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cosec x= 1/2 (x)+1
has root 0<x<1/2 pi
verify by calculation that this root lies between 0.5 and 1
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shivam_bhalla
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Remember
-1 <= sin x <= 1
Therefore
-1 <= cosec x <=1
Now try :)
order
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Still not sure..
shivam_bhalla
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One more question. This is your question , right
\[cosec x= {1 \over 2 }(x)+1\]
order
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Yes
NotSObright
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I think i solved it
shivam_bhalla
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OK. Just put instead of cosec x , 1/2(x) +1
shivam_bhalla
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Ok @NotSObright , let's see what you have done :P
NotSObright
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Say y1=1/2x+1
y2=csc(x)
y1(0)=1
y1(Pi/2)=Pi/4+1
y2(0)=infinty
y2(Pi/2)=1
Since it can be proven using differentiation y1 is inceasing and y2 is decreasing on 0,pi/2
hence in order to attain boundary values graphs must cross each other
order
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So what about the 0.5 and 1?
NotSObright
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sorry i didn't see that
u must show at 0.5 y1<y2
and at 1 y1>y2
u might wanna use a calculator
order
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But how?
NotSObright
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|dw:1336990897568:dw|
order
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in calculation?
NotSObright
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it is in calculation u need to calculate values at o.5 and 1 to show the inequality
order
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So, how exactly?
NotSObright
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i have told everything u need go through the conversation and try to use ur intelligence
order
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hmm. ok..