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iit jee Grade: 12
```        lim         square root[1 - square root(sin 2x)]
x -- pi/4   pi - 4x```
7 years ago

## Answers : (1)

16 Points
```										Dear iit jee ,    square root[1 - square root(sin 2x)]      pi/4 - xmultiplying both numerator and denominator with    sq.rt (1+sq.rt sin2x)we getmod(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/4mod(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/4Please 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 ExpertsChaitanya Sagar
```
7 years ago
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