Results 1 to 6 of 6

Thread: Find the value of k:

  1. #1
    Member great_math's Avatar
    Joined
    Oct 2008
    Posts
    132

    Find the value of k:

    Find the values of $\displaystyle k$ for which the equation

    $\displaystyle \sin^2 x-|k-2|\sin x=\cos^2 x+|k-2|-1$

    is true
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Quote Originally Posted by great_math View Post
    Find the values of $\displaystyle k$ for which the equation

    $\displaystyle \sin^2 x-|k-2|\sin x=\cos^2 x+|k-2|-1$

    is true
    Hi

    You can first get a quadratic in $\displaystyle \sin x$ by substituting $\displaystyle \cos^2 x - 1$ with $\displaystyle -\sin^2 x$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by great_math View Post
    Find the values of $\displaystyle k$ for which the equation

    $\displaystyle \sin^2 x-|k-2|\sin x=\cos^2 x+|k-2|-1$

    is true

    $\displaystyle sin^2x - \mid k-2 \mid sinx = cos^2 x + \mid k-2 \mid -1 $

    $\displaystyle sin^2 x - \mid k-2 \mid sinx = -sin^2x + \mid k-2 \mid $

    $\displaystyle 2sin^2x = (\mid k-2 \mid) (sinx +1 ) $

    $\displaystyle \frac{2sin^2x }{sinx +1 } = \mid k-2 \mid $

    $\displaystyle k-2 = \frac{2sin^2x }{sinx +1 } \Rightarrow k=\frac{2sin^2x }{sinx +1 }+2$

    $\displaystyle k-2 = \frac{-(2sin^2x) }{sinx +1 }\Rightarrow k=\frac{-(2sin^2x) }{sinx +1 }+2$

    $\displaystyle x \ne \frac{3\pi}{2} + 2n\pi $ since if you do not put this condition the denominator will equal zero and k will be infinity so


    $\displaystyle k=\frac{2sin^2x }{sinx +1 }+2 \Rightarrow 2\leq k < \infty $

    $\displaystyle k=\frac{-(2sin^2x) }{sinx +1 }+2 \Rightarrow -\infty< k \leq 2 $

    so

    $\displaystyle -\infty < k < \infty $ but I am not sure
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by Amer View Post
    $\displaystyle sin^2x - \mid k-2 \mid sinx = cos^2 x + \mid k-2 \mid -1 $

    $\displaystyle sin^2 x - \mid k-2 \mid sinx = -sin^2x + \mid k-2 \mid $

    $\displaystyle 2sin^2x = (\mid k-2 \mid) (sinx +1 ) $

    or

    $\displaystyle 2sin^2x = \mid k-2 \mid (sin(x))+\mid k-2 \mid $

    $\displaystyle 2sin^2x -\mid k-2 \mid (sin(x))-\mid k-2 \mid =0 $

    let u=sinx

    $\displaystyle 2u^2 -\mid k-2 \mid (u) -\mid k-2 \mid =0$

    $\displaystyle u=\frac{\mid k-2 \mid \mp \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}}{4} $

    you know that $\displaystyle -1\leq sinx\leq 1 $

    $\displaystyle -1\leq \frac{\mid k-2 \mid - \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}}{4}\leq 1 $

    $\displaystyle -1\leq \frac{\mid k-2 \mid + \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}}{4}\leq 1 $


    $\displaystyle -4\leq (\mid k-2 \mid - \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )})\leq 4 $


    $\displaystyle -4\leq (\mid k-2 \mid + \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )})\leq 4 $


    $\displaystyle 0\leq [\mid k-2 \mid + \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}]\times$$\displaystyle [\mid k-2 \mid - \sqrt{ (\mid k-2 \mid )^2 -4(2)(\mid k-2 \mid )}]\leq 16 $

    $\displaystyle 0\leq (\mid k-2 \mid)^2 - [(\mid k-2 \mid)^2 -8\mid k-2 \mid] \leq 16 $

    $\displaystyle 0 \leq 8 \mid k-2 \mid \leq 16 $

    $\displaystyle 0 \leq \mid k-2 \mid \leq 2 $

    $\displaystyle \mid k-2 \mid \leq 2 $

    $\displaystyle -2\leq k-2 \leq 2 $

    $\displaystyle 0 \leq k \leq 4 $

    hmm
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, great_math!

    A puzzling problem . . . What is meant by "is true"?


    Find the values of $\displaystyle k$ for which the equation: .$\displaystyle \sin^2\!x-|k-2|\sin x=\cos^2\!x+|k-2|-1$ .is true.
    Replace $\displaystyle \cos^2\!x$ with $\displaystyle 1 - \sin^2\!x$:

    . . $\displaystyle \sin^2\!x - |k-2|\sin x \;=\;1 - \sin^2\!x +|k-2| - 1$

    . . $\displaystyle 2\sin^2\!x - |k-2|\sin x - |k-2| \;=\;0$


    Quadratic Formula: . $\displaystyle \sin x \;=\;\frac{|k-2| \pm\sqrt{|k-2|^2 + 8|k-2|}}{4} $



    Now it all depends on what they're asking for . . .


    . . The discriminant must be nonnegative: .$\displaystyle |k-2|^2 + 8|k-2| \;\geq \;0$

    . . Since $\displaystyle |\sin x| \:\leq \:1\!:\quad \left|\frac{|k-2| \pm\sqrt{|k-2|^2 + 8|k-2|}}{4}\right| \;\leq \;1$


    Good luck!

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Putting $\displaystyle t=\sin\ x$.What we have is the quadratic

    $\displaystyle 2t^2-|k-2|t-|k-2|=0$ where $\displaystyle t\in[-1,1]$

    Discriminant = $\displaystyle |k-2|^2+8|k-2|\geq 0$

    Thus we will get real values of $\displaystyle t$ for all real values of $\displaystyle k$.

    But we are required to ensure that these values must lie on the interval $\displaystyle [-1,1]$

    Let $\displaystyle f(t)=2t^2-|k-2|t-|k-2|=0$

    CASE 1

    If exactly one root of $\displaystyle f(t)=0$ lies on (-1,1),then we must have

    $\displaystyle f(-1)f(1)<0$

    Now,$\displaystyle f(-1)=2>0$

    $\displaystyle \therefore f(1)<0$,thus $\displaystyle 2-2|k-2|<0$

    $\displaystyle |k-2|>1$

    $\displaystyle k-2<-1$ or $\displaystyle k-2>1$

    $\displaystyle k<1$ or $\displaystyle k>3$i.e.$\displaystyle k\in(-\infty,1)\cup(3,\infty)$

    CASE 2

    If both roots on the interval $\displaystyle (-1,1)$ then following three conditions must be satisfied:
    (1)$\displaystyle f(-1)>0$

    (2)$\displaystyle f(1)>0$

    (3)$\displaystyle -1<\frac{\alpha+\beta}{2}<1$;$\displaystyle \alpha$ and $\displaystyle \beta $ being the rootsof the equation $\displaystyle f(t)=0$

    $\displaystyle f(-1)=2>0$

    $\displaystyle f(1)>0$ implies that $\displaystyle 2-2|k-2|>0$.

    $\displaystyle \therefore |k-2|<1$

    Thus $\displaystyle -1<k-2<1$.

    $\displaystyle k\in (1,3)$

    $\displaystyle -1<\frac{\alpha+\beta}{2}<1$ implies that $\displaystyle -1<\frac{|k-2|}{4}<1$

    (Recall that if $\displaystyle \alpha $and $\displaystyle \beta $are roots of $\displaystyle ax^2+bx+c=0$ then $\displaystyle \alpha+\beta=-\frac{b}{a}$)

    On solving for $\displaystyle k$ we get $\displaystyle -2<k<6$

    Taking intersection of solutions of (1),(2) and (3),we get $\displaystyle 1<k<3$


    Finally taking union of Case 1 and Case 2 and also checking for $\displaystyle k=1,3$, we arrive at the conclusion that the given equation is true for all real $\displaystyle k$.

    $\displaystyle
    \therefore k\in R
    $ must be the answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: Mar 22nd 2011, 04:57 PM
  2. Replies: 2
    Last Post: Jul 5th 2010, 08:48 PM
  3. Replies: 1
    Last Post: Feb 17th 2010, 03:58 PM
  4. Replies: 0
    Last Post: Jun 16th 2009, 12:43 PM
  5. Replies: 2
    Last Post: Apr 6th 2009, 08:57 PM

/mathhelpforum @mathhelpforum