{d x}\) (2)
(ii) If y = x = at² and y = 2at, find \(\frac{d y}{d x}\) (2)
(iii) The radius of a circle is icnreasing uniformly at the rate of 5 cm.sec. Find the rate at which the area of the circle is increasing when the aradius is 8 cm. (2)
Answer:
(i) y = x^x
log y = x.log x
Differentiating with respect to x,
\(\frac{1}{y} \frac{d y}{d x}\) = x × \(\frac{1}{x}\) + log x = 1 + log x
\(\frac{d y}{d x}\) = y(1 + log x) = x^x (1 + log x)

(ii) x = at² y = 2at
\(\frac{d x}{d t}\) = 2at \(\frac{d y}{d t}\) = 2a
\(\frac{d y}{d x}\) = \(\frac{d y / d t}{d x / d t}\) = \(\frac{2 a}{2 a t}\) = \(\frac{1}{t}\)

(iii) \(\frac{d r}{d t}\) = 5 cm / sec
A = πr²
\(\frac{d A}{d t}\) = 2πr.\(\frac{d r}{d t}\)
= 2πr × 5
= 10πr/
\(\left.\frac{d A}{d t}\right]_{\mathrm{r}=8}\) = 10 × π × 8 = 80 π cm²/sec

Question 19.
(i) \(\int \frac{1}{x^2-a^2} d x\) = ____________ (1)
(ii) Find : \(\int \frac{1}{x^2+4 x-5} d x\) (2)
(iii) Evaluate : \(\int_2^3 \frac{x}{1+x^2} d x\) (3)
Answer:
(i) \(\int \frac{1}{x^2-a^2} d x\) = \(\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\)

(ii) x² + 4x – 5 = x² + 4x + 4 – 4 – 5
=(x + 2)² – 9

= \(\frac{1}{2}\) log (\(\frac{10}{5}\))
= \(\frac{1}{2}\) log 2

Question 20.
Solve the LPP graphically: (6)
Maximize
Z = 250x + 75y
Subject to
5x + y ≤ 100
x + y ≤ 60
x ≥ 0
y ≥ 0
Answer:
5x + y = 100

x	0	20
y	100	0

x + y = 60

x	0	60
y	60	0

Z has max
When x = 10, y = 50
Max, Value = 6250