MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 2,200 off
USE CODE: chait6

				   

sir... i use to do algebra nd trigo by solving examples in das gupta nd then mcq... but some complex nd lengthy ques irritates me nd reduce my confidence..... guide me any useful way... m appearing in jee2010... ll b greatly thankful 2 u...

6 years ago

Share

Answers : (1)

										

Dear jainish


The complexity of the IIT-JEE arises from the fact that the questions are always unique in their context, and solving them often requires a combination of concepts from across different chapters.


I understand how u feel and what kind of pressure is upon you. First thing u should notice that all the students who clear JEE are also students like u and dont think that for clearing JEE u need to be a real genius or something like that.

First of all u be clear in ur conceptual understanding and then concentrate on solving various questions from different study materials. Dont loose ur confidence. Work hard and utilize the time available effectively. Also utilize the faculties of the coaching
centres to know about the techniques of solving the problems and also the what kind of study materials can give the problems which are more related to the JEE exams


i think you have completed your syllabus. so from now you should give more stress to those topics in which you feel difficulty. or in which you are not much confident.and at the same time you should keep revising other topic also.Try to solve new question .
best method is to first read the theory part completly from the standard text book and study material and then solve "solved examples" ,and them compair your methodology with the solution given in the text book .and see which is less time consuming.
many a time student an solve the problem but the time taken by them is much more which certainly decreses their rank.
 and after solving these questions refer question of that chapter from previous  papers. and follow these procedure chapter by chapter.and time to time solve sample papers .

and  do not waste your time in solving question that are very difficult .Many peaople frame such a question which are not a part of  syllabus.and it is  just a wastage of time.

Just stay calm and focus on understanding rather than mugging. Be regular in class and stay with concepts




Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  preparation.

All the best.
 
Regards,
Askiitians Experts
Badiuddin

6 years ago

Post Your Answer

Other Related Questions on Algebra

If alpha is a real root of the equation ax 2 +bx+c and beta is a real root of equation -ax 2 +bx+c. Show that there exists a root gama of the equation (a/2)x 2 +bx+c which lies between alpha...
 
 
 
Ajay 6 months ago
 
Small Mistake in last para posting again..............................................................................................................
 
Ajay 6 months ago
 
We have Similarly, So if P(x) = a/2 x 2 +bx +c, then and are off opposite sign and hence there must exist a root between the two numbers.
 
mycroft holmes 6 months ago
In the listed image can you tell me how beta*gamma = 2 ….. . . .. ??
 
 
The value of gamma is still not correct, either printing mistake or you gave me wrong value. The correct value of gamma is below
 
Ajay 5 months ago
 
Thankyou so much............................. …......................................................................!
 
Anshuman Mohanty 5 months ago
 
Yes sorry..... . . . .it is not so clear.. ok the values are beta = α + α^2 + α^4 and gamma = α^3 + α^5 + α^7
 
Anshuman Mohanty 5 months ago
if |z - i| Options: a*) 14 b) 2 c) 28 d) None of the above
 
 
If |z-i| = ?? PLs complete the question
  img
Nishant Vora one month ago
 
Got it! [z + 12 – 6 i ] can be rewritten as [ z – i + 12 – 5 i] => | z – i | and => |12 – 5 i | = sqrt ( 12^2 + 5^2) = 13......................(2) => | z + 12 – 6 i | => | z + 12 – 6 i |...
 
Divya one month ago
 
I tried posting the question several times, it kept cutting off the rest of the question. Here: If | z-1| Options: a*) 14 b) 2 c) 28 d) None of the above
 
Divya one month ago
sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°......15 solve it Urgent
 
 
Ajay, the complete qution isSolution is sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°..... upto 15 terms. sin 78°=0 sin 42°+sin 54°+ sin 66°+ + sin 18° sin 6°+ where )=0.5 (your required answer),...
 
Kumar 3 months ago
 
Not any people get my answer why. You can no give answer my question I am join this site
 
Vivek kumar 5 months ago
 
Hello If you want to get the solution quick you should post your question in clear manner. Its not clear what you wnat us to solve, and what does 15 at the end of question means?
 
Ajay 5 months ago
Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2- (tan^2 theta + cot ^2 theta)^2
 
 
What needs to be solved here ? The question is incomplete....................................................................
 
Ajay 6 months ago
 
i don’t know how to do this...............................................................................................
 
Saravanan 2 months ago
 
this is the question :: Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2 - (tan^2 theta + cot ^2 theta)
 
Naveen Shankar 6 months ago
solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
 
 
Let the feet of the altitudes on BC, AC, AB, be D,E,F resp. Let the orthocenter be H. The following can be proved easily: ​1. HDCE and HFBD are cyclic quadrilaterals. Then chord HE subtends...
 
mycroft holmes one month ago
 
Draw which is Isoceles as OB = OC. Now which means . Let D, be the foot of the perp from O on BC ( which is also the midpoint of BC). Then OD = OC sin (OBC) = R cos A. Hence the required...
 
mycroft holmes one month ago
 
a cos A = b cos B 2R sin A cos A = 2R sin B cos B sin 2A = sin 2B Either A = B (isoceles or equilateral) or 2A = 180 o – 2B so that A+B = 90 o .(Right-angled)
 
mycroft holmes one month ago
View all Questions »

  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 6,000 off
USE CODE: chait6

Get extra R 2,200 off
USE CODE: chait6

More Questions On Algebra

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details