- Home
- Standard 11
- Mathematics
If the quadratic equation ${x^2} + \left( {2 - \tan \theta } \right)x - \left( {1 + \tan \theta } \right) = 0$ has $2$ integral roots, then sum of all possible values of $\theta $ in interval $(0, 2\pi )$ is $k\pi $, then $k$ equals
$2$
$3$
$4$
$5$
Solution
Let $\alpha$ and $\beta$ be the roots of
$x^{2}+(2-\tan \theta) x-$ $(1+\tan \theta)=0$
$\alpha+\beta=\tan \theta-2 $ ………$(1)$
$\alpha \beta=-\tan \theta-1$ ……..$(2)$
$(1)+(2) \Rightarrow \alpha+\beta+\alpha \beta=-3 \Rightarrow(\alpha+1)(\beta+1)=-2$
Hence either $\alpha+1=-2 $ and $ \beta+1=1$ or
$\alpha+1=-1 $ and $ \beta+1=2$
i.e. either $\alpha=-3 $ and $ \beta=0$ or $\alpha=-2 $ and $ \beta=1$
i.e. either $\tan \theta=-1$ or $\tan \theta=1$
i.e. either $\theta=\frac{3 \pi}{4}, \frac{7 \pi}{4}$ or $\theta=\frac{\pi}{4}, \frac{5 \pi}{4}$
Hence sum of all possible values of $\theta$ is
$\frac{\pi}{4}+\frac{3 \pi}{4}+\frac{5 \pi}{4}+\frac{7 \pi}{4}=\frac{16 \pi}{4}=4 \pi$
