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Q.

 The range of the function, f(x)=cot1log0.5x42x2+3 is

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a

(0,π)

b

(0,3π/4]

c

[3π/4,π)

d

[π/2,3π/4]

answer is C.

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Detailed Solution

  x4-2 x2+3 = x2-12+2 2 logx4-2x2+3 0.5   log 2 0.5  =-log 2 2 =-1   since xy logx a   logy a  if 0<a<1 

 cot-1 logx4-2x2+3 0.5  cot-1-1  since cot-1x fumction is decreasing     

 y= cot-1logx4-2x2+3 0.5   π -cot-11=3π4        y [ 3π4,π)        since cot-1x 0,π

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