JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    If the roots of the quadratic equation \[{{x}^{2}}+px+q=0\]are tan\[30{}^\circ \]and tan\[15{}^\circ \] respectively, then the value of\[2+q-p\]is     AIEEE  Solved  Paper-2006

    A) 3                             

    B)        0                             

    C)        1                             

    D)        2

    Correct Answer: A

    Solution :

    Since,\[tan30{}^\circ \]and\[tan15{}^\circ \]are the roots of equation \[{{x}^{2}}+px+q=0\] \[\therefore \] \[\tan 30{}^\circ +\tan 15{}^\circ =-p\] and       \[\tan 30{}^\circ \,\tan 15{}^\circ =q\] Therefore, \[2-q-p=2+\tan 30{}^\circ \tan {{15}^{o}}\] \[+(\tan \text{ }30{}^\circ +\tan \text{ }15{}^\circ )\] \[\Rightarrow \]\[2+q-p=2+\tan 30{}^\circ \tan 15{}^\circ \] \[+1-\tan 30{}^\circ \tan 15{}^\circ \] \[\left( \because \tan {{45}^{o}}=\frac{\tan {{30}^{o}}+\tan {{15}^{o}}}{1-\tan {{30}^{o}}\tan {{15}^{o}}} \right)\] \[\Rightarrow \]\[2+q-p=3\]

You need to login to perform this action.
You will be redirected in 3 sec spinner