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Math Help - [SOLVED] Number of solutions of arctan equation?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Number of solutions of arctan equation?

    Number of solutions of the equation arctan [1/(2x^2 + 1)] + arctan [1/(4x^2 + 1) = arctan 2/x^4 is

    A)0
    B)1
    C)2
    D)4
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  2. #2
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    Hello, fardeen_gen!

    Number of solutions of the equation: . \arctan\left(\frac{1}{2x^2 + 1}\right) + \arctan\left(\frac{1}{4x^2 + 1}\right) \;= \;\arctan\left(\frac{2}{x^4}\right) is

    . . A)\;0 \qquad B)\;1 \qquad C)\;2 \qquad D)\;4

    Let: . A = \arctan\left(\frac{1}{2x^2+1}\right),\; \;B = \arctan\left(\frac{1}{4x^2+1}\right),\;\;C = \arctan\left(\frac{2}{x^4}\right) .[1]

    . . Hence: . \tan A = \frac{1}{2x^2+1}\quad \tan B = \frac{1}{4x^2+1} \quad \tan C = \frac{2}{x^4} .[2]



    Substitute [1] and we have: . A + B \;=\;C


    Take the tangent of both sides:

    . . \tan (A + B) \;=\;\tan(C)

    . . \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} \;=\;\tan(C)

    . . \tan(A) + \tan(B) \;=\;\tan C - \tan(A)\tan(B)\tan(C)


    Substitute [2]:

    . . \frac{1}{2x^2+1} + \frac{1}{4x^2} \;=\;\frac{2}{x^4} - \frac{1}{2x^2+1}\cdot\frac{1}{4x^2+1}\cdot\frac{2}  {x^4}


    Multiply by the LCD, x^4(2x^2+1)(4x^2+1)

    . . x^4(4x^2+1) + x^4(2x^2+1) \;=\;2(2x^2+1)(4x^2+1) - 2


    This simpifies to: . 6x^6 - 14x^4 - 12x^2 \:=\:0

    . . which factors: . 2x^2(3x^2 + 2)(x^2 - 3) \;=\;0


    Then: . \begin{Bmatrix}<br />
2x^2\:=\:0 & \Rightarrow & x \:=\:0 & \text{extraneous root} \\ \\[-4mm]<br />
3x^2+2 \:=\:0 & \Rightarrow & x^2 \:=\:-\frac{2}{3} & \text{no real roots} \\ \\[-4mm]<br />
x^2-3 \:=\:0 & \Rightarrow & x \:=\:\pm\sqrt{3} & \text{Yes!} \end{Bmatrix}


    There are two solutions . . . answer (C).

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