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Math Help - Differentiation Questions

  1. #1
    Junior Member cupid's Avatar
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    Differentiation Questions

    Questions:


    How do i do the part sec^{-1}




    should the angle in sec^{-1} be inverse of what it is because answer is (b) and it is only possible when y is some constant which will be using the identity sin^{-1}x + cos^{-1}x

    Some Help please ...
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  2. #2
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    Hello, cupid!


    There's something wrong with the problem
    . . or it is a "trick question".


    . . \sec^{-1}\!\left(\dfrac{2x}{1+x^2}\right) \:\text{ and }\:\sin^{-1}\left(\dfrac{x-1}{x+1}\right)\:\text{ are }constants.


    \text{Let }\:\theta \;=\;\sec^{-1}\!\left(\dfrac{2x}{1+x^2}\right) \quad\Rightarrow\quad \sec\theta \;=\;\dfrac{2x}{1+x^2} \;=\;\dfrac{hyp}{adj}


    Hence, \,\theta is in a right triangle with: adj = 1+x^2,\;hyp = 2x


    Since the hypotenuse is the longest side of the right triangle:

    . . adj \:\le\:hyp \quad\Rightarrow\quad 1+x^2 \:\le \:2x

    . . 1 - 2x + x^2 \:\le \:0 \quad\Rightarrow\quad (1-x)^2 \:\le \:0

    The inequality is true only if x = 1.


    Hence: . \sec^{-1}\!\left(\dfrac{1+x^2}{2x}\right) \;=\;\sec^{-1}(1) \;=\;0

    Then: . \sin^{-1}\!\left(\dfrac{x-1}{x+1}\right) \;=\;\sin^{-1}(0) \;=\;0

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  3. #3
    Junior Member cupid's Avatar
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    Hi Soroban
    i think you are right but book says that correct answer is (d) None of these
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  4. #4
    Member pflo's Avatar
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    Quote Originally Posted by Soroban View Post

    . . adj \:\le\:hyp \quad\Rightarrow\quad 1+x^2 \:\le \:2x

    Couldn't you look at it this way:

    . x^2 - 2x + 1 \:\le \:0 \quad\Rightarrow\quad (x-1)^2 \:\le \:0

    In this case, the inequality would be true if x \le 1 and would essentially result in the domain restriction given in the problem.

    When I take the derivative I get \frac{x+1}{x-1}. I must go to class now, but will post my work this afternoon if nobody else gets to it first - I'll check my work too (I could be wrong at this point).
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  5. #5
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    Quote Originally Posted by pflo View Post
    Couldn't you look at it this way:

    . x^2 - 2x + 1 \:\le \:0 \quad\Rightarrow\quad (x-1)^2 \:\le \:0

    In this case, the inequality would be true if x \le 1 and would essentially result in the domain restriction given in the problem.

    When I take the derivative I get \frac{x+1}{x-1}. I must go to class now, but will post my work this afternoon if nobody else gets to it first - I'll check my work too (I could be wrong at this point).
    (x-1)^2 \:\le \:0 is true only for x=1 as well. Something squared can never be less than zero. . .
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  6. #6
    Member pflo's Avatar
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    Yes, yes. Of course you're right. Perhaps the authors of this problem made the same mistake I did in overlooking this.
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  7. #7
    Junior Member cupid's Avatar
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    No i'm pretty sure that 0 is not the solution.

    it may be so for sec-1 term

    but sin-1 (acc. to Soroban's post) gives that x+1 > x-1 \implies 1>-1 which is true for all x.
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