Hello !
Today was my Pre-RMO . I was stuck in the following question please help .
If
sin−1x+sin−1y+sin−1z=23π
Then find the value of :
x9+y9+z9+x9y9z91
Well , I had got the following result from this : ( Might be wrong )
x+y+z=−1
but I couldn't manage to do it any further . Please help .
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The range of sin−1 is [−21π,21π]. Since sin−1x+sin−1y+sin−1z=23π, we know the values of sin−1x,sin−1y,sin−1z straight away (what three numbers between −a and a add to 3a?). Thus we know x,y,z.
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Thank You !
Very simple one! As, sin−1x→[2−π,2π]
So, it leads us to the conclusion that x=1,y=1,z=1
Then, required value becomes 4
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i thought same . quite easy.
max value of the following equation is 3pi/2 and hence each angle should be pi/2. if angle is pi/2 then x=y=z=1 then 1 + 1 + 1 + 1 = 4. That's it
What is Pre-RMO? Is the full paper available somewhere?
Thanks!
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Well this time , the government has allowed every state to set their own papers for pre-RMO . That is the qualification stage for RMO level 1 .
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In Telangana region,there is no pre-RMO....lucky for me I can directly write RMO paper without qualifying.... BTW....The RMO exam is not so easy as pre-RMO....I was dying to solve some of them...
In Mumbai region there is Pre-RMO, as the no. of students taking part was much more. While the other regions in the country have RMO directly (including other-than-Mumbai Maharashtra part. We are going to give it on 1st December. Let's discuss some good problems thereafter...!!!
Promo paper
Priansh, can I get the copy of your PRMO paper??
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Well they didn't allow us to take it back home :/