11th Class Mathematics Complex Numbers and Quadratic Equations Question Bank Critical Thinking

  • question_answer
    The number of real values of \[a\] satisfying the equation \[{{a}^{2}}-2a\sin x+1=0\] is

    A) Zero

    B) One

    C) Two

    D) Infinite

    Correct Answer: C

    Solution :

    Given equation  \[{{a}^{2}}-2a\sin x+1=0\] \[\therefore \] \[a=\frac{2\sin x\pm \sqrt{4{{\sin }^{2}}x-4}}{2}\] \[=\sin x\pm \sqrt{-(1-{{\sin }^{2}}x)}\] \[a=\sin x\pm i\cos x\] If \[x=\frac{\pi }{2}\]Þ \[a=1,\]\[x={{270}^{o}}\]Þ \[a=-1\].

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