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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    Statement 1: The system of linear equations                 \[x+(\sin \alpha )y+(\cos \alpha )z=0\]                 \[x+(\cos \alpha )y+(\sin \alpha )z=0\]                 \[x-(\sin \alpha )y-(\cos \alpha )z=0\]                 has non-trivial solution of only one value of a lying in the interval \[0,\frac{\pi }{2}.\]                 Statement 2 : The equation in \[\alpha \]                 \[\left| \begin{matrix}    \cos \alpha  & \sin \alpha  & \cos \alpha   \\    \sin \alpha  & \cos \alpha  & \sin \alpha   \\    \cos \alpha  & -\sin \alpha  & -\cos \alpha   \\ \end{matrix} \right|=0\]                              has only one solution lying in the interval  \[\left( 0,\frac{\pi }{2}. \right)\]     JEE Main Online Paper ( Held On 23  April 2013 )

    A)                  Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

    B)                                         Statement 1 is true; Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

    C)                                         Statement 1 is true; Statement 2 is false.

    D)                                         Statement 1 is false; Statement 2 is true.

    Correct Answer: C

    Solution :

                     \[{{\Delta }_{1}}=\left| \begin{matrix}    1 & \sin \alpha  & \cos \alpha   \\    1 & \cos \alpha  & \sin \alpha   \\    1 & -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right|\] \[=\left| \begin{matrix}    0 & \sin \alpha -\cos \alpha  & \cos \alpha -\sin \alpha   \\    0 & \cos \alpha +\sin \alpha  & \sin \alpha -\cos \alpha   \\    1 & -\sin \alpha  & \cos \alpha   \\ \end{matrix} \right|\] \[={{(\sin \alpha -\cos \alpha )}^{2}}-({{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha )\] \[={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha -2\sin \alpha ,\cos \alpha -{{\cos }^{2}}\alpha \] \[+{{\sin }^{2}}\alpha \] \[=2{{\sin }^{2}}\alpha -2\sin \alpha .\cos \alpha \] \[=2\sin \alpha (\sin \alpha -\cos \alpha )\] Now, \[\alpha -\cos \alpha =0\]for only\[\alpha =\frac{\pi }{4}\]in\[\left( 0,\frac{\pi }{2} \right)\] \[\therefore \]\[{{\Delta }_{1}}=2(\sin \alpha )\times 0=0,\] since value of sin a is finite for \[\alpha \in \left( 0,\frac{\pi }{2} \right)\] Hence non-trivivial solution for only one value of \[\alpha \] in\[\left( 0,\frac{\pi }{2} \right)\] \[\left| \begin{matrix}    \cos \alpha  & \sin \alpha  & \cos \alpha   \\    \sin \alpha  & \cos \alpha  & \sin \alpha   \\    \cos \alpha  & -\sin \alpha  & -\cos \alpha   \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix}    0 & \sin \alpha  & \cos \alpha   \\    0 & \cos \alpha  & \sin \alpha   \\    2\cos \alpha  & -\sin \alpha  & -\cos \alpha   \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[2\cos \alpha ({{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha )=0\] \[\therefore \]\[\cos \alpha =0\]or\[{{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha =0\] But \[\cos \alpha =0\] not possible for any value of\[\alpha \in \left( 0,\frac{\pi }{2} \right)\] \[\therefore \]\[{{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha =0\Rightarrow \sin \alpha =-\cos \alpha ,\]which is also not possible for any value of\[\alpha \in \left( 0,\frac{\pi }{2} \right)\] Hence, there is no solution.


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