काठमाडौँमा भएको विगत दुई दिनको आन्दोलनले देशलाई गम्भीर संकटतर्फ धकेलेको संकेत देखिएको छ। आन्दोलनका क्रममा २४ जनाले ज्यान गुमाएका छन् भने खरबौँ मूल्यबराबरको निजी तथा सार्वजनिक सम्पत्ति नष्ट भएको अनुमान गरिएको छ।

विगतका जनआन्दोलन, मधेश आन्दोलन, सरकार–माओवादी युद्ध वा २०७२ सालको विनाशकारी भूकम्पजस्ता घटनाको अनुभवले पनि यस्तो प्रकृतिको अराजकता देखाएको थिएन। तर पछिल्ला घटनाले नेपालको राजनीतिक स्थायित्व मात्र होइन, सामाजिक एकता र अर्थतन्त्र समेत गहिरो संकटमा पर्ने आशंका बढाएको छ।

भौतिक सम्पत्तिको ठूलो क्षति

राजधानीसहितका विभिन्न शहरमा भएका आगजनी र तोडफोडबाट खरबौँ बराबरको पूर्वाधार ध्वस्त भएको छ। बजार, यातायात साधन, सरकारी भवनदेखि निजी घरपरिवारसम्म क्षतिग्रस्त भएका छन्। यस क्षतिको असर निकट भविष्यमा अर्थतन्त्रलाई थप जटिल बनाउने विशेषज्ञहरूको ठम्याइ छ।

हिंसाको दुवै रूप चिन्ताजनक

प्रदर्शनमा राज्यको बल प्रयोगले ज्यान गुमाउनु र आन्दोलनकारी भीडबाट वृद्ध नेताहरू कुटपिट हुनु—दुवै घटनाले आक्रोश मात्र होइन, विभाजन र वैमनस्यताको बीउ रोपेको देखिन्छ। विशेष गरी सामाजिक सञ्जालमा युवापुस्ताले हिंसा र आगजनीलाई समर्थन वा रमाइलोको रूपमा प्रस्तुत गरिरहेको प्रवृत्तिले दीर्घकालीन अस्थिरताको जोखिम देखाएको छ।

नेतृत्व र दिशाको अस्पष्टता

आन्दोलन सुरुमा जायज आक्रोशको स्वरूपमा देखा परे पनि हाल यसको नेतृत्व र दिशाबारे स्पष्टता देखिएको छैन। आन्दोलनपश्चात् संक्रमणकालीन राजनीतिक मार्गचित्र कोर्ने बेला सक्षम, दृष्टिवान् नेतृत्व उदाएन भने देश अझ ठूलो अनिश्चिततामा फस्ने चिन्ता बढेको छ।

राजनीतिक दलहरूको भूमिकामाथि प्रश्न

विश्लेषकहरूका अनुसार वर्तमान संकटमा सरकारको अहंकार, प्रमुख दलहरूको सत्ता–केन्द्रित राजनीति र प्रतिपक्षको गैरजिम्मेवार भूमिकाले पनि योगदान गरेको छ। एमाले अध्यक्ष केपी ओलीदेखि सत्तासाझेदारीमा रहेका कांग्रेस र अन्य दलहरू सबै समान रूपमा आलोचनाको घेरामा परेका छन्।

अत: दुई दिनमा देखिएको अराजकता र ठूलो मानवीय क्षतिले मुलुकलाई गम्भीर मोडमा पुर्याएको छ। हिंसा र प्रतिहिंसाको संस्कृतिलाई रोक्न, अराजकतालाई नियन्त्रण गर्न र विवेकशील नेतृत्व खडा गर्न सकिएन भने नेपालको राजनीतिक यात्रामा अझ गहिरो संकट देखा पर्ने विज्ञहरूको चेतावनी छ।

Here’s an essay tailored for a Class 10 board exam (SEE), with around 300 words to match the depth and structure expected at that level:

Discipline is one of the essential trait that helps people achieve success, self-control, and responsibility. To put it simply, discipline is the process of putting bad deeds behind oneself, growing in moral character, being ethical, thinking positively, being professional, charitable, and acting morally. Discipline is crucial for young people, particularly students, as it helps them form positive habits and gets them ready for the future. In order to make students understand the values of diligence, curiosity, intelligence, punctuality and academic success requires discipline. A well-behaved student pays attention in class, studies consistently, and completes his/her assignment on time. Disciplined students prevent themselves from last-minute stress  before tests by practicing time management efficiently. Maintaining discipline helps students learn more and more which in turn increases their confidence. Additionally, discipline in academics helps students learn how to prioritize the task which is most essential in adult life.

Discipline is equally important in personal life too. It enhances the person's ability to resist distractions. Practicing discipline helps everyone to follow healthy routines like exercising regularly, eating balanced diet, and get required sleep. This supports in physical health as well as mental and social health.

In societal context, discipline is the master key to create a better environment. By following social norms and ideologies, a person can contribute to safety and harmony. For instance, in a neighborhood if people manage the garbage properly, it reduces the risk of health issues and tension between people. Discipline fosters cooperation between people and helps in societal development.

In conclusion, discipline is a guiding principle that leads to a fulfilling and balanced life. It is essential for personal growth, academic success and societal harmony. Developing discipline from a young age equips individuals to face life’s challenges, ultimately leading them to a successful and meaningful life.

Sarah and Jamie stood on their tiptoe and gave the card from their school to the tall man behind the counter of a farm.The man frowned and made a face.

"So you're looking for some work, And who are you?" "I'm Sarah. I'm twelve. This is my brother Jamie; he's eleven."

"Shouldn't you be at school?"

"We are on holiday. And would like to earn some money and support our parents." The man lifted his hat and scratched his head. "Working in the farm is a quite tough job for kids like you. I need somebody big and strong to work here," he said.

"There must be some work for us in this big farm. Could you please find one?" Sarah pleaded.

"And right. Let's see. Do you know what a Cabbage White is?" the man asked.

"Yes, it's a beautiful white butterfly that lays its eggs on cabbages. And those eggs change into caterpillars.", answered Sarah.

"And do you know what the caterpillars do?"

"They eat the cabbage leaves!" shouted Jamie.

(a) Where were Sarah and Jamie? 

→Sarah and Jamie were in front of the counter of a farm/in the farm/at the farm.

(b) Why did Sarah and Jamie go to the tall man?

→They went to the tall man in the farm looking for some work/in search of some work/ to find some work.

(c) Why did the man scratch his head?

→The man scratched his head because he could not think of the type of job he could give to the kids.

(d) What do you mean by Cabbage White?

→It's a beautiful white butterfly that lays it's eggs on cabbage.

(e) What do you think their work would be there?

→I think Jamie and Sarah would remove the cabbage White. 

When you want to make your home secure, you lock the door. When you want to prevent your bike from being stolen, you use a bike lock. When you want to protect your valuable possessions, you keep them in a safe. But what do you do to protect your identity online?

Modern life is fundamentally intertwined with the internet. Nearly every daily task now has the opportunity for online integration, and almost everyone owns multiple devices, including laptops, phones, tablets, smart watches, smart TVs and many more. The more accounts and devices you have online, the greater the potential is for criminals to access your personal information and take advantage of you.

Online safety is important no matter your age or life stage, but there are particular concerns for certain vulnerable groups like children, teenagers and senior citizens.

When it comes to internet safety, it's best to start with the basics. Here are some tips to guide you as you evaluate your online habits and make small changes to ensure your privacy and security on the internet.

While creating a new password, pay attention to strong password requirements. Use different user IDs or password combinations for different accounts, Make the password more complicated by combining letters, numbers, and special characters and create hard-to-guess passwords, and keep them private. Change your passwords quite often. Don't share your passwords with other people. Make sure passwords and password hints are stored securely. Record passwords in an encrypted file on your computer, or select another secure password storage method. Avoid writing them down. Prevent "auto-login", uncheck "Keep me signed in" and "Remember my password" options.

A. Write "TRUE" for true and "FALSE" for the false statement. (5x1=5)

(i) We use door lock to save our bike from being stolen. FALSE

(ii) Internet has become an essential part in our daily life. TRUE

(iii) There is a high chance of cyber crimes if we have many accounts and devices. TRUE

(iv) It is needless to ensure our privacy on the Internet. FALSE

(v) We should write passwords to remember in the future. FALSE

B. Answer the following questions.(5x1 =5)

(i) How do you save your valuable things?

→We save our valuable things by keeping them in a safe.

(ii) Who are prone to be victimized on cyber world? 

→Vulnerable groups like children, teenagers and senior citizens are prone to be victimized on cyber world.

(iii) What do you need to do for your cyber safety? 

→We need to learn the basic internet safety tips and make small changes to ensure privacy and security.

(iv) Who is this text for?

→This text is intended for all internet users.

(v) Who do you think is responsible for cyber crime? How?

→The internet user himself/herself is responsible for cyber crime because of his/her carelessness.

Lumbini is located in Rupandehi district of Southern Nepal close to the Indian border and inscribed on the UNESCO world Heritage list. It is renowned for being the official birth place of Siddhartha Gautam who became the Buddha.

The main site is known as Lumbini Development Zone which is sealed in by a perimeter fence. There are lots of beautiful places to visit some of them are listed as:

• Lumbini Temple/ Temple of Mayadevi : Walk inside the temple built over the spot Buddha was born.

• Mayadevi Pond : Where Buddha's mother is said to have bathed before giving birth to him.

• Lumbini Museum : Historic manuscripts and item from the area.

• Bodhi Tree : impressive tree covered in prayer flags beside the sacred pond.

• Ashoka Pillar : stone pillar dedicated to emperor Ashokan.

• Buddhist Temples : 2nd century BC temples to modern day temples.

• Crane Sanctuary : This natural reserve has been put aside as a safe area for local birds, including the Saris Cranes.

• We can get access to Lumbini by flight or by bus. One can fly to Bhairahawa from Kathmandu; then take a bus, taxi or rickshaw to Lumbini which is about 30 minutes away. By bus, it is cheaper and more frequent route to Lumbini. You can approach Lumbini in a reasonable amount of time by bus. From Kathmandu it's 6-9 hours; from Pokhara it's 6-9 hours and from Chitwan it's 5-6 hours.

A. Fill in he blanks with the correct words from the text above.(5x1 =5)

(a) Lumbini lies in Rupandehi district of Southern Nepal.

(b) Lumbini is well-known for being official birth place of Siddhartha Gautam.  

(c) Siddhartha Gautam's mother took bath in mayadevi  pond before giving birth to him.

(d) Ashoka Pillar is dedicated to emperor Ashokan.

(e) It takes about 30 minutes to get to Lumbini from Bhairahawa by bus.

B. Answer the following questions. (5x1=5)

(a) Write down any two attractions of Lumbini.

→Lumbini temple and Mayadevi pond are two attractions of Lumbini.

(b) How is Lumbini Development Zone sealed in?

→Lumbini development zone is sealed in by a perimeter fence.

(c) How much time does a bus take to get to the birth place of Gautam Buddha from Kathmandu?

→It takes 6 to 9 hours by a bus to get into birth place of Gautam Buddha from Kathmandu.

(d) What does the word 'manuscripts' refer to in the text? 

→'Manuscripts' refers to the  texts,documents or book written by hand.

(e) Do you think Lumbini is one of the world's major tourist-inviting places? Why?

→Yes, because it is birth place of Gautam Buddha and also listed in UNESCO world heritage site.

Marie was born in 1867 in Warsaw, Poland, where her father was a professor of physics. At an early age, she displayed a brilliant mind and a blithe personality. Her great exuberance for learning prompted her to continue with her studies after high school. She became disgruntled, however, when she learned that the university in Warsaw was closed to women. Determined to receive a higher education, she defiantly left Poland and in 1891 entered the Sorbonne, a French University, where she earned her Master's degree and doctorate in Physics.

Marie was fortunate to have studied at the Sorbonne with some of the greatest scientists of her day, one of whom was Pierre Curie. Marie and Pierre were married in 1895 and spent many productive years working together in the Physics laboratory. A short time after they discovered radium, Pierre was killed by a horse-drawn wagon in 1906. Marie was stunned by this horrible misfortune and endured heart breaking anguish. The fact that she had two young daughters to raise by herself greatly increased her distress.

Marie was asked to succeed her husband as a Physics professor at the Sorbonne. She was the first woman to be given a professorship at the world-famous university. In 1911 she received the Nobel Prize in chemistry for isolating radium. Although she eventually suffered a fatal illness from her long exposure to radium, she never became disillusioned about her work.

A. Match the words in column 'A' with the words having similar meanings in column B. (5x1=5) 

Column A                     Column B

Anguish→Mental and physical pain

Displayed   → showed 

Fortunate   → lucky

Eventually → finally

Brilliant      → exceptionally talented

B. Write 'TRUE' for the true statement and 'FALSE' for the false statement (5x1=5)

(a) Marie was the daughter of a professor. TRUE 

(b) She got her doctorate in physics from Poland. FALSE

(c) Pierre Curie was a great scientist of his day. TRUE 

(d) Marie brought up her two young daughters after her husband's death. TRUE 

(e) Marie got the Nobel Prize in Physics in 1911. FALSE

C. Answer the following questions. (5x1=5)

(a) When was Marie born?

→Marie was born in 1867.

(b) Why did Marie leave Poland?

→Marie left Poland in order to receive higher education.

(c) How was Marie's husband killed?

→Marie's husband was killed by horse - drawn wagon.

(d) What are two different reasons greatly distressed her according to the text?

→Death of her husband and rearing of children were reasons to distress her.

(e) How does this lesson motivate you?

→It motivates us by giving a lesson that determination and hard work pays off.

Clues : walking - physical activity - minimizes heart diseases - connects one with people and places - mental health - sound mind in sound body - controls obesity

→Walking is a good form of physical activity. It is a cardiovascular exercise that has a number of advantages. Firstly, it helps reduce the risk of heart diseases by keeping the heart healthy and active. It reduces the excessive body fat. Walking allows us to connect with both people and places by promoting social interactions and a sense of belonging. It gives us an opportunity to visit new places that greatly fosters mental health. Consequently, we have a sound mind in a sound body. Lastly, regular walking definitely controls obesity by burning calories and maintaining a healthy weight. All these benefits make walking simple but effective way to nurture overall health and well-being. Walking is therefore an excellent and no cost physical activity for people of all ages and fitness levels.

Clues: Mr. ABC Rai ............. a six (6) years child of XYZ school ............ returning home from school through the jungle ........ lost in the jungle ........... after 3 days rescued by a group of woodcutter ............ parents heartily thanked the woodcutters.

→Lost Child Rescued After 3 Days

By a local correspondent, Jhapa

1 June, 2025

A 6-year-old child (named Mr. ABC Rai) who went missing while he was going home from his school through a forest road, was ultimately rescued by a group of woodcutters.

The child was reportedly playing with his classmates during the lunch break when he wandered off unnoticed. Local authorities, along with concerned residents, launched a comprehensive search operation for three days to locate the child. The community was urged to remain vigilant and report any suspicious activity to the authorities.

A miracle happened after 3 days. A group of woodcutters rescued the child from the dense forest where he is reported to have taken the shortcut before went missing. The parents as well as the local authorities and the members of public greatly thanked the woodcutters for rescuing the child.

This incident reminds the school staff to be vigilant all the time in the school by keeping the main gate always locked.

→Hospital Rules for Patients

• Patients should always carry their previous reports, prescription slips, health insurance card and old OPD cards. It saves time and helps the doctor understand the case.
• Whether in a government hospital or a private clinic, patient should take the given medicines on time and come for follow-up if told.

• If the patient feel any discomfort or side effects from the prescribed medication, he/she should speak up to the doctor and don't keep it within him/herself.

• Patients must listen to the instructions given by hospital staffs and don't break the queue while registration. 

• The patients and visitors should not smoke or chew tobacco or spit anywhere within the hospital periphery.

• Patients should come to emergency services for real emergencies only — not for regular check-ups.

• One shouldn't offer or give bribe to hospital staffs to get faster service.

• Patient's shouldn't enter the ICU, X-Ray room, maternity ward or operation theatre without permission.

→Buddhashanti — 2, Jhapa

1 June, 2025

Dear Ankit,

I am sure this letter finds you well and in good spirits. I wanted to express my reflections on an issue that has been occupying my thoughts recently - the movement of young people and professionals from Nepal to foreign countries.

Every year, a large number of our countrymen leave Nepal in search of better opportunities overseas. Although this migration can foster personal and professional advancement for those individuals, it also has several extensive consequences for our home nation. On the positive side, the remittances sent back to Nepal by these emigrants make a substantial contribution to our economy. Nevertheless, the adverse effects cannot be overlooked. The brain drain resulting from the departure of skilled professionals causes a shortage of talent in essential sectors such as healthcare, education, and engineering. This impedes our nation's growth and advancement.

In order to tackle these challenges, I think it is essential to develop more opportunities within Nepal. The government ought to allocate resources towards education, healthcare, and infrastructure to foster a supportive atmosphere for progress. Furthermore, promoting entrepreneurship and aiding small enterprises can generate employment and lessen the urge to migrate. We, as members of society, should likewise cultivate a sense of patriotism and inspire our young people to take part in the advancement of the nation.

I'm eager to hear your opinions on this issue. Let's collaborate to discover solutions that will benefit our cherished Nepal. Stay safe and take care.

Warm regards — Your friend Anjan

(a) Let me play, ............... ? (supply the correct question tag)

→Let me play, will you?

(b) Has the man had a bath? (change into statement)

→The man has had a bath.

(c) She has been working here for five years. (Change into 'How long' question)

→How long has she been working here for?

(d) Don't make a noise ! I .......... (read) (Use the correct tense of the verb read)

→Don't make a noise! I am reading.

(e) "Do not touch that paper, "said the inspector. (Change into indirect speech)

→The inspector told us not to touch that paper.

(f) Did they finalize the date? (Change into passive voice)

→Was the date finalized by them?

Once there was a well in a forest. One day, a fox was walking by ......... (a) ......... (a/ an/ the/ no article) well. Suddenly, he slipped .......... (b) .......... (into/ on/ to/ from) the well. The well was very deep, ......... (c) ........ (so/ however/ therefore/ so that) he could not come out of the well. He made lots of efforts, ........... (d) ......... (didn't he/ don't he/ doesn't he/ isn't he) ? At the same time, a thirsty goat .............. (e) ........ (has passed/ pass/ was passing/ is passing) by the well and saw the fox in the well. The goat asked the fox ......... (f) ......... (what he was doing/was he doing/ what he did/what he does) there. The clever fox replied that he had been there to drink water. He further said, the water there was very good. Hearing that, the goat ..... (g) ..... (jumped/ was jumped/ is jumped/ jump) into the well at once. The fox stepped onto the back of the goat and jumped out of the well easily. The poor goat ......... (h) .........(was/ were/ is/ are) left to die there in the well. The persuasive words of the clever fox made the goat …….. (i) ……… (jump/ to jump/jumped/had jumped) into the well. If the goat had thought properly, it .......... (j) ......... would not have died/ would die/ will die/ to die) there.

THE END 

SEE Optional Mathematics New Specification Grid 2080

MODEL QUESTION 2080

Subject: Optional Mathematics

Class: 10

Full Marks: 75

Time: 3 hrs

All the questions are compulsory.

Note: Due to limitation of blogger, I was unable to use the vector symbol on question no. 9, 17 and 32. So, add vector symbols accordingly while you practice this set.

Group A [10 × 1 = 10]

1. Define trigonometric function.

   ✎ Trigonometric function is the function of an angle expressed as the ratio of two sides of a right-angled triangle. The most common trigonometric functions are sine, cosine, and tangent.

   For example:

   sin θ = P/h, cos θ = b/h, and tan θ = P/b

2. What is arithmetic mean between two numbers a and b?

   ✎ The arithmetic mean (AM) between two numbers a and b is:

   AM = (a + b) / 2

3. Write the name of the set of numbers which is continuous.

   ✎ The set of real numbers is continuous.

4. If matrix A = [[a, b], [c, d]], what is the value of |A|?

   ✎ The value of |A| is:

   |A| = ad - bc

5. If the angle between two straight lines is θ and their slopes are m₁ and m₂ respectively, write the formula for tan θ.

   ✎ Here, the formula for tan θ is:

   tan θ = ±(m1 - m2) / (1 + m1m2)

6. Which geometric figure will be formed if a plane intersects a cone parallel to its base?

   ✎ A circle will be formed if a plane intersects a cone parallel to its base.

7. Express sin 2A in terms of tan A.

   ✎ sin 2A can be expressed in terms of tan A as:

   sin 2A = (2 tan A) / (1 + tan² A)

8. Define angle of elevation.

   ✎ The angle of elevation is the angle between a horizontal line from the observer and the line of sight to an object that is above the horizontal line.

   
9. What is the scalar product of two vectors a and b if the angle between them is θ?

   ✎ Here, the scalar product of given vectors is:

   a · b = |a||b|cosθ

10. If P' is image of P and r is radius of inversion circle with center O in an inversion transformation, write the relations of OP, OP', and r.

    ✎ The relations of OP, OP', and r is:

    OP × OP' = r²

Group ‘B’ [8 × 2 = 16]

11. If 2x³ − 7x² + x + 10 = (x − 1)Q(x) + R, find the remainder R and quotient Q(x).

    ✎ Here, 2x³ − 7x² + x + 10 = (x − 1)Q(x) + R

    Comparing (x − 1) with (x − a), we get a = 1.

    Now, using Synthetic Division:

                
    x³   x²   x     x°
    2   −7    1    10
    ↓    2   −5    −4    
    2   −5   −4     6 = R
    x²   x    x°
    
                

    Now,

    Quotient = Q(x) = 2x² − 5x − 4

    Remainder = R = 6

12. Write down the inequality represented by the shaded region in the adjoining figure.

    ✎ From the given figure, the boundary line passes through the points (4, 0) and (0, −3). So, the equation of boundary line is:

    y − y₁ = (y₂ − y₁) / (x₂ − x₁) (x − x₁)

    or, y − 0 = (−3 − 0) / (0 − 4) (x − 4)

    or, y = (3/4)(x − 4)

    or, 4y = 3x − 12

    ∴ 3x − 4y = 12

    Since, the half plane with the boundary line contains origin, so the inequality represented by the shaded region is:

    3x − 4y ≤ 12

    

13. Find the determinants D and D₂ of coefficients of x and y by using Cramer's rule from the equations 4x − 5y = 2 and 3x + 4y = 48.

    ✎ Here,

    4x − 5y = 2  ...(i)

    3x + 4y = 48  ...(ii)

    Now, arranging the coefficients and constant term, we get:

    Coeff. of x	Coeff. of y	Constant
    4	-5	2
    3	4	48

    Now,

    D:

    | 2 -5 |

    | 48 4 |

    = 2×4 − 48×(−5) = 8 + 240 = 248

    D₂:

    | 4 2 |

    | 3 48 |

    = 4×48 − 3×2 = 192 − 6 = 186

14. Find the slopes of two straight lines 3x + 4y = 5 and 6x + 8y + 7 = 0 and write the relationship between them.

    ✎ Slope of 3x + 4y = 5 is m₁ = -coeff. of x / coeff. of y = -3 / 4

    Slope of 6x + 8y + 7 = 0 is m₂ = -coeff. of x / coeff. of y = -6 / 8 = -3 / 4

    Since m₁ = m₂, the given lines are parallel.

15. Convert sin6A.cos4A into sum or difference of sine or cosine.

    ✎ We know that, 2sinAcosB = sin(A+B) + sin(A−B)

    Now,

    sin6Acos4A = (1/2) × 2sin6Acos4A

    = (1/2)[sin(6A + 4A) + sin(6A − 4A)]

    = (1/2)[sin10A + sin2A]

16. If 2sinθ = √3, find the value of θ. (0° ≤ θ ≤ 180°)

    ✎ Here,

    2sinθ = √3

    or, sinθ = √3 / 2

    or, sinθ = sin60°

    ∴ θ = 60°

    Again,

    sinθ = sin60° = sin(180° − 60°) = sin120°

    ∴ θ = 120°

    Thus, the values of θ are 60° and 120°.

17. O is the origin in the given figure. If a and b are the position vectors of the points A and B, show that the position vector of the point P is p = 1/2 (a + b).

    ✎ Here,

    
    

    OA = a

    ,

    OB = b

    , and

    OP = p

    
    
    

    Now,

    
    

    AP = PB

    
    or,
    

    AO + OP = PO + OB

    
    or,
    

    -OA + OP = -OP + OB

    
    or,
    

    -a + p = p + b

    
    or,
    

    p + p = a + b

    
    or,
    

    2p = a + b

    
    or,
    

    p = (a + b)/2

    
    ∴

    p = (a + b)/2

18. In a series, the first quartile (Q₁) = 35 and third quartile (Q₃) = 75, find the quartile deviation and its coefficient.

    ✎ Here, Q₁ = 35 and Q₃ = 75
    Now,
    Quartile Deviation (QD) = (Q₃ - Q₁) / 2
    = (75 - 35) / 2
    = 40 / 2
    = 20
    
    Coefficient of Quartile Deviation = (Q₃ - Q₁) / (Q₃ + Q₁)
    = (75 - 35) / (75 + 35)
    = 40 / 110
    = 0.36

    Group ‘C’ [11 × 3 = 33]

19. If two functions are f(x) = (2x + 5) / 8 and g(x) = 3x - 4, find (f ∘ g)^-1(3).

    ✎ Here, f(x) = (2x + 5) / 8 and g(x) = 3x - 4
    Now,
    f ∘ g(x) = f(g(x))
    = f(3x - 4)
    = [2(3x - 4) + 5] / 8
    = (6x - 8 + 5) / 8
    = (6x - 3) / 8
    Let f ∘ g(x) = (6x - 3) / 8 = y
    
    To find (f ∘ g)^-1(x), interchanging the roles of x and y and solving for y,
    6y - 3 / 8 = x
    or, 6y - 3 = 8x
    or, y = 8x + 3 / 6
    or, (f ∘ g)^-1(x) = 8x + 3 / 6
    So, (f ∘ g)^-1(3) = 8 x 3 + 3 / 6 = 9 / 2
    

20. Solve by graphical method: 2x² + x - 6 = 0

    ✎ Here, 2x² + x - 6 = 0
    or, 2x² = 6 - x
    Let y = 2x² = 6 - x
    Then, y = 2x²   ... (i)
    y = 6 - x   ... (ii)
    
    Equation (i) represents the parabola with vertex at origin and equation (ii) represents a straight line. The point of intersection of the parabola and straight line gives the solution.
    
    From equation (i), y = 2x²
    

    x	-2	-1	0	1	2
    y	8	2	0	2	8

    
    

    Plotting these points on the graph and joining them, we get a parabola with vertex at origin.

    
    
    From equation (ii), y = 6 - x
    

    x	0	2
    y	6	4

    
    

    Plotting these points on the graph and joining them, we get a straight line.

    
    
    
    

    From the graph, the points of intersection of the parabola and straight line are (-2, 8) and (1.5, 4.5).
    ∴ x = -2, 1.5

21. For a real valued function f(x) = 2x + 3, find the values of f(2.99), f(3.01), and f(3). Is this function continuous at x = 3?

    ✎ Here, f(x) = 2x + 3
    Now,
    f(2.99) = 2 × 2.99 + 3 = 8.98
    f(3.01) = 2 × 3.01 + 3 = 9.02
    f(3) = 2 × 3 + 3 = 9
    

    When x = 2.99 → 3⁻, f(x) = 8.98 → 9

    ∴ lim_x→3⁻ f(x) = 9
    When x = 3.01 → 3⁺, f(x) = 9.02 → 9
    ∴ lim_x→3⁺ f(x) = 9
    When x = 3, f(3) = 9
    ∴ lim_x→3⁻ f(x) = lim_x→3⁺ f(x) = f(3) = 9
    Thus, function f(x) is continuous at x = 3.

22. By using the matrix method solve the following systems of equations:

    3x + 5y = 11, 2x - 3y = 1
    
    ✎ Here,
    3x + 5y = 11
    2x - 3y = 1
    
    Writing in matrix form, x

    

23. Find the equations of pair of lines represented by equation 6x² - xy - y² = 0 and also find the angle between them.

    ✎ Here,
    6x² - xy - y² = 0
    or, 6x² - (3 - 2)xy - y² = 0
    or, 6x² - 3xy + 2xy - y² = 0
    or, 3x(2x - y) + y(2x - y) = 0
    or, (2x - y)(3x + y) = 0
    
    Thus, the two separate equations are: 2x - y = 0 and 3x + y = 0

    Comparing 6x² - xy - y² = 0 with ax² + 2hxy + by² = 0, we get:

    a = 6, 2h = -1 → h = -1/2 and b = -1

    Now, The angle between the pair of lines is given by:

    tan θ = ± 2√(h² - ab) / (a + b)

    Substituting values:

    tan θ = ± 2√((-1/2)² - 6(-1)) / (6 - 1)
    = ± 2√(1/4 + 6) / 5
    = ± 2√(25/4) / 5
    = ± (2 × 5 / 2) / 5
    = ± 1

    Taking Positive Sign

    tan θ = 1
    θ = tan⁻¹(1)
    θ = 45°

    Taking Negative Sign

    tan θ = -1
    θ = tan⁻¹(-1)
    θ = tan⁻¹(tan (180° - 45°))
    θ = 135°

    Thus, the angle between them is 45° and 135°.

24. Prove that: tan A + 2tan 2A + 4cot 4A = cot A
    ✎ Solution

    LHS:

    tan A + 2tan 2A + 4cot 4A
    = tan A + 2tan 2A + 4 / tan 4A
    = tan A + 2tan 2A + 4 / tan 2(2A)
    = tan A + 2tan 2A + 4 / (2tan 2A / (1 - tan² 2A))
    = tan A + 2tan 2A + (2(1 - tan² 2A)) / (tan 2A)
    = tan A + (2tan² 2A + 2 - 2tan² 2A) / (tan2A)
    = tan A + (1 - tan² A) / tan A
    = (tan² A + 1 - tan² A) / tan A)
    = 1 / tan A
    = cot A
    = RHS
    Proved.

25. If A + B + C = π, prove that: sin²A − sin²B + sin²C = 2sinA cosB sinC

    ✎ Solution:

    Here:
    A + B + C = π
    ⇒ A + B = π − C
    ⇒ sin(A + B) = sin(π − C)
    ⇒ sin(A + B) = sin C

    Now:
    LHS = sin²A − sin²B + sin²C
    = (1/2)[2sin²A − 2sin²B] + sin²C
    = (1/2)[(1 − cos2A) − (1 − cos2B)] + sin²C
    = (1/2)[−cos2A + cos2B] + sin²C
    = (1/2)[cos2B − cos2A] + sin²C
    = (1/2). 2sin[(2A + 2B)/2] sin[(2A − 2B)/2] + sin²C
    = sin(A + B) sin(A − B) + sin²C
    = sin C sin(A − B) + sin²C
    = sinC [sin(A − B) + sinC]
    = sinC [sin(A − B) + sin(A + B)]
    = sinC . 2sinA cosB
    = 2sinA cosB sinC
    = RHS
    Proved.

26. From a place at the ground level in front of a tower, the angles of elevation of the top and bottom of a flagstaff 6m high situated at the top of the tower are observed to be 60° and 45°, respectively. Find the height of the tower and the distance between the base of the tower and the point of observation.

    ✎ Solution:

    
    Let BD be the tower, AD is the flagstaff, and BC is the distance between the base of the tower and the point of observation C.
    Given:
    AD = 6m
    ∠BCA = 60° (angle of elevation of the top of the flagstaff)
    ∠BCD = 45° (angle of elevation of the bottom of the flagstaff)
    BD = x (say) Now, in ∆ DCB:
    tan 45° = x / BC
    ⇒ 1 = x / BC
    ⇒ BC = x

    Again, in ∆ACB:
    tan 60° = (x + 6) / x
    ⇒ √3 = (x + 6) / x
    ⇒ 1.732x = x + 6
    ⇒ 1.732x − x = 6
    ⇒ 0.732x = 6
    ⇒ x = 6 / 0.732
    ⇒ x = 8.20

    Thus, the height of the tower is 8.20 m, and the distance between the base of the tower and the point of observation is 8.20 m.

27. Find a 2×2 matrix which transforms the unit square

    

28. Find the mean deviation from mean and its coefficient from given data:

    Marks Obtained	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
    No. of students	2	3	6	5	4

    ✎ Here,

    Marks	No. of students (f)	Mid-value (x)	fx	|D| = |x – x̄|	f|D|
    0 – 10	2	5	10	23	46
    10 – 20	3	15	45	13	39
    20 – 30	6	25	150	3	18
    30 – 40	5	35	175	7	35
    40 – 50	4	45	180	17	68

    N = Σf = 20      Σfx = 560

    Now,

    ∴ Mean = x̄ = Σfx / N = 560 / 20 = 28

    ∴ Mean Deviation = M.D. = Σf|D| / N = 206 / 20 = 10.3

    ∴ Coefficient of Mean Deviation = M.D. / x̄ = 10.3 / 28 = 0.37

29. Find the standard deviation and coefficient of variation from given data:

    ✎ Here, taking a = 25 and h = 10

    Age	No. of persons (f)	Mid-value (x)	d = (x – a) / h	fd	fd²
    0 – 10	4	5	-2	-8	16
    10 – 20	6	15	-1	-6	6
    20 – 30	10	25	0	0	0
    30 – 40	20	35	1	20	20
    40 – 50	6	45	2	12	24
    50 – 60	4	55	3	12	36

    N = Σf = 50      Σfd = 30      Σfd² = 102

    Now,

    Standard Deviation (σ) = h √(Σfd² / N - (Σfd / N)²)

    = 10 √(102 / 50 - (30 / 50)²)

    = 10 √(2.04 - 0.36)

    = 10 √1.68

    = 10 × 1.296

    = 12.96

    Group ‘D’ [4 × 4 = 16]

30. The sum of three terms in an arithmetic series is 24. If 1, 6, and 18 are added to them respectively, the results are in geometrical series, find the terms.

    ✎ Let the three consecutive terms in an arithmetic be a – d, a, a + d.

    It is given that the sum of these numbers is 24.

    a – d + a + a + d = 24

    or, 3a = 24

    or, a = 24 / 3

    ∴ a = 8

    If 1, 6, and 18 are added to these terms respectively, then a – d + 1, a + 6, a + d + 18 are in geometric series.

    Substituting a = 8, we get:

    a – d + 1 = 8 – d + 1 = 9 – d

    a + 6 = 8 + 6 = 14

    a + d + 18 = 8 + d + 18 = 26 + d

    i.e., 9 – d, 14, 26 + d are in geometric series.

    So, 14² = (9 – d)(26 + d)

    or, 196 = 234 + 9d – 26d – d²

    or, 196 = 234 – 17d – d²

    or, d² + 17d + 196 – 234 = 0

    or, d² + 17d – 38 = 0

    or, (d + 19)(d – 2) = 0

    Either, (d + 19) = 0 ⇒ d = –19

    Or, (d – 2) = 0 ⇒ d = 2

    When d = –19:

    a – d = 8 – (–19) = 27

    a = 8

    a + d = 8 – 19 = –11

    When d = 2:

    a – d = 8 – 2 = 6

    a = 8

    a + d = 8 + 2 = 10

    Thus, the required three terms are 27, 8, –11 or 6, 8, 10.

31. In the given figure, X and Y are the center of circles A and B respectively. Circle A passes through center Y of circle B. If the equation of circle B is x² + y² - 4x + 6y - 12 = 0 and the coordinate of X is (-4, 5), find the equation of circle A.

    

    ✎ The equation of circle B is:

    x² + y² - 4x + 6y - 12 = 0 ... (i)

    Comparing equation (i) with x² + y² + 2gx + 2fy + c = 0, we get

    2g = -4, 2f = 6, c = -12

    ∴ g = -2, f = 3

    Center of circle B = Y(-g, -f) = Y(2, -3)

    Center of circle A = X(-4, 5) = (h, k)

    Now,

    Radius of circle A (r) = XY

    r = √((x₂ - x₁)² + (y₂ - y₁)²)

    r = √((2 + 4)² + (-3 - 5)²)

    r = √(36 + 64)

    r = √100

    r = 10

    The equation of circle A is:

    (x - h)² + (y - k)² = r²

    or, (x + 4)² + (y - 5)² = 10²

    or, x² + 2.x.4 + 4² + y² - 2.y.5 + 5² = 100

    or, x² + y² + 8x - 10y + 16 + 25 - 100 = 0

    ∴ x² + y² + 8x - 10y - 59 = 0

32. By using vector method, prove that the quadrilateral formed by joining the midpoints of adjacent sides of a quadrilateral is a parallelogram.

    ✎ Given:

    
    ABCD is a quadrilateral. The midpoints of sides AB, BC, CD, DA are P, Q, R, S respectively.

    To prove: PQRS is a parallelogram.

    Construction: Join A and C.

    Proof:

    Statements	Reasons
    1. PQ = 1/2 AC and PQ // AC	1. In △ABC, the line segment joining the midpoint of two sides of triangle is half of third side and parallel to third side.
    2. SR = 1/2 AC and SR // AC	2. In △ADC, the line segment joining the midpoint of two sides of triangle is half of third side and parallel to third side.
    3. PQ = SR and PQ // SR	3. From statements (1) and (2).
    4. PQRS is a parallelogram	4. From statement (3), being opposite side of quadrilateral equal and parallel.

    Proved

33. The image of the triangle A is A' and image of A' is A'' in the given graph.

    1. a) By what transformation the image of the triangle A is A'? Write with reason.
    2. b) By what transformation the image of the triangle A' is A''? Write with reason.
    3. c) Write the name of transformation which denotes the combined transformation of the above two transformations? Write with reason.
    

    ✎ The vertices of triangle A are: (1, 0), (5, 1) and (5, 4).

    The vertices of triangle A' are: (-1, 0), (-5, -1) and (-5, -4).

    The vertices of triangle A'' are: (0, 1), (1, 5) and (4, 5).

    a) When triangle A is transformed into A':

    • (1, 0) → (-1, 0)
    • (5, 1) → (-5, -1)
    • (5, 4) → (-5, -4)

    The transformation of triangle A into triangle A' is the rotation through 180° about origin because it follows the rule of the rotation through 180° about origin i.e., (x, y) → (-x, -y).

    b) When triangle A' is transformed into A'':

    • (-1, 0) → (0, 1)
    • (-5, -1) → (1, 5)
    • (-5, -4) → (4, 5)

    The transformation of triangle A' into triangle A'' is the reflection on y = -x because it follows the rule of the reflection on y = -x i.e., (x, y) → (-y, -x).

    c) When triangle A is transformed into A'' (combined transformation):

    • (1, 0) → (0, 1)
    • (5, 1) → (1, 5)
    • (5, 4) → (4, 5)

    The combined transformation is the reflection on y = x because it follows the rule of the reflection on y = x i.e., (x, y) → (y, x).

    THE END