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JEE Main & Advanced Sample Paper JEE Main Sample Paper-9

  • question_answer
    Passage (Q. - 90) Let \[{{S}_{1}}\] be the set of all those solutions of the equation\[(1+a)cos\theta cos(2\theta -b)=\]\[(1+acos2\theta )cos(\theta -b)\]which are independent of a and b and \[{{S}_{2}}\] be the set of all such solutions which are dependent on a and b. Then All the permissible values of b if a = 0 and \[{{S}_{2}}\] is a subset\[(0,\pi )\]:

    A) \[b\in (-n\pi ,2n\pi ),n\in Z\]

    B) \[b\in (-n\pi ,2-n\pi ),n\in Z\]

    C) \[b\in (-n\pi ,n\pi ),n\in Z\]

    D)  none of these

    Correct Answer: B

    Solution :

    \[a=0\Rightarrow \sin (2\theta -b)=0\] \[2\theta -b=n\pi ,n\in Z\] \[\theta =\frac{n\pi +b}{2},n\in Z:{{S}_{2}}=\left[ \frac{n\pi +b}{2};n\in Z \right]\] \[{{S}_{2}}\] must be subject of \[(0,\pi )\] \[0<\frac{n\pi +b}{2}<\pi ,n\in Z\] \[0<n\pi +b<2\pi \Rightarrow -n\pi <b<-n\pi +2\pi \] \[\Rightarrow \]\[b\in (-n\pi ,2\pi -n\pi ),n\in Z.\]


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