JEE MAIN - Mathematics (2006 - No. 18)
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
2
3
0
1
Explanation
$${x^2} + px + q = 0$$
Sum of roots $$ = \tan {30^ \circ } + \tan {15^ \circ } = - p$$
Products of roots $$ = \tan {30^ \circ }.\tan {15^ \circ } = q$$
$$\tan {45^ \circ } = {{\tan {{30}^ \circ } + \tan {{15}^ \circ }} \over {1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{ - p} \over {1 - q}} = 1$$
$$ \Rightarrow - p = 1 - q \Rightarrow q - p = 1$$
$$\therefore$$ $$2 + q - p = 3$$
Sum of roots $$ = \tan {30^ \circ } + \tan {15^ \circ } = - p$$
Products of roots $$ = \tan {30^ \circ }.\tan {15^ \circ } = q$$
$$\tan {45^ \circ } = {{\tan {{30}^ \circ } + \tan {{15}^ \circ }} \over {1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{ - p} \over {1 - q}} = 1$$
$$ \Rightarrow - p = 1 - q \Rightarrow q - p = 1$$
$$\therefore$$ $$2 + q - p = 3$$
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