Buddhism ssc cgl notes

         BUDDHISM (600BC)

• Buddhism was founded by  Gautama Buddha.
• Buddha was born on the Vaisakh  Purnima day in 563 B.C.
• He belonged to the Sakya clan of Kshatriyas.
• His father was Suddhodana, the ruler of Kapilvastu.
• He was born in Lumbini in Kapilvastu.
• His mother was Mahamaya of the Kosala dynasty.
• Buddha got Nirvana at the age of 35 years.
• Buddha got Nirvana at Uruvela on the bank of river Niranjana
• Buddha gave his first Sermon at Sarnath.
• Buddha's first sermon is called as "Dhama Chakraparivartana
• Buddha died in 483 B.C. at Kushinagar. Kushinagar has been identified with village Kasia in Deoria dis- trict of U.P.
• Buddha's last words were "All com- posite things, strive diligently".
• Buddha was brought up by his stepmothef Gautami.
• After seeing an old man, a sick man, a corpse and an ascetic Buddha decided to become a wanderer. Asvajit,Upali, Mogallana, Sariputra and Ananda weré five disciples of Buddha.

FIVE GREAT EVENTS OF BUDDHA'S LIVE AND THEIR SYMBOLES 

1. Birth: Lotus and Bull
2. Great Renunciation: Horse
3. Nirvana: Bodhi tree
4. First Sermon: Dharmachakra or wheel
5. Parinirvana or Death: Stupa

FOUR NOBLE TRUTH'S 

1. The world is full of sorrows.
2. Desire is root cause of sorrow.
3. If desire is conquered, all sorrows can be removed.
4. Desire can be removed by following the eight-fold path.

EIGHT FOLD PATH

(1)Right understandingg

(2) Right speech

3) Right livelihood

(4)Right mindfulness

(5) Right thought

(6)Right action

(7)Right effort and

(8)Right concentration

THE RATNAS 

1) bhuddhas

2) Dharmha

3) Sangha

Code of Conduct:
(1) Do not covet the property of otherss
(2) Do not commit violence
(3) Do not speak a lie
(4) Do not indulge in corrupt practices

BUDDHIST COUNCLs

• The First Council was held in 483 BC at Saptaparni cave near Rajagriha to compile the Dhamma Pitaka and Vinaya Pitaka. Chairman Mahakassapa, Patron: Ajatshatru

• The Second Council was held at Vaisali in 383 BC. The monks of Vaisali wanted some change in the rites. Schism into Shaviravadins and Mahasanghikas. Chairman : Sabakami, Patron Kalashoka

• The Third Council was held at Pataliputra duríng the reign of Ashoka 236 years after the death of Buddha. It was held under the Presidentship of Moggliputta Tissa to revise the scriptures.

• The Fourth Council was held during the reign of Kanishka in Kashmir under the Presidentship of Vasumitra, who was helped by Aswvaghosha and resulted in the division of Buddhists into Mahayanists and Hinayanists.

SECTS OF BUDDHuSM

Hinayana:

(a) Its followers belleved in the

original teachings of Buddha

b) They sought Individual sal

vation through self-discipline

and meditation.

(c) They did not believe in idol

worship

(d) Hinayana, like Jainism. Is a

religlon without God, Kama

taking the place of God

Mahayana:

(a) Its followers believed in the

heavenliness of Buddha and

sought the salvation of all

through the grace and help

of Buddha and Bodhisatvas

b) Believes in idolworship.

c)Believes that Nirvana is not a

negattve cessation of misery

but a positve state of bliss.

C) Mahayana had two chicf philo

sophical schools: the

Madhyamika and the

Yogachara.

BUDDHIST LITERATURE

The Sutta Pitaka is divided into

five Nikayas.

The five Nikayas are Digh

Nikaya, Majjhima Nikaya.

Samyutta Nikaya, Anguttara

Nikaya and Khuddaka Nikaya.

The Khuddaka Nikaya consists of

large number of miscellanceous  worker.

BUDDHIST ARCHITECTUREE

Buddhist architecture developed

essentially in three forms, viz.

(a) Stupa (relics of the Buddha or

some prominent Buddhist

monks are preserved)

b) Chaitya (prayer hall

(c)Vihara (resldence)

NCRT math class 9th polynomial exersice 2.3

                       EXERSICE 2.3

Question 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1
(ii) x – 
(iii) x
(iv) x + π
(v) 5 + 2x
     Solution:
     Let p(x) = x3 + 3x2 + 3x +1
(i) The zero of x + 1 is -1.
∴  p(x=-1) = (-1)^3 + 3(-1)^2 + 3(-1) +1
    = -1 + 3- 3 + 1 = 0
   Thus, the  remainder = 0

(ii) The zero of  is 

     Thus, the  remainder = 

iii) The zero of x is 0.
∴ p(0) = (0)^3 + 3(0)^2 + 3(0) + 1
= 0 + 0 + 0 + 1 = 1
Thus, the  remainder = 1.

(iv) The zero of x + π is -π.
       p(-π) = (-π)^3 + 3(- π)^2  + 3(- π) +1
       = -π3 + 3π2 + (-3π) + 1
       = – π3 + 3π2 – 3π +1
       Thus, the required remainder is                        -π3 + 3π2 – 3π+1.

(v) The zero of 5 + 2x is  .


     Thus, the remainder is  .

Question 2.
Find the remainder when x3 – ax2 + 6x – a is divided by x – a.
Solution:
We have, p(x) = x3 – ax2 + 6x – a and zero       of x – a is a.
  ∴ p(a) = (a)3 – a(a)2 + 6(a) – a
  = a3 – a3 + 6a – a = 5a
 Thus, the  remainder is 5a.

Question 3.
Check whether 7 + 3x is a factor of 3x3+7x.
Solution:
We have, p(x) = 3x3+7x. and zero of 7 + 3x is .

Since,( ) ≠ 0
i.e. the remainder is not 0.
∴ 3x3 + 7x is not divisib1e by 7 + 3x.
Thus, 7 + 3x is not a factor of 3x3 + 7x.

NCRT CLASS 9TH MATH CHAPTER 2( POLYNOMIALS EXERSICE 2.2 SOLUTION )

              EXERSICE 2.2 

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
 *    let  p(x) = 5x – 4x2 + 3
(i) so, p(x=0) = 5(0) – 4(0)2 + 3

                       = 0 – 0 + 3  =3

     Thus, p(x=o) is 3 . 

(ii) p(x=-1) = 5(-1) – 4(-1)2 + 3
                   = – 5x – 4x2 + 3 

                   = -9 + 3 = -6

     hence , the value of 5x – 4x2 + 3 at 

     x =   -1 is -6.

(iii) p(2) = 5(2) – 4(2)2 + 3 

               = 10 – 4(4) + 3
               = 10 – 16 + 3 = -3

                hence , the value of 5x – 4x2 + 3 at x = 2 is – 3.

Question 2.
Find p (0), p (1) and p (2) for each of the following polynomials.
(i) p(y) = y2 – y +1
(ii) p (t) = 2 +1 + 2t2 -t3
(iii) P (x) = x3
(iv) p (x) = (x-1) (x+1)
Solution:
(i) Given that p(y) = y2 – y + 1.

    so,
∴  P(y=0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1
    p(y=1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1
    p(Y=2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3
(ii) Given that p(t) = 2 + t + 2t2 – t3
∴ p(t=0) = 2 + 0 + 2(0)2 – (0)3

    = 2 + 0 + 0 – 0=2
   P(t=1) = 2 + 1 + 2(1)2 – (1)3
   = 2 + 1 + 2 – 1 = 4
  p(t=2) = 2 + 2 + 2(2)2 – (2)3
   = 2 + 2 + 8 – 8 = 4
(iii) Given that p(x) = x3
∴ p(x=0) = (0)3 = 0, p(1) = (1)3 = 1
    p(x=2) = (2)3 = 8
(iv) Given that p(x) = (x – 1)(x + 1)
∴ p(x=0) = (0 – 1)(0 + 1) = (-1)(1) = -1
   p(x=1) = (1 – 1)(1 +1) = (0)(2) = 0
   P(x=2) = (2 – 1)(2 + 1) = (1)(3) = 3

Question 3.

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1,x = –
(ii) p (x) = 5x – π, x = 
(iii) p (x) = x2 – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x2, x = 0
(vi) p (x) = 1x + m, x = – 
(vii) P (x) = 3x2 – 1, x = – ,
(viii) p (x) = 2x + 1, x = 
Solution:
(i) We have , p(x) = 3x + 1

(ii) We have, p(x) = 5x – π
∴ 
(iii) We have, p(x) = x2 – 1
∴ p(1) = (1)2 – 1 = 1 – 1=0
Since, p(1) = 0, so x = 1 is a zero of x2 -1.
Also, p(-1) = (-1)2 -1 = 1 – 1 = 0
Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1.

(iv) We have, p(x) = (x + 1)(x – 2)
∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0
Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).
Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0
Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x2
∴ p(o) = (0)2 = 0
Since, p(0) = 0, so, x = 0 is a zero of x2.

(vi) We have, p(x) = lx + m

(vii) We have, p(x) = 3x2 – 1

(viii) We have, p(x) = 2x + 1
∴ 
Since,  ≠ 0, so, x =  is not a zero of 2x + 1.

Question 4.
Find the zero of the polynomial in each of the following cases
(i) p(x)=x+5
(ii) p (x) = x – 5
(iii) p (x) = 2x + 5
(iv) p (x) = 3x – 2
(v) p (x) = 3x
(vi) p (x)= ax, a≠0
(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.
Solution:
(i)  p(x) = x + 5. 

     Since, p(x) = 0
⇒ x + 5 = 0
⇒ x = -5.
    Thus, zero of x + 5 is  -5.

ii) P(x)=x - 5

    Since, p(x)=0.

     x-5=0 ⇒x=5.

     hence, zero of x - 5 is 5.

 iii)p(x) = 2x+5

      Since, p(x)=0.

      2x+5=0⇒2x=-5

       x=-5/2.

      hence zero of 2x+5 is -5/2.

iv)  p(x)= 3x-2

      Since, p(x)=0

       3x-2=0⇒x=2/3

       hence zero of 3x-2  is 2/3.

v)   p(x)= 3x

       Since, p(x)=0

       3x=0 ⇒x=0

       hence zero of 3x is 0.

vi)   p(x) =ax

        since ,p(x) =0

        ax=o ⇒x=0

         hence zero of ax is 0.

vii)  p(x)=cx+d

        since ,p(x) =0

        cx+d =0 ⇒ cx=-d

       x=-d/c

        hence zero of cx+d is -d/c

    

NCRT 9th class chapter=2 polynomials exercise 2.1

                  Exercise 2.1

Question :1)

Which of the following expressions are polynomials in one variable and which are not?

State reasons for vour answer.

 i)   4x²-3x +7

  Answer: 

      In equation 4x²-3x +7 , the one variable 'x' is given and has a power in whole number  , so the above equation is in one variable .

ii)   y²+√2

      solution 

     In equation y²+√2 , the one variable 'x' is 

      given , so the above equation is in one 

      variable .

iii) 3√t+√2 t

       Answer:

      In equation 3√t+t√2 , the power of √t =t^1/2      so the power is in fraction thats why we            can say that above expression  is not in one             variable .

iv) y+2/y

Answer :

     In equation y+2/y , the power of y is (-1).

     it is not in whole number i.e the above                 equation is not in one variable .

  

v)  x10+  y3 + t50

Answer:

     In equation x10+  y3 + t50, every variable is 

     in whole number but there is different                   variable

    (x,y,z) is given . so the above expression is          not in one variable.

 Question : 2 )

Write the coefficients of x2 in each of the following

(i) 2 + x2 + x
(ii) 2 – x2 + x3
(iii)  x2 + x
(iv) √2 x – 1
Solution:
(i)   The given polynomial is 2 + x2 + x.
       The coefficient of x2 is 1.
(ii)  The given polynomial is 2 – x2 + x3.
       The coefficient of x2 is -1.
(iii) The given polynomial is  + x.
       The coefficient of x2 is .
(iv) The given polynomial is √2 x – 1.
       The coefficient of x2 is 0.

 Question: 3.)

       Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
(i)   Abmomial of degree 35 can be                    x35 -4.
(ii)   A monomial of degree 100 can be              5x100.

Question : 4)
       Write the degree of each of the                   following polynomials.
(i) 5x3+4x2 + 7x
(ii) 4 – y2
(iii) 5t – √7
(iv) 3
Solution:
(i) The given polynomial is 5x3 + 4x2 + 7x.
The highest power of the variable x is 3.
So, the degree of the polynomial is 3.
(ii) The given polynomial is 4- y2. The highest
power of the variable y is 2.
So, the degree of the polynomial is 2.
(iii) The given polynomial is 5t – √7 . The highest power of variable t is 1. So, the degree of the polynomial is 1.
(iv) Since, 3 = 3x° [∵ x°=1]
So, the degree of the polynomial is 0.

Question 5.
Classify the following as linear, quadratic and cubic polynomials.
(i) x2+ x
(ii) x – x3
(iii) y + y2+4
(iv) 1 + x
(v) 3t
(vi) r2
(vii) 7x3
Solution:
(i) The degree of x2 + x is 2. So, it is a quadratic polynomial.
(ii) The degree of x – x3 is 3. So, it is a cubic polynomial.
(iii) The degree of y + y2 + 4 is 2. So, it is a quadratic polynomial.
(iv) The degree of 1 + x is 1. So, it is a linear polynomial.
(v) The degree of 3t is 1. So, it is a linear polynomial.
(vi) The degree of r2 is 2. So, it is a quadratic polynomial.
(vii) The degree of 7x3 is 3. So, it is a cubic polynomial.

UART communication protocol (embedded system)

What is UART?

• Universal asynchronous receiver / transmitter

• Protocol for exchanging serial data between

two devices

TX

TX

• Uses only two wires

- TX to RX (each direction)

• Can be simplex, half-duplex, or full-duplex

• Data is transmitted as frames.

Where is UART used?

• UART was one of the earliest serial protocols

Serial (COM) ports, RS-232, modems, etc.

• Popularity of UART is however decreasing:

- SPI and 12C between components

- Ethernet and USB between computers and

peripherals

• Still important for lower-speed, low-throughput

applications

About timing / synchronization

• UART is asynchronous – the transmitter and

receiver do not share a common clock

Common UART

baud rates

• The transmitter and receiver therefore must:

Transmit at the same (known) speed

- Use the same frame structure / parameters.

UART frame format:

• UART frames consist of:
• Start / stop bits
• - Data bits
• Parity bit (optional)
• • High voltage ("mark") = 1, low voltage ("space") = 0

Start and stop bits

• The start bit indicates data is coming

Transition from idle (high) to low

The stop bit(s) indicate data is complete

Stay / return to idle (high)

- Second (optional) stop bit.

Data bit :

• User (useful) data

• Length: 5 to 9 bits (usually 7 or 8)

• Data is typically sent with the least

significant bit (LSB) first.

Example:
- 7-bit ASCII 'S' (0x52) = 1 0 10011

LSB order= 110010 1

Parity bit (optional)

• Used for error detection

• Example:

ASCII 'S' (0x52 = 1010011) four 1's

- If even parity, parity bit is 0 (because

number of 1's already even)

• Even parity: number of 1's must be even

• Odd parity : number of 1's must be odd

- If odd parity, parity bit is 1 (to make the

number of 1's odd)

Summary

• UART = universal asynchronous receiver / transmitter

Simple, two-wire protocol for exchanging serial data

No shared clock – baud rate must be configured the

same on both sides

Start / stop bits used to frame user data

Optional parity bit for detecting single-bit errors

• Widely used serial data protocol, but slowly being

replaced by SPI, 12C, USB and Ethernet.

I2C bus communication protocol

I2C bus communication protocols 

What is 12c:

-Inter Integrated Circuit

-Bidirectional Data Transfer

-Half duplex (have only one data line)

-Synchronous bus so data is clocked with clock signal

-Colock is controlled when data line is changed.

Speed of 12c:

low (under 100 kbps)

Fast (400 kbps)

high speed (3.4 mbps) 12C V2.0

2 wire communication:

SDA and SCL

Vtg high = 1, low = 0.

when SLC =1, it is valid data, when SLC=0, data is changed .

Basic protocol is master slave protocol

-Master controls the clock

-Slave device may hold the clock low to prevent data transfer

-no data transfer is present when clock is low

-It is a kind of wired and connection

-need to put pullup resistor

-default it is a open-drain or open-collector, so that adding pull up resistor is necessary so that it

will have only two states that is 1.floating high and 2.drive low

-Default state is high when no device is pulling it low.

Data Transfer:

• It is byte oriented (bit)
• Ack transmited by recepient of the data
• MSB first
• First byte is address
• First byte is transmitted by master and addressed slave is the recepient
• Next byte is based on the last bit (R/ W)
• 7 bit address
• 1 bit R/ W
• 0-master write 
• First byte is transmitted by master and addressed slave is the recepient
• Next byte is based on the last bit (R/ W)
• 7bit address
• 1 bit R/ W
• 0-master write

1-master receiver.

 

Full I2C transfer:

12C Multi Master:

• It is a multi master bus
• So bus arbitration is required
• 1
• When two device tries to drive SDA to different value
• It is necessary to be sure that is not interfiering with another message
• If a device is trying to send logic one but hears logic 0, it immediately stops transmission and gives
• the other sender priority
• Synch needed in SCL

Advantages:

• Good for comm in On-board devices
• mc
• Easy to link multiple devices because of addressing scheme
• Cost and complexity do not scale up with the num of devices

Disadvantages:

The complexity of supporting software components can be higher than that of scheme (EX. SPI-No

need of address in SPI)