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, \[2-q-p=2+\tan 30{}^\circ \tan {{15}^{o}}\] \[+(\tan \text{ }30{}^\circ +\tan \text{ }15{}^\circ )\] \[\Rightarrow \]\[2+q-p=2+\tan 30{}^\circ \tan 15{}^\circ \] \[+1-\tan 30{}^\circ \tan 15{}^\circ \] \[\left( \because \tan {{45}^{o}}=\frac{\tan {{30}^{o}}+\tan {{15}^{o}}}{1-\tan {{30}^{o}}\tan {{15}^{o}}} \right)\] \[\Rightarrow \]\[2+q-p=3\]
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