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

x -- pi/4   pi - 4x

6 years ago


Answers : (1)


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)
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

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Chaitanya Sagar

6 years ago

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