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order

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

  • one year ago
  • one year ago

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  1. shivam_bhalla
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    Remember -1 <= sin x <= 1 Therefore -1 <= cosec x <=1 Now try :)

    • one year ago
  2. order
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    Still not sure..

    • one year ago
  3. shivam_bhalla
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    One more question. This is your question , right \[cosec x= {1 \over 2 }(x)+1\]

    • one year ago
  4. order
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    Yes

    • one year ago
  5. NotSObright
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    I think i solved it

    • one year ago
  6. shivam_bhalla
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    OK. Just put instead of cosec x , 1/2(x) +1

    • one year ago
  7. shivam_bhalla
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    Ok @NotSObright , let's see what you have done :P

    • one year ago
  8. 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

    • one year ago
  9. order
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    So what about the 0.5 and 1?

    • one year ago
  10. 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

    • one year ago
  11. order
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    But how?

    • one year ago
  12. NotSObright
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    |dw:1336990897568:dw|

    • one year ago
  13. order
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    in calculation?

    • one year ago
  14. NotSObright
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    it is in calculation u need to calculate values at o.5 and 1 to show the inequality

    • one year ago
  15. order
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    So, how exactly?

    • one year ago
  16. NotSObright
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    i have told everything u need go through the conversation and try to use ur intelligence

    • one year ago
  17. order
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    hmm. ok..

    • one year ago
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