If $\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{4}\right)=\log _e k$, then

  • A
    $3 k^2-12 k-1=0$
  • B
    $3 k^2-12 k+1=0$
  • C
    $9 k^2-12 k+1=0$
  • D
    $9 k^2-12 k-1=0$

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