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. Today we will know about the integration formula and at the we will download formula list in pdf file.

**Introduction: **

First you have to know that what is integration? In previous class we uploaded the Integration tutorials you can views that. In generally there have two types of formula that’s are (A) fundamental formula and (B) some standard form on integrals. But some where you can see formula divide on the basis of function like **algebraic function, trigonometric function, exponential function etc**. it depends upon the different teachers.

**Let’s see the Formula list**

**(A) Fundamental Integrals :**

1) ∫ Af(x) dx = A ∫ f(x) dx

2) ∫ dx = x + c

3) ∫ x^{n} dx = \( \frac {x^(n+1)}{n+1} \) +c ; ( n≠1)

4) ∫ e^{mx} dx = \( \frac{e^(mx)}{m} \) + c

5) ∫ (1/x) dx = log|x| + c

6) ∫ e^{x} dx = e^{x} + c

7) ∫ a^{mx} dx = a^{mx} / mlog|a| + c ( a≠1)

8) ∫ a^{x} dx = a^{x} /log|a\ + c

9) ∫ sinmx dx = \( – \frac {cosmx}{m} \) + c

10) ∫ sinx dx = – cosx + c

11) ∫ cosmx dx = \( \frac {sinmx}{m} \) + c

12) ∫ cosx dx = sinx + c

13) ∫ sec^{2}mx dx = \( \frac {tanmx}{m} \) + c

14) ∫ sec^{2} x dx = tanx + c

15) ∫ cosec^{2}mx dx = \( – \frac {cotmx}{m} \) + c

16) ∫ cosec^{2}x dx = – cotx + c

17) ∫ secmx · tanmx dx = \( \frac{secmx}{m} \) + c

18) ∫ secx · tanx dx = secx + c

19) ∫ cosecmx · cotmx dx = \( – \frac {cosecmx}{m} \) + c

20) ∫ cosecx · cotx dx = – cosecx + c

**# Some important forms of Integrals :**

21) If n≠ -1 then, ∫ {f(x)}^{n} · f^{‘}(x) dx = (1/n+1) × {f(x)}^{n+1} + c

22) ∫ f^{‘}(x)/f(x) dx = log|f(x)| + c

23) ∫ e^{x} [ f(x) + f^{‘}(x) ] dx = e^{x}·f(x) + c

24) ∫ tanx dx = log|secx| + c

25) ∫ cotx dx = log|sinx| + c

26) ∫ secx dx = log| secx + tanx | + c

27) cosecx dx = log | cosecx – cotx | + c

**(B) Some Standard Integrals:**

35) ∫ e^{ax}· sinbx dx = \( \frac{e^(ax)}{a^2+b^2} \) ( asinbx – bcosbx ) + c

36) ∫ e^{ax}· cosbx dx = \( \frac {e^(ax)}{a^2+b^2} \) (a cosbx + b sinbx ) + c

**SUMMATION SERIES INTEGRATION SHORTCUT FOR CBSE/JEE MAIN/JEE ADVANCED/IIT JEE/NTA/NDA**

INTEGRATION FORMULA DOWNLOAD IN PDF FILE

INTEGRATION OF SECX | INTEGRATION OF SEC^2X

INTEGRATION OF LOG X | integration of logx

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Hello students, I am Bijoy Sir and welcome to our educational forum or portal.

Today we will discuss about the Integration, but you of all know that very well, Integration is a huge part in mathematics. Then all of the topics of Integration can’t be covered in one class, so we will discuss the all topics but lightly, means we will see all types of Integration in with the some example. We will take more illustration on the topic of Integration in upcoming classes or tutorials.

Let’s start the discussion –

– Students this is the class of Integration so we will start from very beginning. Before starting Integration I want to tell you something that, Integration mainly developed by Newton Sir.

When Newton sir was trying to calculate the value gravitational potential, but the same time had no sufficient formula for proper calculation. Then Newton sir was thinking we need a new type of formula, that can give us a new way to sum. Then Newton sir had given a way to adding some element, which known as summation or Integration. We can add the very small element by the Integration. Which we will discuss very well in the tutorials of Definite Integration or Summation Series

In the earlier section we already learnt the derivatives or differentiation, foe properly understanding the Integration we have to just remind the derivatives or differentiation.

Definition- In generally Integration is the inverse process of derivatives or in simple language integration is anti derivative s or primitive of derivatives.

In mathematically,

Let the derivative of a function [ f(x)] is g(x). Then,

\( \frac {df(x)}{dx}= g(x) \)

Then we can say Integration of g(x) is equal to f(x). [ where derivative of f(x) is g(x).]

So, we can write ,

\( \frac{df(x)}{dx}=g(x) \)

Or, d f(x) = g(x) ·dx

or, f(x) = \( \frac{1}{d} \)g(x)· dx

Hence, f(x) = ∫ g(x)· dx + c [ where ‘c’ is integral constant ]

Note: If we take limit on integration the constant term ‘c’ will be automatically vanished.

# We can write that equation in another way

like, \( \frac {df(x)}{dx}=g(x) \)

Or, d f(x) = g(x) . dx

Now taking the integration on the both side we get,

or, ∫ d f(x) = ∫ g(x)· dx + c

or, \( \frac{1}{d} \) · df(x) = ∫ g(x)· dx + c

∴ f(x) = ∫ g(x)· dx + c

[ where, ∫ = 1/d we already learnt that ]

Now it’s time to see the illustration or time to learning the integration, that how can we deal with the integration.

**Integration is two types – **(**a) Indefinite Integration and (b) Definite Integration. We study both of them but in separate classes.**

** Indefinite Integration :**

Indefinite Integration mainly use learning the integration that we can apply in Definite integration or in higher studies.

Indefinite integration divides in three types according to the solving method – **i) Basic integration ii) By substitution, iii) By parts method, and another part is integration on some special function.**

Downlad Here Integration Formula In Pdf File

**i) Basic Integration :**

Basic integration is simplest type of integration comparison to other types of integration. We just do it by separation the function, means in this case first of all we convert the given function in a simple form and then do it.

Let try to understanding by some illustration.

Illustration :1

Evaluate the given integral \( \int \frac{1}{x^(\frac{-2}{3})} dx \)

This is the most of basic integration like,

∫ x^{n} dx = \( \frac {x^(n+1)}{n+1} \)+ c

Here n = +2/3, hence the required integration

or, \( \int \frac{1}{x^(\frac{-2}{3})} dx \)=\( \frac{x^(\frac{2}{3}+1)}{(\frac{2}{3}+1)} \)+c

or, \( \int \frac{1}{x^(\frac{-2}{3})} dx \) = \( \frac{3}{5}x^\frac{5}{3} \) + c

Illustration :2

Ans: Let, I = ∫ cos^{2} x · dx

or, I = 1/2 [∫ 2·cos^{2}x dx]

or, I = [ ∫ (1 + cos2x) dx]

or, 2I = ∫ dx + ∫ cos2x dx +c

or, 2I = x + sin2x/2 + c

or, I = \( \frac{1}{2}[ x + \frac{sin2x}{2}] \) + c

**II) Substitution Method :**

Definition of substitution method – Integration is made easier with the help of substitution on various variables. We will see a function will be simple by substitution for the given variable. This is one of the most important and useful methods for evaluating the integral.

Now we will take few examples that help us to understanding the concepts.

Example :1

Evaluate the integration of \( \int \frac{1}{x (lnx)^n} dx \), n ≠ 1

Ans: Let, I = \( \int \frac{1}{x (lnx)^n} dx \)

[Now we are trying to understand that how to use substitution methods in integral]

Now, we substitute the (lnx) by replacing ‘z’

Hence, lnx = z

Now differentiate on both sides with respect to ‘x’ we get,

or, \( \frac{d(lnx)}{dx} = \frac{dz}{dx} \)

or, (1/x)· dx = dz

And now putting the value of 1/x and( lnx) we get,

Hence the required integration is equal to \( \frac{(lnx)^(-n+1)}{-n+1} \) + c

Example :2

Evaluate the integration of

Now we will replace the x^{n} – 1 by z^{2}, hence x^{n} – 1 = z^{2}

And now differentiate both sides with respect to ‘x’ we get,

or, \( \frac {d(x^n-1)}{dx} = \frac{dz^2}{dx} \)

or, n · x^{n –} ^{1 }dx = 2 z dz

or, n · \( \frac {x^n}{x} \)· dx = 2 z dz

Now putting the value of x^{n} , from x^{n} – 1 = z^{2}, x^{n} = z^{2} + 1

We get, or, n·\( \frac {z^2 + 1}{x}dx = 2zdz \)

or, \( \frac{dx}{x} = \frac{2z} {n(z^2+1)}dz \)

And now putting the value of dx/x and x^{n-1} in the given integral we get,

III) By parts methods of Integration :

When the integration has given in the form of a product of two functions, in this case the integral can’t be able to solve by transformation or substitution methods, then we use this method to solve the given integral.

# Fundamental rules of By parts Methods –

In this case we use, a product of two functions f(x) and g(x) . So now our Integral looks like in the form = ∫ f(x) · g(x) dx .

* The fundamental formula of By Parts Methods –

I = ∫ f(x) · g(x) dx = f(x) ∫ g(x) dx – \( \int[\frac{df(x)}{dx}\int g(x)dx] dx \)

This is the main fundamental rule to solving by parts methods.

But, there has some problems or confusion that which function we take first f(x) or g(x), to clear this problem, we have a solution that also fundamental rule is the form of ILATE or LIATE, we take the function first according to this name, as which come first.

So, let’s see some example to better understanding.

Example :1

Evaluate the integral ∫ x^{2} e^{x} dx

Ans: Let, I = ∫ x^{2} e^{x} dx and x^{2} = f(x) & e^{x} = g(x)

Then, according to the rule we have to take f(x) first due to algebraic function.

Given I = ∫ x^{2} e^{x }now write the fundamental formula we get,

In the second part, we have to use by parts once again due to product of two functions.

Note: Students, a full class on by parts methods will be coming soon, so there we will discuss briefly about the by parts methods. Then please subscribe our website [jeewithbijoy.com

]]>**Hello students I am Bijoy, completed B.tech from Jadavpur University and this is my educational wesite for Board Exams and JEE. Today I will give you a question set on Wave Motion for IIT JEE, this question set only made for IIT JEE level.**

Question starts from here:

Q. y ( x, t ) = \( \frac {0.8}{[(4x+5t)^2+5]} \) represents a moving pulse where x and y are in metre and t in second. Then choose the correct alternatives.

(a) Pulse is moving in positive x – direction

(b) in 2sec it will travel a distance of 2.5 m

(c) it is maximum displacement is 0.16 m

(d) it is a symmetric pulse

# One end of 12.0 m long rubber tube with a total mass of 0.9 kg is fastened to a fixed support. A cord attached to the other and passes over a pulley and supports an object with a mass of 5.0 kg. The tube is struck a transverse blow at one end. Find the time required for the pulse to reach the other end. ( g = 9.8 m/s ^{2}).

# A uniform rope of mass 0.10kg and length 2.45 m hangs from a ceiling.

(a) Find the speed of transverse wave in the rope at a point 0.5 m distant from the lower end.

(b) Calculate the time taken by a transverse wave to travel the full length of the rope.

# A stretched string is forced to transmit transverse waves by a means of an oscillator coupled to one end. The string has a diameter of 4 mm. The amplitude of the oscillation is 10^{-4} m and the frequency is 10 Hz. Tension in the string is 100 N and mass density of wire is 4.2 ×10^{3} kg/m ^{3}. Find :

(a) The equation of the waves along the string

(b) The energy per unit volume of the wave

(c) The average energy flow per unit time across any section of the string and

(d) Power required to drive the oscillator.

# A simple harmonic wave of amplitude 8 units travels along positive x – axis. At any given instant of time, for a particle at a distance of 10 cm from the origin, the displacement is +6 units, and for a particle at a distance of 25 cm from the origin, the displacement is +4 units, calculate the wavelength of the given harmonic wave.

# Wave passing through a certain medium have a dispersion relation ω (k) = ( ω_{0}^{2} + α^{2}k^{2} )^{1/2}. Here ω_{0} and α are constants. Find phase velocity and group velocity in this medium.

# Consider a wave propagating in the negative x – direction whose frequency is 100 hz. At t = 5 s the displacement associated with the wave is given by, y = 0.5 cos (0.1x). Where x and y are measured in centimeters and t in seconds. Obtain the displacement (as a functioned of x ) at t =10 s. What is the wavelength and velocity associated with the wave?

# A thin string is held at one end and oscillates vertically so that, y(x=0, t) = 8sin 4t (cm). Neglect the gravitational force. The string’s linear mass density is 0.2 kg/m and its tension is 1 N. The string passes through a bath filled with 1 kg water. Due to friction heat is transferred to the bath. The heat transfer efficiency is 50%. Calculate how much time passes before the temperature of the bath rises one degree Kelvin.

# (a) Write the expression for y as a function of x and t for a sinusoidal wave travelling along a rope in the negative x direction with the following characteristics :

A =8.0 cm, wavelength (lamda) = 80.0 cm, f = 3.00 Hz, and y (0, t) = 0 at t = 0

(b) Write the expression for y as function of x and for the wave in part (a), assuming that y (x, 0) = 0 at the point x = 10.0 cm.

# A string of length L consists of two sections : The left half has mass per unit length μ = μ_{0}/2 , whereas the right half has a mass per unit length μ^{‘} = 3μ = 3μ_{0}/2. Tension in the string is T0. Notice from the data given that this string ha same total mass as a uniform string of length L and of mass per unit length μ_{0}.

(a) Find the speeds v and v’ at which transverse wave pulses travel in the two sections. Express the speeds in term of T_{0} and μ_{0 } and also as multiplies of the speed μ_{0} = [T_{0} / μ_{0}]^{1/2}.

(b) Find the time required for a pulse to travel from one end of the string to the other. Give your results as a multiple of t_{0} = [L/v_{0}].

# A block of mass is attached with a string of mass m and length l as shown in figure. The whole system is placed on a planet whose mass and radius is three times the mass and radius of earth. Find

(a) The time taken by a wave pulse to travel from one end A to B of the string.

(b) The ratio of maximum and minimum velocity of wave pulse (g = acceleration due to gravity on earth).

# One end of each two identical springs, each of force – constant 0.5 N/m, are attached on the opposite sides of a wooden block of mass 0.1 kg. The other ends of the springs are connected to separate rigid supports such that the springs are unstitched and are collinear in a horizontal plane. To the wooden piece is fixed a pointer which touches a vertically moving plane paper. The wooden piece, kept on a smooth horizontal table is now displaced by 0.02 m along the line of spring and released. If the speed of paper is 0.1 m/s, find the equation of the path traced by the pointer on the paper and the distance between two consecutive maxima on this path.

# A long string having a cross – sectional area 0.08 mm^{2} and density 12.5 kg/m ^{3}, is subjected to a tension of 64 N along the x – axis. One end of this string is attached to vibrator moving in transverse direction at a frequency of 20 Hz. At t = 0, the source is at a maximum displacement y = 1.0 cm.

(a) Find the speed of the wave traveling on the string.

(b) Write the equation for the wave.

(c) What is the displacement of the particle of the string at x = 50 cm at time t = 0.05?

(d) What is the velocity of this particle at this instant?

Note : Answer of this part will be coming soon.

# MOMENT OF INERTIA ALL FORMULA PDF-1

# MOMENT OF INERTIA FORMULA PDF -2

# INTEGRATION FORMULA LIST IN PDF

# MECHANICS TOP 30 QUESTION AND SOLUTION FOR JEE/NEET | BIJOY SIR

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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.

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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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