Showing posts with label cbse. Show all posts
Showing posts with label cbse. Show all posts

C B S E Date sheet for March 2019 Declared



Date Sheet for March 2019




cbse date sheet 2019 class 12,CBSE(  Central Board of Secondary Education) released 10th and  12th Board Exam 2019 Date Sheet on its  cbse.nic.in site. CBSE 's12th Board Examination will begin from  15th February and Class 10th Board exam will commence from 21st February .


cbse date sheet for march 2019


Date sheet released in cbse.nic.in. website


Date sheet of  10th and  12th classes for March 2019 Exam CBSE (Central Board of Secondary Education) released in its site  cbse.nic.in The exam of 12th class will be starting from 15th February 2019  and Class 10th Board exam will be  starting from 21 th February . cbse board exam date 2019 class 10 can be downloaded from  cbse.nic.in.

 Date Sheet of Board examination is prepared based on the subject combinations opted by the students of Class10th and Class10+2th. 
The core examinations  would commence from March 1, 2019 for Class 10th and 10+2th. CBSE Board has also decided to conduct the examination of skill subjects in latter half of Feb, 2019.
Last paper of 10+2 class is of   MULTIMEDIA & WEB INTRO TO HOSP MGMT on   WEDNESDAY,  03rd APRIL 2019 067


Time of  Exam


Answer books would be distributed to candidates between 10.00 - 10.15 A.M. and Question paper will be distributed at 10.15 A.M. Duration of time for each paper has been indicated in the date-sheet. From 10.15 -10.30 A.M.(15 minutes) candidates shall read the question paper . Start Time of paper will be 10:30 Am and concluding time will be different for different subject .

Complete Date sheet along with cbse date sheet 2019 class 12, can be Downloaded from the link given here  cbse.nic.in.  The official circular released by the board on cbse.nic.in, the board states that, CBSE(Central Board of Secondary Education) This time Date Sheet of Boards examination is prepared in view of  subjects combination opted by the students of these two classes .That was the latest news regarding cbse exam date 2019 class 10,cbse 12th date sheet 2019,


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HOW TO PROVE TRIGONOMETRIC IDENTITIES || TRIGONOMETRY


Proof of trigonometric identities , trigonometric identities problems, proving trigonometric identities formulas,these trigonometric identities of class 10, fundamental trigonometric identities,trigonometric identities class 11 and its formation with the help of some examples.

How to prove Identity


cos 6x = 32cos6 x - 48.cos4 x   + 18.cos2 x  - 6.cos2 x  - 1

Proof

1st of all  rewrite 3x as 3.2x

L.H.S. = cos 6x =  cos (3.2x) 

Now using the result cos 3θ = 4cos3 θ - 3 cos θ  -----(1)

Replacing θ as 2x in (1), we get 

L.H.S. = 4cos3 2x - 3 cos 2x  -----------(2)


Now using the result  1+ cos 2θ = 2 cos2 θ 

                                   ⇒ cos 2θ = 2 cos2 θ -1

Replacing cos 2x = 2 cos2 x -1 in (2), we get 

L.H.S.= 4 {2cos2 x -1}3 - 3 {cos2 x  -1}


Now using the result {a - b }3 = {a}3 - b }3  -  3{a }2 .b   + 3(a). b2

  cos 6x    = 4[ {2cos2 x  }3 - { 1 }3  -  3{2cos2 x  }2 .1   +3.(2cos2 x) .12 ] - 3 . {cos2 x  -1}


Taking the product of powers to simplify it

cos 6x  =   4[ 8cos6 x  - 1 - 12cos4 x  + 6cos2 x]  - 3{2cos2 x-1}

Multiply by 4 in 1st term and multiply by -3 in 2nd term

 cos 6x  = 32cos6 x  - 4 - 48cos4 x  + 24cos2 x  - 6cos2 x + 3

Adding the like powers terms and arranging in descending order

cos 6x   = 32cos6 x - 48cos4 x  + 18cos2 x  - 6cos2 x  - 1

Hence the Proof



Prove the Identity 

tan (2x) =  2tan x  1 - tan2 x 

Proof

We know that 


tan (A+B) =  tan A +  tan B1 - tan A tan B 

Put A = B  = x in above formula . then it becomes

tan (x+x) =  tan x +  tan x1 - tan x tan x 


tan (2x) =  2tan x  1 - tan2 x 
Hence the Proof


Prove that sin 2x = 2sin x cos x

Proof


As we know that sin (A + B) = sin A cos B + cos A sin B..  ...(1)

Put A = B  = x in ...   (1)

sin (x + x) = sin x cos x + cos x sin x

sin (2x) = sin x cos x +  sin x cos x

sin (2x) = 2 sin x cos x

Hence the Proof


Prove that cos 2x = cos2 x - cos2 x

Proof


As we know that cos (A + B) = cos A cos B - sin A sin B..  ...(1)
Put X = A = B in (1) , we get

cos (x + x) = cos x cos x - sin x sin x

cos 2x = cos2 x - sin2 x   

cos 2x = cos2 x - sin2 x   


Hence the Proof


                                                                                                                                                                   

 Using the result 
1+cos 2θ = 2cos2 θ
cos 2θ = 2cos2 θ -1 -------------(1)
Replacing θ with 2x in eq (1)
1+ cos 4x = 2cos2 2x
cos 4x = 2cos2 2x -1

Again using  cos 2θ = 2cos2 θ -1

cos 4x = 2Sq(2cos2 x -1) -1

It is the square of 2cos2 x -1

cos 4x = 2Sq(2cos2 x -1) -1

cos 4x = 2(4cos4 x +1 - 4cos2 x) -1

cos 4x = 8cos4 x +2 - 8cos2 x -1

cos 4x =  8cos4 x - 8cos2 x +1

Hence the Proof


What is the value of sin3x?



To find the value of sin 3x ,  use this formula which contain sin (A+B)
therefore sin (A+B) = sin A cos B cos A sin B——-(1)
put A = 2x and B = x in (1)
then Sin 3x = sin 2x cos x + cos 2x sin x

As we know that cos 2x = 1 - 2sin3 x and sin 2x = 2 sin x cos x


sin 3x = (2 sin x cos x) cos x + (1 - 2sin3 x ) sin x
sin 3 x = 2 sin x cos2 x + sin x -  2sin3 x

As we know that cos2 x = 1sin2 x

sin 3x= 2 sin x (1-sin3 x) + sin x - 2sin3 x
sin 3x = 2 sin x -2 sin3 x + sin x - 2sin3 x
sin 3x = 3 sin x - 4 sin3 x

Similarly we prove that cos 3x= 4 cos3 x - 3 cos x
For learning and memorising more trigonometric formulas

Conclusion




In this post I have discussed trigonometric identities ,trigonometric identities problems, proving trigonometric identities formulas . If this post helped you little bit, then please share it with your friends to benefit them, comment your views on it to boost me and to do better, and also follow me on my Blog .We shell meet in next post till then Bye


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CBSE TO IMPLEMENT NEW PATTERN OF QUESTION PAPERS FOR THESE CLASSES FROM 2020

Hello Friends ! Today we are going to discuss the new proposal of Central Board of School Education (CBSE) to introduce new pattern of question papers for 10th and 12th classes from 2020.


Classes 10th and 12th Question Papers Style


                   There is a big  news coming in  media regarding some changes are being planned by CBSE in the pattern of Question Papers of their board classes 10th and 12th .  The central Board of School Education ( CBSE ) has decided to introduce new pattern of  Question Papers for 10th and 12th classes as a part of revamp that would change examination  schedule for vocational  subjects , It would also be  implementing to other mains subjects.

What is the need of Change in existing Pattern

          
       According to the sources from the The Ministry of Human Resources and Development and some agencies , the new pattern of question papers of classes  of 10th and 12th will  not support the students who are dependent on  rote learning , it will be design to discourages such type of students .

       This pattern  would also stop students from total  blind copying of text books at home and pasting or vomiting of texts in the  examination hall in the Answer Book .

       The new pattern of question Papers of these classes would test students on their analytical skill and reasoning abilities .
         

           The CBSE also thinks that  its new steps  will also increase quality Education and better result of its institutions .


These are the majors changes expected in new pattern of CBSE 


              The new question papers pattern  will be design to check the analytical thinking and art of problems solving of the students in the examination.

          Annual Board exams to be completed before the 15th  of the March  and vocational  course Exams to be completed by the end of  February Every year .

                      More short answers type questions from 1 to 5 marks shall be included in the new pattern of the question paper . More emphasis would  be  on probing the critical thinking and ability of the students.

             CBSE has already submitted its proposal to reform in examination pattern to the concered Ministry for approval .
                    The proposal for examination reforms is still in discussion stage and the CBSE has started working to implement new pattern by March 2020.


Benefits  of proposed  Paper pattern 


           From the  Early completion of the Board  examination ,the examiners would have more time to evaluate  answer books of the students , which will result in early declaration  of Annual Board exams. 
            Renewed paper pattern will focus on simplifies and shorten the rules of affiliations and renewal for the school.

At Last but not the least

        
  I would like to thanks for devoting your valuable time for this post "Regarding proposal of CBSE to introduce new pattern of Question Paper " of my blog . If you liked this  blog/post, Do Follow me on my blog and share this post with your friends . We shall meet again in next post with solutions of most interesting mathematics topics and mind blowing puzzles ,till then Good Bye.

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HOW TO LEARN INTEGRATION FORMULAE/FORMULAS USING TRICKS

Let us learn and remember most Important formulas of Integration , tips and tricks to learn algebraic ,most important differentiation questions for plus 2 maths, indefinite integration tricks and shortcuts trigonometric and by parts formulas in an easy and short cut manners.


Trigonometric Formulae



  1  ∫ sin x dx           =  - cos x +c  


where "c" is called constant of Integration.

The integration of sin x is  - cos x ,then divide it with the derivative of its angle. 


If we have to find the integration of  sin 2x , then we shall find it as


Step1    1st find the integration of sin x which is - cos x .

Step2    Divide it with the derivative of 2x ,which is 2, so 


∫ sin 2x dx     =  - ( cos 2x) 2 + c ,
   ∫ sin 8x dx      =  - ( cos 8x) 8 + c ,

∫ sin  3x4  dx  =   - ( cos  3x4 )   3 4  + c ,
Therefore   ∫ sin nx dx =  - ( cos nx) n +c ,






2  ∫ cos x dx          =    sin x +c  

The integration of cos x is  sin x ,then divide it with the derivative of its angle.
If we have to find the integration of  cos 2x , then we shall find it as

Step 1   1st find the integration of cos x which is sin x .

Step 2   Divide it with the derivative of 2x ,which is 2, so 


∫ cos 2x dx     =  ( sin 2x) 2 + c ,

∫ cos 8x dx     =  ( sin 8x) 8 + c ,

∫ cos  3x4  dx  = ( cos  3x4 )   3 4  + c ,

Therefore   ∫ cos nx dx =   ( sin nx) n +c ,



3  ∫ tan x dx = log |sec x| +c or - log |cos x| + c


The integration of tan x is   log |sec x| +c   or -log |cos x| +c  , then divide it with the derivative of its angle.


If we want to find the integration of tan  x2  . The integration of  tan  x2 is  log |sec x2 | +c , then divide it with the derivative of its angle.

Step 1

Find the integration of  tan  x2 ,which is log |sec  x2 | or  - log |cos  x2 |,


Step 2
 Divide it with the derivative of angle x2 ,which is 2x.

Therefore
∫ tan  x2  dx = -(1/2x) log |cos x2 | +c or (1/2x)log |sec x2 | + c

4  ∫ cot x dx = log |sin x| +c or - log |cosec x| + c


 The integration of cot x is   -log |cosec x| +c   , then divide it with the derivative of its angle.



If we want to find the integration of  cot x2  . Then integration of  cot  x2 is  -log |cosec x2 | +c  or  log |sin x2 | +c  , then divide it with the derivative of its angle.

Step 1 

 Find the integration of  cot  x2 ,which is -log |coec  x2 | or  log |sin x2 |,

Step 2

  Divide it with the derivative of angle x2 , which is 2x.   

Therefore      ∫ cot  x2  dx = -(1/2x) log |cosec x2 | +c or (1/2x ) log |sin x2 | + c



 5  ∫ sec x dx         =  log |sec x - tan x | +c 


If we want to integrate sec√x .Then 1st of all we apply the formula of integration of sec(any angle) then divide with the formula of integration of √x,So we have


 ∫  sec√x dx = ( log |sec√x - tan √x | )(2√x) +c


6 ∫ cosec x dx = - log |cosec x - cot x | +c

If we want to integrate cosec√x .Then 1st of all we apply the formula of integration of cosec (any angle) then formula of integration of √x,So we have



 ∫  cosec√x dx = - (log | cosec√x -co√x | )(2√x) +c
  

7  ∫ sec2 x dx = tan x + c


Because the derivative of tan x is sec 2 x , So the Antiderivative or Integration of sec 2 x  is tan x .




∫ sec 2 √x dx     =  (2√x ) tan √x  + c

∫ cosec 2 dx = - cot x +c



Because the derivative of  cot x is  - cosec 2 x , So the  Anti derivative or Integration of  cosec 2 x is - cot x .









 8 ∫ sec x tan x dx = sec x +c



Because the derivative of sec x is sec x tan x ,Therefore the integration of tan x sec x is sec 
x .


If we want to integrate sec√x .tan √x .Then its  integration  will be sec √x,


    sec √x tan √x   dx      =  √x   sec √x + c  


9 ∫ cosec x cot x dx = - cosec x +c


Because the derivative of cosec x is    - cosec x cot x , Therefore the integration of tan x sec x is sec x .


∫ cosec √x cot √x dx = - ( 2 √x  cosec √x ) +c 


 Also Read   WHAT IS SET, TYPES OF SETS, UNION INTERSECTION AND VENN DIAGRAMS


Algebraic Formulae


1 ∫ (constant) dx = (constant ) x +c

Integration of constant function is the constant function itself multiplied by the variable .

∫ 5 dx   = 5x  +c

2  ∫  xn  dx  = xn+1  n+1dx  + c ,

∫ x3   dx  =  x4  4  + c ,


HOW TO LEARN INTEGRATION  FORMULAE/FORMULAS USING TRICKS

   
To find the integration of function where variable "x" or f(x) has power 'n' , where "n" is any real number, we shall increase the power of "x"  by 1 and divide it with increased power.
e.g 

 ∫  x2  dx   =  {1/(2+1)} x2+1  + c

∫  (x ) 2/3   dx     =   (x ) (2/3)+1  (2/3)+1 + c ,

                  =   3(x ) 5/3  5 + c ,

∫  (ax+b ) n   dx     =   (ax+b ) n+1  a(n+1) + c ,

∫  (3x + 7 ) 2   dx     =   (3x+7 )2+1  3(2+1) + c ,

                              =   (3x+7 )3  9 + c ,

If we have to integrate sum of two functions ,then we shall integrate it separately as follows

4  ∫  [ f(x) + g(x)] dx = ∫f(x)dx +  g(x) dx + c


∫  [{ x2}3  + (2x) ]dx = ∫ { x2}3 dx +  ∫ (2x) dx

  =∫   x6dx +  ∫ 2x dx =  
 x  6+1  6+1 +(2/2)x2  + c

 =   x  7  7 x2  + c

∫  {4x2  + 3x }dx = 4x2   + ∫ 3x dx
                         =  4×(1/3)x3  (3/2)x2 +c


 ∫ [ 6x / 3x2] dx = log |  3x2 | + c

HOW TO LEARN INTEGRATION  FORMULAE/FORMULAS USING TRICKS




Memorize  these integration formulas along with differentiation in Hindi




Integration By Parts 

∫ [ f(x) g(x)] dx = f(x) ∫ g(x) dx -  {f '(x) ∫ g(x) dx}dx + c


    
∫ x sin x dx = x ∫ sin x dx - x' { ∫ sin x dx}dx + c
= x(-cos x) -  (-cos x)dx + c
 = -x cos x - sin x +c

∫ log x dx =  ∫ log x.1 dx 
= log x  - f '(log x) ( x )dx + c
= log x .1  - ∫(1/x)  x dx + c  
= log x  - ∫ 1 dx + c 
= log x  - x  + c


 Integration   Exponential Function

1  ∫  ex  dx   =  ex  + c

2 ∫  ax  dx = ax / log a    + c  

3 ∫ log x dx = x log x - x + c

4 ∫ (1/x ) dx = ln |x | + c




Exponential and Derivative Mixed Formula

   ex  [ f(x) + f '(x)] dx =ex  f(x) + c


  ex  [ sin x + cos x] dx = ex  sin x + c

Conclusion


Thanks for devoting your valuable time for this post "HOW TO LEARN INTEGRATION  FORMULAE / FORMULAS VERY   EASILY" of my blog . If you liked this  blog/post, Do Follow me on my blog and share this post with your friends . We shall meet again with new post  ,till then Good Bye.

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