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
iit jee Grade: 12
        

lim         square root[1 - square root(sin 2x)]


x -- pi/4   pi - 4x

6 years ago

Answers : (1)

AskIitian Expert Chaitanya
16 Points
										

Dear iit jee ,
 
   square root[1 - square root(sin 2x)]
      pi/4 - x
multiplying both numerator and denominator with    sq.rt (1+sq.rt sin2x)we get
mod(cosx-sinx) /sq.rt(1+sq.rt sin 2x)
so ..,
   
  lim         square root[1 - square root(sin 2x)]
  x -- pi/4   pi/4 - x
=lim       mod(cosx-sinx) /sq.rt(1+sq.rt sin 2x)*(pi-4x)
  x->pi/4
mod(cosx-sinx)= cosx-sinx  for x->(pi/4)-
                
              = sinx-cosx  for x->(pi/4)+



  lim         square root[1 - square root(sin 2x)]
  x -- pi/4   pi/4 - x
= lim  sinx-cosx/ sq.rt(1+sq.rt sin 2x)*(pi-4x) for x->(pi/4)+

= infinity
 
lim tends to 1 when x->pi/4-


hence the function is discontinous at pi/4

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 iit jee !!!


Regards,

Askiitians Experts

Chaitanya Sagar

6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 4,500 off
USE CODE: SSP3
Get extra R 320 off
USE CODE: MOB20

Ask Experts

Have any Question? Ask Experts

Post Question

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