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Urgent! This is a question of inverse trigonometry, where we have to find the range of x for the two conditions combined.

Oberoi Veroni , 9 Years ago
Grade 12
anser 1 Answers
Vikas TU

Last Activity: 9 Years ago

arcsinx > pie/2 – arcsinx
2arcsinx > pie/2
arcsinx > pie/4
therefore, x > sin(pie/4)
                1 > x > 1/root2............................(1)
 
Let arctan = t
then t^2 – 3t + 2>0
(t-1) (t -2) > 0
t 2
arctanx
x
 and
arctanx > 2 which is beyond its range pie/2 hence rjected.
 thus Range will be the intersection of both (1) and (2).
 
 

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