Hello students welcome to our educational portal and today we will explore the NTA and NTA Examinations.

So lets start –

**NTA Full Form is NATIONAL TESTING AGENCY**

**NTA RECENT EXAMINATIONS**

1) Joint Entrance Examination (Main) January 2020, Registration date – 2^{nd} september to 30^{th} september 2019.

2) Indian Institute of Foreign Trade (IIFT) MBA Test 2020, Registration date – 9^{Th} september to 25^{th} october 2019.

3) UGC – National Eligibility Test (UGC – NET) – December 2019, Registration date – 9^{Th} september to 9^{th} october 2019.

__NTA History –__

NTA conducts basically the Entrance Examinations like JEE MAIN, NEET – UG, UGC – NET, CMAT, GPAT, DUET. In few earlier years that examinations had been conducted by separate boards, but during conducting the mentioned above examinations in that time, previous boards have been facing some major problems because this all of the exams held in all over the India, they told, they faced the problems in managing the huge amount of students, so they have been decreasing their quality in Examination to Education and also there was some problems to conduct the Examinations by the separate boards. After that the Government of India (with HRD Minister) makes the decision to establish the **NTA (NATIONAL TESTING AGENCY)** for maintaining the quality and good in Education.

**NTA – NATIONAL TESTING AGENCY**

NTA (NATIONAL TESTING AGENCY) is a Govt. testing agency established in November 2017 and now NTA is the biggest testing agency for entrance examination in INDIA. NTA conducts that’s types of examinations – **JEE MAIN, NEET – UG, UGC – NET, CMAT, GPAT, DUET.**

**NTA OFFICIAL WEBSITE – https://nta.ac.in**

__Why the NTA ?__

NTA – NATIONAL TESTING AGENCY mainly developed for maintaining the quality and good in Examination and Education that to help our students in all over the India by reach to them.

__What is NTA (NATIONAL TESTING AGENCY) ? __

**National Testing Agency** (NTA) has been developed like as premier with specialization which autonomous and self sustained testing agency to operating the Entrance Examinations like Jee Main, Neet – UG, UGC – NET, and etc. for admission and fellowship in higher studies.

Read More – nta.ac.in

__OBJECTIVE of NTA__

* Conducting the entrance test which is international level, transparent and efficiently test to admission and requirement purpose.

* Identifying the experts and Institutions.

Read More – nta.ac.in

**NTA CONDUCTING EXAMINATIONS**

NTA basically conducts the seven types of examinations, which are Jee Main, NEET – UG, UGC – NET, CMAT, GPAT, DUET, Hotel Management and Hospitality.

**NTA EXAMINATIONS**

NTA is now mainly popular for Engineering Entrance Examinations which is known as Jee Main. So we start from Jee Main.

**Engineering Exam – (NTA Jee Main)**

Jee Main in previously had been conducting by the CBSE (CENTRAL BOARD SECONDARY EDUCATION) till 2019 and for increasing the quality in good , now jee main conducting the NTA from 2019.

JEE MAIN 2020 (JANUARY SESSION) CLICK HERE TO APPLY –

DOWNLOAD INFORMATION BULETTIN JANUARY JEE MAIN – 2020

JEE Main is applicable for admission in NITs, IIITs and CFTIs, which are participating though Central Seat Allocation Board. And the condition to admission that institutes that the candidate should have secured at least 75% marks in the 12th class examination, or be in the top 20 percentile in the 12th class examination conducted by yours examination boards. For SC/ST candidates the qualifying marks to taking admission are 65% in the 12^{th} class examination.

**Subject combinations required in the qualifying examination to taking admission in B.E./B.Tech. & B.Arch. and B.Planning Courses in NITs, IIITs, and other CFTIs is as under:-**

Required Criteria for B.E/B.TECH candidate must Passed 10+2 examination with Physics and Mathematics as compulsory subjects along with one of the Chemistry or Biotechnology or Biology or Technical Vocational subject.

and for B.ARCH./B.PLANNING candidate must have been Passed 10+2 examination with Mathematics with previous condition.

The above mentioned policy could also be taken by other Technical Institutions which are participating in counseling through JoSAA/CSAB. In case a State opts to admit students in the engineering Colleges affiliated by State Universities, then the state Universities give separate rank list according to them.

**More details** at – https://nta.ac.in/Engineeringexam

**Jee Main Free study Materials –**

**VECTOR INTRODUCTION FOR NTA / JEE / NEET / CBSE / JEE MAIN / IIT JEE / JEE ADVANCED / CLASS 12TH**

**H.C VERMA CHAPTER -3|KINEMATICS SOLUTION PART-2 |JEE/CBSE/ISC**

**INTEGRATION PART SHORTCUT TRICKS FOR CBSE / JEE MAIN / IIT JEE / JEE ADVANCED /NDA /BITSAT**

**Medical Exam – (NTA NEET)**

Second most popular entrance examination under The NTA is Medical Entrance Examination. That is known as The National Eligibility Cum Entrance Test (**NEET) UG**, now conducting the NTA from 2019 onwards. **NEET** also was being conducted by the Central Board of Secondary Education (CBSE) till 2018. A candidate can be taken admission to MBBS/BDS Courses in India in Medical/Dental Colleges run with the approval of Medical Council of India/Dental Council of India under the Union Ministry of Health and Family Welfare, except the institutions which established through an Act of Parliament of Government of India i.e. AIIMS and JIPMER Puducherry.

NTA has the most valuable the responsibility that the NTA will limited to the conduct of the entrance examination, announcement of result and giving All India Rank to the Directorate General Health Services and Government of India operating of counseling for few parts of total candidates (15%) all India Quota Seats and for supplying the result to state or other Counseling Authorities.

**More details** at – https://nta.ac.in/medicalexam

**Free NEET – UG Study materials –**

**MECHANICS PART -2 [NEWTON LAWS OF MOTION] FOR WBJEE | BIJOY SIR**

*Management Exam – NTA CMAT*

**In management entrance examination NTA mainly conducts CMAT examination (Common Management Admission Test),** which is a national level entrance examination for entry into management programmes.In the earlier period CMAT had been conducting the AICTE according to the directions of Ministry of Human Resource Development (MHRD), Government of India till 2018. CMAT is now conducting the NTA from 2019.CMAT is a three hour totally computer based online test (CBT) which is conducted in a single session to check or judging the student’s that the of ability of them to across the various segments like Quantitative Technique, Logical Reasoning, Language Comprehension and General Awareness.

This test facilitates Institutions to select suitable graduate students for admission in all Management programs.

The CMAT score is valid for all AICTE-Approved Institutions/University Departments/Constituent Colleges/Affiliated Colleges.

Eligibility Criteria for CMAT

(i) candidate must be a citizen of India.

(ii) Any Graduates student in any discipline can apply for CMAT exam.

(iii) In the final year running students of Graduate Courses in the system of 10+2+3 whose result declares before commencement of admission who can also apply for CMAT online exam.

**More details** at – https://nta.ac.in/Managementexam

**Pharmacy Exam – NTA GPAT**

**GRADUATE PHARMACY APTITUDE TEST (GPAT)** is a national level entrance examination for entry into M.Pharm programmes also conducting the NTA from 2019, before that, in the earlier time GPAT had been conducting by the AICTE (All India Council for Technical Education) according to the directions of Ministry of Human Resource Development (MHRD) of Government of India.

This test give a chance to institutions to select suitable Pharmacy graduates for admission into the Master’s (M.Pharm) program. The GPAT examination process same as the management examination that is a three hour computer based online test which is conducted in a single session. The GPAT score is valid for all AICTE-Approved Institutions and University Departments or Constituent Colleges or Affiliated Colleges. Some scholarships and other financial helps in that field of Pharmacy are also given on the basis of the GPAT score.

**Eligibility Criteria for GPAT**

(i) candidate must be a citizen of India.

(ii) They have must be Bachelor’s degree holders in Pharmacy (4 years after 10+2, including with the lateral entry candidates).

(iii) Final year running students of Graduate Courses in B. Pharmacy whose result will be declared before commencement of admission can also apply for GPAT online exam.

(iv) B.Tech or B.tech in Pharmaceutical and fine chemical technology) or Equivalent Students are not eligible to appear for GPAT examination.

**More details** at – https://nta.ac.in/Pharmacyexam

**College/University Teaching & Fellowship Exam – NTA NET**

UGC-NET in the earlier time was conducted by CBSE, on 84 subjects at the 91 cities across the India, but now in present time UGC – NET is conducting by NTA from 2018 onwards.

On the behalf of UGC, the UGC-NET (National Eligibility Test) is conducted to find the eligibility of the students, only for Assistant Professor and for Junior Research Fellowship and Assistant Professor both in Indian Universities and Colleges.

To the chance for Junior Research Fellowship and Eligibility for Assistant Professor both or Eligibility for Assistant Professor that depends on the performance of the student in both the papers of UGC-NET in aggregate. In case a candidate qualifying for ‘Assistant Professor only’ is not to be considered for award of JRF.

**More details** at (https://nta.ac.in/collegeexam)

**SWAYAM Test by NTA **

To achieve the three cardinal principles of an Education Policy viz., access, equity and quality, Ministry of HRD of India, has assigned on a major and new initiative, called SWAYAM. SWAYAM does provide the best On-line e-learning teaching learning resources to all. It pursues to bridge, the digital divide for students, who has hitherto remained unknown by the digital revolution and has not been able to join the mainstream of the knowledge of economy. SWAYAM provide the integrated platform and a portal for online courses for your Education, using of information technology and communication technology, covering of learners from Schooling class 9 to 12 and Under Graduates & Post Graduates, in all subjects. Now in present time more than 2,000 MOOCs are hosted on SWAYAM, prepared by specially chosen faculty and teachers, across over the Country.The courses are interactive and also are available in free of cost, for the learners.

Read more at (https://nta.ac.in/Swayamexam)

**Hotel Management Exam**

The **National Council for Hotel Management Joint Entrance Examination** is now conducting the NTA from 2019 onwards (NCHM JEE-2019) in CBT mode for admission to the B.Sc. Course in Hospitality and Hotel Administration (B.Sc.HHA) across all the Country.

More details at – https://nta.ac.in/HotelManagementexam

**Delhi University Entrance Test ( DUET )**

Delhi University Entrance Test (DUET) is now conducting the NTA from 2019 onwards.

NTA official website provides you all of the things regarding to the examinations like Mock Test, Question Papers, Information Bulletin and etc.

If you have any queries then you can contact on that following numbers 9453827203,9455874491,9455874492,9455874494 from **9:00am to 6:00pm, Monday to Saturday. You can mail at **DUET.NTA@gmail.com

More details at – https://nta.ac.in/DuetExam

**Formula Chart **

__# Integration by substitution__

Basic rule of substitution

∫ f(x) dx = ∫ f {Φ(z)·Φ^{‘} (z)} dz ; Substitute x =Φ(z)

__# Integration by parts __

If u and v be two functions of x then ∫ uv dx = u ∫ v dx – \( \int (\frac{du}{dx}\int v dx)dx \) is

known as formula for integration by parts.

Note : In finding Integrals by this method proper choice of u and v is essential. In general the funcyion as u is taken which comes first in the world **ILATE. **

Where**, I – Inverse circulated function**

** L – Logarithmic function**

** A – Algebraic function**

** T – Trigonometric function**

** E – Exponential function**

**Note :**

* if both the function s are trigonometric, take that function as v whose integral is simpler.

** If both functions are algebraic take that the function as ‘u’ whose d.c is simpler.

__# ____Integration of rational function :__

If integrand is of the P(x)/Q(x), where P(x) and Q(x) are polynomials of x.

Case I. If P(x)/Q(x) is proper rational function ( i.e. highest power of x in Q(x) > P(x) integration by partial function is done.

Case II. If P(x)/Q(x) is improper rational function (i.e. highest power of x in Q(x) < P(x)). Then it is made proper rational function by simple division and then partial fraction is used.

f(x) is integrated by general rule and h(x)/Q(x) is integrated by partial fraction.

__# Integration of certain irrational expressions : __

(i) If the integrand is a rational function of a fractional powers of an independent variable x i.e. the function R (x, x^{p1/q1} ……x^{pk/qk} ). Substitute x = t^{m} is L.C.M. of q_{1} ,q_{2} …….q_{k}.

(ii) If the integrand is rational function x and fractional powers of a linear fractional function of the form

**# Integration of the form ∫R (x, √ax**^{2}** + bx + c) dx**

(i) Integrals of the form

Substitute \( x + \frac {b}{2a}=t \) and reduce it to the form

M_{1 } , N_{1} , k are constants.

(ii) Integrals of the form I = ∫ [ P_{m} (x) / √ax^{2} + bx + c] dx

P_{m} (x) is a polynomial of degree m.

Where P_{m-1} (x) is polynomial of degree (m-1) and k is constant. Coefficient of is determined by the method of undetermined coefficients.

(iii) Integrals of the form ∫ dx / ( x – a_{1} )^{m} √ax^{2} + bx + c

Substitute ( x – a_{1} ) = \( \frac {1}{t} \)

**Note :** For the integration of the form ∫ √ax^{2} + bx + c dx or ∫ dx/ √ax^{2} + bx + c or ∫ dx /ax^{2} + bx + c.

First convert ax^{2} + bx + c in the form A^{2} + X^{2} , X^{2} – A^{2} , or A^{2} – X^{2} where A is constant and x is px + q.

**Following results is used**

**# Integration of the form of ∫ x ^{m} ( a + bx^{n}) P dx ; m, n, p are rational number**

Case 1. If P is a positive integer then integrand is expanded by binomial expansion.

Case 2. If P is negative integer substitute x = t^{k} where k is L.C.M. of denominator of m and n.

Case 3. \( \frac {m+1}{n} \) is an integer, put a + bx^{n} = t^{α} where α is denominator of fraction P.

Case 4. \( \frac {m+1}{n} + P \) is an integer, put a + bx^{n} = t^{α} x^{n} where α is denominator of fraction P.

**INTEGRATION TRICKS IN 2 SEC FOR CBSE / JEE / IIT JEE / JEE MAIN / NDA /JEE ADVANCED/WBJEE**

Hello students I am Bijoy and welcome to our educational forum. And today we discuss about **class 10 maths sample paper**, in this section we will see solution hints and at the end download the sample paper.

This is the official **NCERT sample paper** by **CBSE**. Students now I am giving official paper and our official sample paper coming soon by **jeewithbijoy.**

Before starts solving the sample paper, covered that topics which mentioned in below with the hints. That’s help you to learning the mathematics properly

So let’s start the discussion about **ncert class 10 mathematics sample paper**

**VSA: 1 mark**

1. Application of finding n – th term of an A.P , Use t_{n} = a + ( n-1 )·d [where a = first term, d = common difference ]

2. Question no 2.0 is the application of **Height and Distance**, where given the ratio of Base and Height is √3:1 , hence you can use cotθ = √3:1.

3. In question no 3.0 you have to find the **surface area** of a cone.

Hints : First find radius of base of the given cone then apply pythagoras theorem.

4. Hints : **Area of triangle** given by Δ = 1/2 | x_{1} ( y_{2} – y_{1} ) + x_{2} ( y_{3} – y_{1} ) + x_{3} ( y_{1} – y_{2} ) | sq^{2} unit.

**NCERT CLASS 10 MATHS SAMPLE PAPER PDF**

**SA – I : 2 MARKS**

1. Finding the roots of a given **quadratic equation**.

Hints : Use Shreedhar Acharya’ formula

2. Finding the n – th term of an A.P

Hints : Find common difference -11 – (-8) = – 3,

Then use, t_{n} = a + (n-1).d and put t_{n} =49, then find the value of t_{n-4}

3. In the question no 3.0 just use theorem.

4. Finding the area of a required sector.

Hints : Area of a sector given by A_{θ} = ½ (θ r^{2})

5. Just use theorem of **probability**

Hints : Probability = no of chance to outcomes/ total no of events

6. In question no 6.0 has given coin tossing problem in **probability**, hence use same formula. In this case total no of events is 4 and outcomes are 2.

**CLASS 10 SAMPLE PAPER IN PDF**

**SA – II : 3 MARKS**

1. In this question has given a **quadratic equation** in rational form.

Hints : first reduce in quadratic form and then use Shreedhar Acharya’s formula.

2. Hints : First find common difference, then use formula of t_{n} = a + ( n-1 ).d. Now find the value of ‘n’

3. The question no 3.0 is the application of geometry.

Hints : Just follow the steps, first draw two **perpendicular** from R to O and Q to O respectively. Hence ∠MRO = ∠LQO= 90^{0} and then find ∠SQO and ∠SRO then add ∠SOR and ∠SRO and minus from 180 that will be yours answer.

4. Application of Height and Distance.

Hints : Use this diagram

5. Finding the **area of a sector**

Hints : use A_{θ} = 1/2 (θ r^{2})

6. Find the **area of a circle**, A = π·r^{2} In this case you have to find area of the path.

Hints : first find the area of the Pond without Path A_{1}, then add the radius and width of path and now find total area A_{2} . Then do A_{2} -A_{1} you get area of path.

7. Hints : Find the total **area of Cylinder** A_{1} then find the area of two cones A_{2} and A_{3} , now add them A_{1}+A_{2}+A_{3}. That will be your answer.

8. Hints : First compare the **surface area of sphere** with 5 times of **surface area of Cone** ( A_{1}= 5·A_{2} ), now you get an equation and find the height of cone from this equation. Then use V = 1⁄3 ×π×r^{2}×h.

9. Hints : Apply divisional or inter sectional formula both Internally and Externally one by one and use m=1, n=1.

10. Question no 10 is the application of coordinate geometry.

Hints: Write a second degree equation of Circle with the help of any one point ( ) then put the value of x and y from another point.

**LA : 4 MARKS**

This section will be coming soon in video lectures, if you need so please comment.

**CLASS 10TH RELATED TOPICS**

Hello students I am Bijoy and welcome to our educational forum. Today we will discuss about the ncert class 10 mathematics syllabus, chapter content etc. and at the end of this page you will get a link that you can download the book in pdf ( in hindi version ) & if you want any other ncert books then please comment.

Q. How to good score in mathematics?

Ans. Students if you want to be expert in mathematics, then you have to follow some simple steps. That you can get a good score in mathematics .First analysis the whole book deeply and then start studying.

# Now question is how could you do that?

Answer is here first note down the syllabus, then analysis the all chapters and then note down the all topics that covered in the chapters, that is most important for a to a good student in any subject. And last make schedule to cover that topics, time to time with concentration.

Now we will discuss syllabus and topics covered in ncert class 10 maths.

**CHAPTER CONTENT**

**CHAPTER1 – REAL NUMBER**

**Topics covered** – Introduction to Real Number, Euclid’s division Lemma and H.C.F, The Fundamental Theorem of Arithmetic, Revisiting Irrational Number ( advance to irrational number ), Revisiting Rational Number and Their decimal Expansion, how to find rational and irrational number between two numbers etc. Summery.

**Chapter2 – POLYNOMIALS**

**Topics covered** – Introduction to Polynomials some advance level, Geometrical meaning of the zeroes of a Polynomials, how to find no of zeroes by the graph of y = p(x), Relationship between Zeroes and Coefficients of a polynomial, proving the relationship between the roots and their coefficients, Division Algorithm for polynomials, Condition for factors of given Polynomial, Summery.

**Chapter3 – Pair of Linear Equation in Two Variables**

**Topics Covered** – Introduction to Linear Equation in Two Variables ( ax + by = c ), Graphical Method of Solution of a Pair of Linear Equations ( How to solve two equations by graph, finding two parallel lines by the graph ), Algebraic Methods to solve the Linear Equations ( 1. Substitution Method 2. Elimination Method 3. Cross – Multiplication Method ), Separation on pair of equations or Equations reducible to a pair of Linear Equations in Two Variables, Summery.

**Chapter4 – Quadratic Equations**

**Topics Covered** – Introduction to Quadratic Equation, Second degrees Polynomial Equation as Quadratic Equation, Study of quadratic equations, Solution of a quadratic Equation by Factorisation or Completing the Square form, Study about their roots, Nature of roots, Summery.

**Chapter5 – Arithmetic Progressions**

**Topics Covered** – introduction to Arithmetic Progressions, Study Arithmetic Progressions, Study of nth Term of an AP, Sum of First n terms of an AP, Summery.

**Chapter6 – Triangles**

**Topics Covered** – Introduction of Triangles, Study of Similar Figures, Properties of Similarity of Triangles, Criteria to Similarity of Triangles, Area of Similar Triangle as study of application, Pythagoras Theorem and its application, Summery.

**Chapter7 – Coordinate Geometry**

**Topics Covered** – Study of Coordinates in quadrant, Distance Formula and its application, Section Formula as an Internally or Externally, Area of a triangle and its application i.e condition of co-linearity, Summery.

**Chapter8 – Introduction to Trigonometry**

**Topics Covered** – Introduction to Angles and Study of Angles, Trigonometric Ratios as the Study of sinx, cosx, tanx, cosecx, secx, and cotx, Trigonometric Ratios of some Specific Angles at 0, 30, 45, 60 and 90 degrees, Trigonometric Ratios of Complementary Angles as an example cox ( 90 – 30 ) = sin 60, Trigonometric Identities, Summery.

**Chapter9 – Some Application of trigonometry**

**Topics Covered** – Introduction as study angle of Elevation and Angle of Depression, an application on Height and Distances, Summery.

**Chapter10 – Circles**

**Topics Covered** – Introduction to a Circle, Study of Tangent to a Circle, understanding the Number of Tangents from a Point on a Circle, Summery.

**Chapter11 – Construction**

**Topics Covered** – Introduction of some special drawing, Study of drawing of Division of a Line Segment, drawing of Tangents on a Circle, Summery.

**Chapter12 – Area Related to Circles**

**Topics Covered** – Introduction as study of circle’s all dimension, Study of Perimeter and Area of a Circle and its application, finding the Sector and Segment of a Circle, Area of Combinations of Plane Figures, summery.

**Chapter13 – Surface Area and Volumes**

**Topics Covered** – Introduction of Surface Area and Volumes, Study of Surface Area of Combination of Solids, Study of Conversion of Solid from a given shape to another shape, Study of a Cone, Summery.

**Chapter14 – Statistics**

**Topics Covered** – Introduction, Finding Mean mode and Median from a given Grouped data, Representation of Cumulative Frequency Distribution in Graph, Summery.

**Chapter15 – Probability**

**Topics Covered** – introduction to Probability, Theoretical study on Probability, Summery.

**Download link is given in below –**

Class 10 mathematics book chapter – 1

Class 10 mathematics book chapter – 2

Class 10 mathematics book chapter – 3

Class 10 mathematics book chapter – 4

Class 10 mathematics book chapter – 5

Class 10 mathematics book chapter – 6

Class 10 mathematics book chapter – 7

Class 10 mathematics book chapter – 8

Class 10 mathematics book chapter – 9

Class 10 mathematics book chapter – 10

Class 10 mathematics book chapter – 11

Class 10 mathematics book chapter – 12

Class 10 mathematics book chapter – 13

Class 10 mathematics book chapter – 14

]]>Hello fiends I am Bijoy and welcome to our Informatics Forum. We of all use mobile phone and some we need some help for few problem on our phone, that we try to find authorised and nearest service center, so today I give you the list of authorised samsung service center and nearest to you decide one of which is nearest and just visit there for repair your phone thank you.

FOR MORE INFORMATION VISIT SAMSUNG OFFICIAL WEBSITE JUST CLICK HERE

SAMSUNG SERVICE CENTER NEAREST TO KOLKATA

1. Authorised Service Center

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**Address : **Dutt House 1st Floor, 94 C/R/Avenue, 700012

Contact no – 03340625094

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Contact no – 03340628246

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Time-O-Service

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Contact no – 9830264105

S-Care

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**Address : **51B, Garcha Road, Opposite-Hazra Law, Law College, 1st Floor, 700019

Contact no – 9804908080

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Kolkata, 7.09 km

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B2X Service Solutions India Pvt Ltd

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Fast Information Centre

Kolkata, 10.23 km

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Time-N-Services

Howrah, 10.62 km

**Address : **430 Gt Road, Bally, 711201

Contact no – 8479903229

Kolkata, 11.4 km

**Address : **90, Diamond Harbour Road(1St Floor), 700008

Contact no – 7595065757

Babli Media Center

Kolkata, 12.05 km

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Shivam Enterprise *(Collection Point)*

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Dey Xerox *(Collection Point)*

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**SAMSUNG SERVICE CENTER NEAREST TO HOWRAH**

Time-N-Services

Howrah, 0.9 km

**Address : **430 Gt Road, Bally, 711201

Sanchar Infocom

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**Address : **20, 1st Floor, Shantiniketan , Apartment, Round Tank Lane , Howrah, 711101

Time-O-Service

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Uma Enterprise *(Collection Point)*

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Contact no – unavailable

**SAMSUNG SERVICE CENTER NEAREST TO HOOGHLY**

Sandhya Mobile *(Collection Point)*

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Shree Krishna Services (** SAMSUNG SERVICE CENTER SERAMPORE AND SAMSUNG SERVICE CENTER DANKUNI )**

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Mitra Traders

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Arindam Enterprise

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]]>Hello students I am Bijoy and welcome to my educational forum/portal. Today we will discuss about * NCERT SOLUTIONS FOR CLASS 10 MATHS* , so let’s start.

-:*POLYNOMIALS:-*

What is polynomial ?

A mathematical expression on various types of function like algebraic, exponential, trigonometric etc. which maintain the any n-th degrees in variable, that expression is called as **polynomial** on that function. Where ‘n’ is an integer.

Note – but in the class of ten (X) we only discuss the algebraic polynomial on 1th, 2^{nd}, 3^{rd} degree polynomials. Ex – x^3+ 2x^2-3

**EXERCISE – 2.1**

Using formula –** Graphical representation**

A given polynomial p(x) with degree of n, then the number of zero or zeroes is/are represented by the no of intersecting point / points on x – axis of the graph of y = p(x).

1. The **graphs** of y = p(x) are given in fig 2.10 below for some polynomials p(x). Find the number of zeroes of p(x), in each case.

i) Ans. **no of zeroes = no of intersected point on x – axis, graph of y = p(x)**

Hence according to question, the first polynomial has no zero, because in this case has no intersecting point on x -axis or here the graph does not intersect anywhere on the x – axis.

ii) No of zero is equal to ‘1’ (one)

Note – Students, in this case you can see have no intersected point of the graph of y = p(x) at x – axis, but remember just one thing that no of zeroes is equal to no of points of intersecting only on x – axis, does not count from y – axis.

iii) In the no of question 3^{rd} the value of zeroes are 3 (three). Because here the graph of y = p (x) cut x – axis at the three points.

iv) In this case no of zeroes are equal to ‘2’ (two).

Note – here the graph of y = p(x) intersects the x- axis at the two points.

v) Students you just observe that in the question no (v) the graph of y = p(x) intersects the x -axis at the four points.

Hence no of zeroes are equal to ‘4’ (four).

vi) In the question (vi) the graph of y = p(x) touches the x – axis at the three points, hence the no of zeroes are ‘3’ (three).

**EXERCISE – 2.2**

Using formula :-

i) find the value of zeroes

y = P(x) = 0

ii) Required polynomial by the given sum and product of zeroes = x ^{2}– (α+β) x + α.β

solution :

1. **Find the zeroes** of the following **quadratic polynomials** and verify the relationship between the **zeroes and the coefficients**.

i) Given, x^{2} -x-8

Now let y = p(x) = x^{2} -x-8

hence p(x) = 0

or, x^{2} -x-8 = 0

Now factorize the polynomial and splitting 2x we get,

Or, x^{2} -(4-2)x-8=0

note :- first divide the ‘8’ in two numbers that after minus this two numbers just get 2 (i.e 4-2=2).

Or, x^{2} -4x+2x-8=0

Or, x(x-4) + 2(x-4)=0

Now taking common (x-4), we get

Or, (x-4) (x+2) =0

Hence the required zeroes are

(x-4)=0 or (x+2)=0

∴ X=4 or ∴ x=-2

∴ The required zeroes of given polynomial are 4 and -2.

How to verify the relationship between the zeroes and the coefficients ?

Let, the zeroes a = 4 and b = -2

∴ a+b = 4+(-2 )= 2 = – coefficient of x/coefficient of x^{2}

∴ a.b = 4.(-2) = -8 = constant term/coefficient of x^{2}

ii) Given, polynomial 4s^{2} – 4s + 1

hence, 4s^{2} – 4s + 1= 0

now split the 4 as (2+2) we get,

or, 4s^{2}– (2+2) s + 1= 0

or, 4s^{2 }-2s -2s +1 = 0

or, 2s(2s-1)-1(2s-1)=0

now taking common (2s-1)

or,(2s-1) (2s-1)=0

hence the required zeroes are

(2s-1)=0 or (2s-1)=0

∴ S =1/2 or, ∴ s=1/2

∴ The required zeroes of given polynomial are 1/2.

Now check the relationship between the coefficients and zeroes

a=1/2 and b=1/2

Hence, a+b=1/2+1/2=1= -(coefficient of x)/coefficient of s^{2}

a.b= \( \frac{1}{2}*\frac{1}{2}=\frac{1}{4} \)= (constant term)/coefficient of s^{2}

iii) Given, p(x)=6x^{2}-3-7x

hence 6x^{2}-3-7x=0

or, 6x^{2}-7x-3=0

now split the 7 as (9-2=7)

or, 6x^{2}-(9-2)x-3=0

or, 6x^{2}-9x+2x-3=0

or, 3x(2x-3)+1(2x-3)=0

or, (2x-3) (3x+1)=0

hence the required zeroes are

(2x-3)=0 or, (3x+1)=0

Therefore x=3/2 or, x=-1/3

∴ The required zeroes of given polynomial are 3/2 and -1/3.

Now checking the relation between coefficients and zeroes

a=3/2 and b= -1/3

Hence a+b=3/2+(-1/3)=7/6= -(coefficient of x )/coefficient of x^{2}

a.b= \( \frac{3}{2}*\frac {-1}{3}=\frac{-1}{2} \)= constant term/coefficient of x^{2}

iv) given, p(x) = 4u^{2}+8u

hence, 4u^{2}+8u+0=0

now taking common 4u and we get

4u(u+2)=0

Therefore the required zeroes are

4u=0 or, (u+2)=0

∴ U=0 or, ∴ u=-2

∴ The required zeroes of given polynomial are 0 and -2.

Now check the relationship between the coefficients and zeroes

a = 0 and b=-2

a+b=0-2=-2=-(coefficient of u)/coefficient of u^{2})

a.b=0.(-2)=0= constant term/coefficient of u^{2}

v) given, p(t)=t^{2}-15

hence t^{2}+0-15=0

or, t^{2} – (√15)^{2} = 0

or, ( t – √15 ) (t – √15 ) = 0

Hence the zeroes of given polynomial are

( t + √15 ) = 0 or ( t – √15 ) = 0

∴ t = – √15 or, t = √15

∴ The required zeroes of given polynomial are – √15 and √15.

Now check the relationship between the coefficients and zeroes

therefore a = – √15 and b = √15

∴ a + b = – √15 + √15 = 0 = -(coefficient of t)/coefficient of t^{2})

vi) given, p(x) = 3x^{2}-x-4

hence, 3x^{2}-x-4=0

now splitting the 1.x as (4-3=1)

we get, 3x^{2} – (4-3)x -4=0

or, 3x^{2} -4x+3x-4=0

or, x ( 3x – 4) + 1 (3x-4) = 0

now taking common by ( 3x – 4 ) we get,

or, ( 3x – 4 ) ( x + 1 ) = 0

hence the required zeroes are

( 3x – 4 ) = 0 or ( x + 1 ) = 0

∴ x = 4/3 or ∴ x = -1

∴ The zeroes of given polynomial are 4/3 and -1.

Now check the relationship between the coefficients and zeroes

Therefore a = 4/3 and b = -1

A+b = 4/3 + (-1) = 1/3 = – ( coefficient of x ) / ( coefficient of x^{2} )

And a.b = \( \frac{4}{3}*-1= -\frac{4}{3} \) = constant term / coefficient of x^{2}

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Using formula : –

The required polynomial = x^{2} – ( sum of zeroes )·x + ( product of zeroes )

Solution –

(i) Given, sum of zeroes = \( \frac{1}{4} \) and product of zeroes = -1

Hence the required polynomial

= x^{2} – \( \frac{1}{4}x \) + (-1)

= 4x^{2} – x + 4

(ii) Given, sum of zeroes = √2 and product of zeroes = \( \frac{1}{3} \)

Hence the required polynomial is

= x^{2} – ( √2 ) x + \( \frac{1}{3} \)

= 3x^{2} – 3 √2 x + 1

(iii) Given, sum of zeroes = 0 and product of zeroes = √5

Hence the required polynomial is

= x^{2} – (0).x + √5

=x^{2} + √5

(iv) Given, sum of zeroes = 1 and product of zeroes = 1

Hence the required polynomial is

= x^{2} – (1) x + 1

= x^{2} – x + 1

(v) Given, sum of zeroes = -1/4 and product of zeroes = 1/4

Hence the required polynomial is

= x^{2} – (\( \frac{-1}{4} \)) x + \( \frac{1}{4} \)

= 4x^{2} + x + 1

(vi) Given, sum of zeroes = 4 and product of zeroes = 1

Hence the required polynomial is

= x^{2} – 4x + 1

**Exercise – 2.3**

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :

(i) p(x) = x^{3} – 3x^{2} + 5x – 3, and g(x) = x^{2} – 2

Hence the quotient is (x-3) and remainder is ( 7x – 9 )

(ii) Given, p(x) = x^{4} – 3x^{2} + 4x + 5 and g(x) = x^{2} + 1 – x

∴ The quotient is x^{2} + x – 3 and remainder is 8

(iii) Given, p(x) = x^{4} – 5x + 6, and g(x) = 2 – x^{2}

Hence the quotient is – x^{2} – 2 and remainder is – 5x + 10.

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :

(i) t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t 12

so, the remainder is zero (0), hence the first is the factor of second polynomial.

(ii) x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

Since the remainder is zero, first polynomial is the factor of second poynomial.

(iii) x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1

In this case remainder is 2 not zero, the the first polynomial is not factor of second polynomial.

3. Obtain all other zeroes of 3x^{4} + 6x^{3} – 2x^{2} 10x – 5, if two of its zeroes are √5/3 and -√5/3.

Solution : Given the zeroes are √5/3 and – √5/3. Hence the two factors of given polynomial are ( x -√5/3 ) and ( x + √5/3 ).

Therefore , ( x + √5/3 ) ( x -√5/3 ) = ( x^{2} – \( \frac{5}{3} \) ) is also the factor of 3x^{4} + 6x^{3} – 2x^{2} 10x – 5.

Now divide the polynomial 3x^{4} + 6x^{3} – 2x^{2} 10x – 5 by ( x^{2} – \( \frac{5}{3} \) ) we get,

therefore the quotient is 3x^{2} + 6x + 3, hence 3x^{2} + 6x + 3 also the factor of 3x^{4} + 6x^{3} – 2x^{2} 10x – 5.

So, if we want to find another zeroes of the polynomial we need to do 3x^{2} + 6x + 3 = 0.

Now by solving the quadratic we get,

Or, 3x^{2} + 6x + 3 = 0

Or, 3x^{2} + 3x + 3x + 3 = 0

Or, 3x ( x + 1 ) +3 ( x + 1) = 0

Or, ( x + 1 ) ( 3x + 3 ) = 0

Hence the required zeroes are,

( x + 1) = 0 or, ( 3x + 3) = 0

∴ x = -1 or, ∴ x = – 1

Therefore the zeroes of 3x^{2} + 6x + 3, hence 3x^{2} + 6x + 3 are √5/3, -√5/3, -1 and -1

4. On dividing x^{3} – 3x^{2} +x + 2 by a polynomial g(x), the quotient and remainder are x – 2 and – 2x + 4, respectively. Find g(x).

Solution :

We know, polynomial or dividend = divisor × quotient + remainder

Given, polynomial = x^{3} – 3x^{2} +x + 2, quotient = x – 2 and remainder = – 2x + 4, and now putting the value we get,

here divisor = g(x)

x^{3} – 3x^{2} +x + 2 = g(x) × (x – 2) + 4 – 2x

∴ g(x) = (x^{3} – 3x^{2} +3x – 2)/(x – 2), hence g(x) must be quotient when we divide x^{3} – 3x^{2} +x + 2 by ( x – 2 ).

Therefore the g(x) = x^{2} – x + 1

5. Give an examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0

Solution :

(i) p(x) = 2x^{5} – 6x^{3} + 4x^{2} + 10

∴ q(x) = ( x^{5} – 3x^{3} + 2x^{2} + 3 ), divisor = 2 and remainder [ r(x) ] is 4.

Therefore the degree of p(x) and q(x) are equal.

(ii) Assume p(x) = x^{3} – 2x +1 and g(x) = x^{2}, now if we divide p(x) by g(x) then quotient must be = x and remainder will be -2x + 1. means q(x) = x and r(x) = -2x + 1.

Hence the deg of q(x) = deg of r(x) .

(iii) deg r(x) = 0, means remainder must be a constant term and that example already in example no (i).

]]>Hello students I am Bijoy and welcome to my * educational forum*. Today we will deal with the

Lets start, the evaluating the **integration of log x**

Let, I = \( \int log|x| dx \)

In previous section of **integration** we have learnt basic **integration**, **substitution method** and **by parts** method for **integrating the functions**. but here students if try to observe then you can understand easily that we can not **integrate logx** by basic integration or substitution, so we have only one option that to **integrate log x**, with by parts that to.

Now I want just recap to all of you the by parts method or rule, like **LIATE or ILATE rule**, just remind that.

so, lets start I = \( \int log|x| dx \)

students if you observe , you can see two **function logx** and 1, now we can apply **LIATE rule** and we get,

or, I = \( logx \int dx – \int (d(logx)/dx \int dx)dx \)

or, I = \( logx .(x) – \int (\frac {1}{x}).x. dx \)

or, I = \( x.logx – \int dx \)

or, I = x . logx – x + c [ where ‘c’ is integral constant }

Hence, \( \int log|x| dx \)= x.logx – x + c = x ( logx – 1 ) + c

Now we will try to evaluate another type of integration that is **integration of log ax**.

so, our integration

I = \( \int log|ax|dx \)

students same as the above, two functions logax and 1, now applying LIATE and get,

or, I = \( log|ax| \int dx – \int(\frac{dlog|ax|}{dx} \int dx)\int dx \)

or, I = \( log|ax|.x – \int (\frac{a.x}{x})dx \)

or, I = x log|ax| – x + c [ where ‘c’ is integral constant ]

So required answer equal to x log|ax| + c

Our required integration, I = \( \int(\frac{1}{xlogx}) dx \) ———- (1)

This is the one of the simple on logx problems, in this case you have to a simple substitution on log x.

so, assume that log x = z, now differentiating on both side with the respect to ‘x’ and we get,

\( \frac{d(logx)}{dx} = \frac{dz}{dx} \)

or, \( \frac{1}{x} dx = dz \)

Now putting the value in equation no (1) and we get,

or, I = \( \int \frac{1}{x} dx *logx \)

or, I = \( \int \frac{dz}{z} \)

or, I = log|z| + c [ where ‘c’ is integral constant ]

or. I = log|log|x|| + c

Hence, **integration of 1/xlogx** equal to log|log x|| + c

so we have to find the value of, I = \( \int (logx)^2 dx \)

Students, if try to observe then you can see the two functions one is (logx)^{2} and another one is 1.

So, now applying LIATE rule and we get,

I = (logx)^{2} ∫1.dx – \( \int [\frac{d(logx)^2}{dx} \int 1.dx]dx \)

or, I = x ⋅ (log x)^{2} – \( \int \frac{2.logx}{x}.x dx \)

or, I = x⋅ (log x)^{2} – 2 ∫ (logx) dx

Now using by parts once again in second part and we get,

or, I = x⋅ (log x)^{2} -2x logx + 2x + c [ Where ‘c’ is integral constant]

∴ I = x (logx)^{2} – 2 x (logx – 1) + c

**Ex. Evaluate the integration of log(1/x).**

Ans. Let, I = \( \int log(\frac{1}{x}) dx \) ———— (1)

or, I = log (1/x) ∫dx – \( \int [(\frac{dlog(1/x)}{dx})\int dx]dx \)

or, I = x⋅log(1/x) – \( \int( -\frac{1}{x^2} x.x dx \)

or, I = x⋅log(1/x) + ∫ dx

∴ I = x⋅log(1/x) + x + c [ where ‘c’ is integral constant ]

Hence, the **integration of log(1/x)** equal to x⋅log(1/x) + x + c.

Ex.2

What is the value of \( \int [\frac{1}{log x}- \frac{1}{(logx)^2}] \) ?

Ans. For solving that integration first we have to start from,

Let, I = \( \int \frac{1}{logx} dx \) ——— (1)

Now using by parts and we get,

or, I = \( \frac{1}{logx} \int dx – \int [(\frac{d(\frac{1}{logx})}{dx})\int dx]dx \)

or, I = \( x \frac{1}{log x} – \int[-\frac{1}{(log x)^2}* \frac{x}{x} dx] \)

or, I = \( \frac{x}{logx} +\int \frac{dx}{(log x)^2} + c \)

[ where c is integral constant ]

Now applying the side changing rule on second part of given integration and we get,

or, I – \(\int \frac{dx}{(log x)^2} + c = x⋅\frac{1}{log x}\)

Now put the value of I, from equation no (1) we get,

\( \int \frac{dx}{log x} – \int \frac{dx}{(logx)^2}= x\frac{1}{log x} + c \)

Ex.3

**What is the value of integration of x⋅(log x) ^{2 }?**

Ans. Let, I = ∫ x ⋅ (log x)^{2} dx —————- (1)

Students first we use substitution method .

now assume, log x = z and **differentiating** on both side with respect to x.

and we get, 1/x dx = dz, and dx = x⋅dz

again log x = z , ∴ x = e^{z}

Now putting the value in equation (1)

we get, I = ∫ e^{z} ⋅z^{2}⋅e^{z} dz

or, I = ∫ e^{2z} ⋅ z^{2} dz

Now, by using by parts as LIATE rule, we get

I = z^{2} ∫ e^{2z} dz – \( \int [\frac{dz^2}{dz} \int e^(2z) dz]dz \)

or, I = \( \frac{e^(2z) . z^2}{2} – \int \frac{2z}{2} * e^(2z) dz \)

or, I= e^{2z} ⋅ \( \frac{z^2}{2}- \) ∫ e^{2z} z dz

once again apply the LIATE rule,

or, I = z^{2}. e^{2z}/2 – [ z ∫ e^{2z} dz- ∫ {dz/dz ∫ e^{2z} dz} dz

or, I= z^{2}. e^{2z}/2 – [ z⋅ e^{2z}/2 – e^{z}/4] + c [ where c is integral constant ]

or, I = \( \frac{z^2}{2} (e^z)^2 – \frac{z}{2}(e^z)^2 + \frac{1}{4} (e^z)^2 + c \)

or, I = \( \frac{1}{2} \) [ (log x)^{2} ⋅ x^{2} – x^{2} ⋅ (log x) + \( \frac{1}{2} \) x^{2} + c ]

∴ I = \( \frac{1}{2} x^2 \) [ (logx)^{2} – (logx) + 1/2 + c]

Hence, the **integration of x ⋅(log x) ^{2}** equal to \( \frac{1}{2} x^2 \) [ (logx)

**Ex. Find the value of integration of log (1 + y ^{2} ).**

Ans. Let, I = ∫ log( 1 + y^{2} ) dy

Now using ILATE rule and we get,

or, I = log( 1 + y^{2} ) ∫ dy – \( \int [\frac{dlog(1+y^2)}{dy} \int dy] dy \)

or, I = y ⋅ log( 1 + y^{2} ) – \( \int \frac{2y}{1+y^2} ydy \)

or, I = y ⋅ log( 1 + y^{2} ) – \( 2\int \frac{y^2}{1+ y^2} dy \)

or, I = y ⋅ log( 1 + y^{2} ) – \( 2\int \frac{1+y^2-1}{1+y^2}dy \)

Now separate the second part and get,

or, I = y ⋅ log( 1 + y^{2} ) – \( 2[\int \frac{1+y^2}{1+y^2}dy – \int \frac{dy}{1+y^2}] \)

or, I = y ⋅ log( 1 + y^{2} ) – 2 [ x – tan^{-1}y ] + c [ where c is integral constant ]

This is your required answer, so students if you want to more illustration then please comment and thank you. Just visit on INTEGRATION OF SECX | INTEGRATION OF SEC^2X.

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**Integration of sec x** is so important for beginner students, because Integration of **secx** does learn to you about the how to do separation and how to do operation on **trigonometric function** for integrate various type of function.

In this session we will try to evaluate the **Integration of sec x** ; \( \int secxdx \)

So, students lets starts the solving, this is the so simple method to solve it, just in there have a some separation on secx.

We get,

Now multiplying secx with the numerator ( secx + tanx ) and separate them

And we get,

let secx + tanx = z

now differentiating both side with the respect to x

we get,

Now putting the value and we get,

[where c is Integral Constant]

Hence,

I = log|secx + tanx|+ c

Further you can separate that answer or solution you also get,

I = log|secx + tanx|+ c

Now replacing secx by 1/cosx and tanx by sinx/cosx

we get,

[where c is Integral Constant]

Now dividing with the cos(x/2) on both of numerator and denominator

We get,

This is the final answer of the Integration of sec x. As your according you can use both of the answers.

And with the same way you can evaluate the integration of cosecx.

Now we will try to evaluate the **integration of sec x tanx**

\( \int \)secx.tanx dx

Assume, secx = z

Now differentiating on both side with respect to x

We get,

Or, secx.tanx dx = dz

Now putting the value and we get,

\( \int(dz) \)

hence, I = z + c

Or, I = secx + c

Alternative method,

I =\[ \int secx.tanx dx \]

assume cosx = z

now differentiating on both side with respect to x

and we get,

-sinx dx = dz

Now putting the value on given integration

[ Where c is integral constant ]

Hence, I = secx + c

So students this is final answer of **integration of sec^x tan^x** and as you can see in both of the method answer will be same is secx + c

Let try to solving, in this case we solve the \( \int(sec^2x) dx \)

Once again, I = \( \int(sec^2x) dx \)

Now multiply tanx/tanx with the sec^2x

And we get,

Assume that tanx = z

Now differentiating on both side with respect to x

sec^{2}x dx = dz

Hence, I = tanx + c

So, tanx + c is the answer of** integration of secant squared x or sec^2x.**

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**JADAVPUR UNIVERSITY, KOLKATA**

**INSTITUTE OF ENGINEERING & MANAGEMENT, SALT LAKE, KOLKATA**

**TECHNO INDIA UNIVERSITY**

**UNIVERSITY OF CALCUTTA**

**KALYANI GOVT. JALPAIGURI GOVT ENGINEERING COLLEGE **

**JALPAIGURI GOVT ENGINEERING COLLEGE**

**TECHNO INDIA, SALT LAKE, SECTOR-V**

**HERITAGE INSTITUTE OF TECHNOLOGY, KOLKATA**

**UNIVERSITY OF KALYANI (NADIA)**

**MAKAUT (WBUT)**

**GOVT. COLLEGE OF ENGINEERING & LEATHER TECHNOLOGY, KOLKATA**

**GOVT. COLLEGE OF ENGINEERING & TEXTILE TECHNOLOGY, BERHAMPUR**

**HALDIA INSTITUTE OF TECHNOLOGY, HALDIA, PURBA MEDINIPUR**

**MEGHNATH SAHA INSTITUTE OF TECHNOLOGY, KOLKATA**

**GOVT. COLLEGE OF ENGINEERING & CERAMIC TECHNOLOGY, KOLKATA**

**GOVT. COLLEGE OF ENGINEERING & TEXTILE TECHNOLOGY, SERAMPORE**

**ACADEMY OF TECHNOLOGY, ADISAPTAGRAM, HOOGHLY**

**NETAJI SUBHAS ENGINEERING COLLEGE, GARIA, KOLKATA**

**ASANSOL ENGINEERING COLLEGE, ASANSOL, BURDWAN**

**COLLEGE OF ENGINEERING AND MANAGEMENT, KOLAGHAT**

**BIDHAN CHANDRA KRISHI VISWA VIDYALAYA**

**UTTAR BANGA KRISHI VISWA VIDYALAYA**

**MCKV INSTITUTE OF ENGINEERING, LILLUAH, HOWRAH**

**BUDGE BUDGE INSTITUTE OF TECHNOLOGY, BUDGE BUDGE**

**DURGAPUR INSTITUTE OF ADVANCED TECHNOLOGY & MANAGEMENT**

**B.P.PODDAR INSTITUTE OF MANAGEMENT & TECHNOLOGY, KOLKATA**

**DR.BC ROY**

**CIEM, TOLLYGUNGE**

**NSHM KNOWLEDGE CAMPUS**

**SAROJ MOHAN INSTITUTE OF TECHNOLOGY, GUPTIPARA, HOOGHLY**

**ABACUS INSTITUTE OF ENGINEERING & MANAGEMENT, MOGRA, HOOGHLY**

**REGENT EDUCATION AND RESEARCH FOUNDATION, BARASAT, KOLKATA**

**RCC INSTITUTE OF INFORMATION TECHNOLOGY, KOLKATA**

**SKFG, MANKUNDU, HOOGHLY**

**NSHM KNOWLEDGE CAMPUS**

**JIS GROUP**

** 2019**

** ENGLISH (Second Language)**

** NEW SYLLABUS**

** SECTION- A**

** READING COMPREHENSION (SEEN)**

**Read the passage carefully and answer the question that follow:**

“What have you written, Father ?” Swami asked apprehensively.

“Nothing for you. Give it to your headmaster and go to your class.”

“Have you written anything about our teacher Samuel ?”

“Yes. Plenty of things.”

“What has he done, Father ?”

“Everything is there in the letter. Give it to your headmaster.”

Swami went to school feeling that he was the worst boy on earthy. His consciences bothered him. He wasn’t at all sure if his description of Samuel had been accurate. He felt he had mixed up the real and the imagined.

Swami stopped on the roadside to make up his mind about Samuel. Samuel was not such a bad man after all. Personally he was much more friendly than the other teachers. Swami also felt Samuel had a special regard for him.

Swami’s head was dizzy with confusion he could not decide if Samuel really deserved the allegations made against him in the letter. The more he thought of Samuel, the more Swami grieved for him. To Samuel’s dark face, his thin moustache, unshaven cheek and yellow coat filled Swaminathan with sorrow.

**write the correct alternative to complete the following sentences :**

(a) Swami’s father wrote the letter to Swami’s ______________

(i) friends (ii) headmaster (iii) teacher (iv) mother

(b) The letter made Swami feel ______________________________________

(i) happy (ii) sad (iii) excited (iv) worried

(c) Swami stopped on his way to school to decide whether Samuel was a/am _______________

(i) bad person

(ii) good and friendly person

(iii) honest person

(iv) dishonest person

(d) Swami felt dizzy because he was ________________________________

(i) excited

(ii) indifferent

(iii) afraid

(iv) puzzled

(e) The more Swami thought of Samuel the more _______________________

(i) aggrieved he felt

(ii) cheerful he felt

(iii) disappointed he felt

(iv) angry he felt

**Complete the following sentences with information from the text :**

(i) Swami’____________________________________________________________________

___________________________________________________________________________.

(ii) While going to school Swami felt ________________________________________________

___________________________________________________________________________.

(iii) Other teachers were not as ____________________________________________________

____________________________________________________________________.

**Answer the following question :**

(i) Why did Swami’s conscience bother him ? _____________________________________________

______________________________________________________________________________.

(ii) How did Swami recall Samuel’s appearance ? ___________________________________________

__________________________________________________________________________.

**Read the poem and answer the question that follow :**

Once I crept in an oakwood – I was looking for a stag.

I met an old woman there – all knobbly stick and rag.

She said: ‘I have your secret here inside my little bag.’

Then she began to cackle and I began to quake.

She opened up her little bag and I came twice awake –

Surrounded by a starting tribe and me tied to a stake .

They : said ‘We are the oak-trees and your own family.

We are chopped down, we are torn up, you do not blink an eye.

Unless you make a promise now – now you are going to die.

‘Whenever you see an oak-tree felled, swear now you will plant two.

Unless you swear the black oak bark will wrinkle over you

And root you among the oaks where you were born but never grew.’

This was my dream beneath the boughs, the dream that altered me.

When I came out of the oakwood, back to human company,

My walk was the walk of a human child, but my heart was a tree.

- Write the correct alternative to complete the following sentences :

(a) The young child met ___________________________________________

(i) a forester

(ii) an old man

(iii) an old woman

(iv) a stag

(b) The old woman was _______________________________________________

(i) very smart

(ii) looking nervous

(iii) richly dressed

(iv) unsteady and poorly dressed

** READING COMPREHENSION (UNSEEN)**

**Read the following passage and answer the questions that follow :**

It is a known fact that children generally emulate the adults. When a child imitates an elder

Smoking by holding a pencil between his/her tiny fingers, most of us find it quite cute. We fail to

Realise that this harmless imitation often results in actual smoking when they grow up. So it is time

To take a pause and think about its consequences. According to WHO data, tobacco kills more than

Seven million people each year. While more than six million of those deaths are the result of direct

Tobacco use, many non-smokers die because they are to second -hand smoke.

Over eighty percent of individuals, who start smoking during adolescence, will continue to smoke in

Adulthood, while one-third of them will die prematurely due to smoking related disease like cancer,

Heart disease and chronic lung disease. Experimentation by children can rapidly escalate into heavy

Smoking.

Experts say that unless urgent action is taken immediately, the annual death toll could rise to more than eight million by 2030. This sure is a gloomy and cheerless world that we are leaving behind to our future generations. So it is time motivate youngsters to stay away from this deadly habit.

Dr. Shyama Chona, the educationist says, “The best way to motivate them stay away from smoking is to involve them in some activities.” She suggests the schools can organize events like sports, drama and story-telling sessions. Such participatory learning is very effective because it dissuades kids from smoking, when they themselves propagate its bad effects.

Parents have a big role to play when it comes to influencing their kids. Theatre director Arvind Gaur says that a smoking parent can never tell his/her child not to smoke because that would be too hypocritical.

*[Adapted from an article published in The Times of India dated March 14, 2018]*

**Tick the right answer**:

(i) We find it cute when

(a) a child imitates cute when

(b) an adult imitates a child

(c) a child copies an elder smoking by holding a pencil between his/her tiny fingers

(d) a child actually smokes

(ii) According to WHO data, each year direct use of tobacco kills –

(a) seven million people

(b) more than seven million people

(c) less than six million people

(d) more than six million people

(iii) In the opinion of experts urgent action should be taken

(a) immediately

(b) after one year

(c) by 2030

(d) by the parents

(iv) Dr. Shyama Chona is

(a) a medical practitioner

(b) a psychologist

(C) a social worker

(d) an educationist

(v) ‘Deadly habit’ in paragraph 3 refers to

(a) sports

(b) smoking

(c) play-acting

(d) story-telling

(vi) A smoking parent can

(a) always tell his child to smoke

(b) tell his child to be hypocritical

(c) never tell his child not to smoke

(d) tell his child to be honest and truthful

**Each of the following statements is either ‘True’ or ‘False’. Write ‘T’ for True and ‘F’ for False in the boxes on right5-hand side. Also pick out suitable line(s) or phrase(s) from the passage in support of your answers :**

(i) A child never imitates an elder smoking

Supporting sentence : ______________________________________________________________

(ii) Many non-smokers die because of passive smoking.

Supporting sentence : _____________________________________________________________

(iii) Our future generation will be left in a dark and hopeless world due to smoking.

Supporting sentence : ______________________________________________________________

**Answer the following Question**:

(i) How does a child imitate an elder smoking ?

Ans: __________________________________________________________________________________________________________________________________________________________________________

(ii) Name of some of the smoking-related disease.

Ans: __________________________________________________________________________________________________________________________________________________________________________

(iii) Why should urgent action be taken immediately to stop smoking ?

Ans: __________________________________________________________________________________________________________________________________________________________________________

(iv) How can children be motivate to stay away from smoking ?

**SECTION-B**

**GRAMMER AND VOCABULARY **

**4.Write the correct alternative to fill in the blanks : **

When Mrs. Gupta ___________(return/returned/was returning) home she noticed at once that thieves ____________ (enter/have entered/had entered) her house during her absence. The local police station ___________ (was informing/had informed/was informed) immediately.

- Do as directed :
- Fill in the blanks with appropriate articles and prepositions :

Sunil, __________ young driver jumped __________ the river and saved the child ____________ drowning.

- (i) He has worked out the answer carefully. (Change the voice)

Ans: ________________________________________________________________________________________________________________________________________________.

(ii) You do not take care of your health. You may fall ill.

(Join into a single sentence using an adverb clause)

Ans: ________________________________________________________________________________

_________________________________________________________________________________

(iii) “What have you written, sachin ?” The teacher said. (Change into Indirect Speech)

Ans: ________________________________________________________________________________

________________________________________________________________________________

**Choose the correct phrasal verbs from the list given below to replace the words underlined. Write the correct phrasal verbs in the boxes on the right hand side changing the form where necessary. There is one extra phrasal verb in the list.**

(i) Kshudiram __sacrificed__ his life for his motherland.

(ii) Police is __investigating__ the case.

(iii) I am __trying to find__ a job.

[look into, come by, look for, give up]

**Given below are the meanings of four words which you will find in the passage in Question No.3. Find out the words and write them in the appropriate boxes on the right-hand side :**

(i) imitate

(ii) increase rapidly

(iii) discourage

(iv) spread and promote an idea widely

** SECTION – C**

** WRITING **

**Write a story (within 100 words) using the given hints. Give a title to the story. [10]**

Three children collecting green mangoes near railway track – one sees a crack on rails – hears a train coming – senses danger – tears his red shirt – three pieces – waves like flags – driver stops – accident avoid – children rewarded.

**Write a letter (within 100 words) to the editor of an English daily stating reasons why plastic carry bags be banned**. [10]

Your letter may include the following points :

Large scale use of plastics carry bags – choking sewerage system – water logging on roads – drains overflowing – pollution of land and ocean – must be banned to save environment – use of substitutes like jute bags and paper bags.

**Your school is going to host an inter-school drama competition. Suppose you are the Cultural Secretary of your school. Write a notice (within 100 words) calling students to participate in the competition. Mention the time, date and venue of the competition. Your notice should be countersigned by the Headmistress/Headmaster of your school.**

** **

** SECTION – D**

** (SUPPLEMENTARY)**

** [FOR EXTERNAL CANDIDATES ONLY]**

**Fill in the blanks with appropriate forms of the words in brackets :**

(a) Mrs. Bhimani is well known for her ___________________ (kind).

(b) ___________ (honest) is the best policy.

(c) Sumona completed the race __________ (success).

(d) The gentleman is famous for his _________ (wise).

(e) Mr. Soren is an ____________ (honour) man.

**Add prefix with the following words to make antonyms :**

** (a) **Happy ______________________ .

(b) Please ______________________ .

(c) Movable ____________________ .

(d) Rational _____________________ .

(e) Logical _______________________ .

[dis______, un _______, ir _______, im _______, il ________]

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