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JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    If a is the common positive root of the equation \[{{x}^{2}}-ax+12=0,{{x}^{2}}-bx+15=0\]and\[{{x}^{2}}-(a+b)x+36=0\] and \[\cos x+\cos 2x+\cos 3x=\alpha ,\]then \[\sin x+\sin 2x+\sin 3x\]

    A)  3                            

    B)  -3     

    C)  0                            

    D)  None

    Correct Answer: C

    Solution :

    Since \[\alpha \] is a common root \[{{\alpha }^{2}}-a\alpha +12=0\]                                             (1) \[{{\alpha }^{2}}-b\alpha +15=0\]                                             (2) \[{{\alpha }^{2}}-(a+b)\alpha +36=0\] \[(1)+(2)-(3)\]\[\Rightarrow \]\[{{\alpha }^{2}}-9=0\Rightarrow \alpha =3\] Given, \[\cos x+\cos 2x+\cos 3x=\alpha =3\] This possible only when \[\cos x=1,\cos 2x=1,os3x=1\] \[\therefore \]\[\sin x=0,\sin 2x=0\And \sin 3x=0\] Hence \[\sin x+\sin 2x+\sin 3x=0\]


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