JEE Advance - Mathematics (2016 - Paper 1 Offline - No. 13)
Let $$ - {\pi \over 6} < \theta < - {\pi \over {12}}.$$ Suppose $${\alpha _1}$$ and $${\beta_1}$$ are the roots of the equation $${x^2} - 2x\sec \theta + 1 = 0$$ and $${\alpha _2}$$ and $${\beta _2}$$ are the roots of the equation $${x^2} + 2x\,\tan \theta - 1 = 0.$$ $$If\,{\alpha _1} > {\beta _1}$$ and $${\alpha _2} > {\beta _2},$$ then $${\alpha _1} + {\beta _2}$$ equals
$$2\left( {\sec \theta - \tan \theta } \right)$$
$$2\,\sec \,\theta $$
$$ - 2\tan \theta $$
$$0$$
Explanation
Given, first equation $x^2-2 x \sec \theta+1=0$
Using quadratic equation formula we get,
$$ \begin{aligned} x & =\frac{-(-2 \sec \theta) \pm \sqrt{(-2 \sec \theta)^2-4}}{2} \\\\ \Rightarrow & x=\frac{2 \sec \theta \pm \sqrt{4 \sec ^2 \theta-4}}{2} \\\\ \Rightarrow & x=\frac{2 \sec \theta \pm 2 \tan \theta}{2} \\\\ \Rightarrow & x=\sec \theta \pm \tan \theta \text { as } \theta \in\left(\frac{-\pi}{6}, \frac{-\pi}{2}\right) \\\\ \Rightarrow & \alpha_1=\sec \theta-\tan \theta \text { and } \beta_1=\sec \theta+\tan \theta \end{aligned} $$
Now, for the equation $x^2+2 x \tan \theta-1=0$
$$ \begin{aligned} & x=\frac{-2 \tan \theta \pm \sqrt{4 \tan ^2 \theta+4}}{2} \\\\ \Rightarrow & x=\frac{-2 \tan \theta \pm 2 \sec \theta}{2} \\\\ \Rightarrow & x=-\tan \theta \pm \sec \theta \\\\ \Rightarrow & x=(\sec \theta-\tan \theta),-(\sec \theta+\tan \theta) \end{aligned} $$
Given, that $\alpha_2$ and $\beta_2$ are the roots of equation and $\alpha_2>\beta_2$
$$ \begin{aligned} &\Rightarrow \alpha_2 =\sec \theta-\tan \theta, \beta_2=-(\sec \theta+\tan \theta) \\\\ &\Rightarrow \alpha_1+\beta_2 =(\sec \theta-\tan \theta)-(\sec \theta+\tan \theta) \\\\ & =\sec \theta-\tan \theta-\sec \theta-\tan \theta \\\\ &\Rightarrow \alpha_1+\beta_2 =-2 \tan \theta \end{aligned} $$
Using quadratic equation formula we get,
$$ \begin{aligned} x & =\frac{-(-2 \sec \theta) \pm \sqrt{(-2 \sec \theta)^2-4}}{2} \\\\ \Rightarrow & x=\frac{2 \sec \theta \pm \sqrt{4 \sec ^2 \theta-4}}{2} \\\\ \Rightarrow & x=\frac{2 \sec \theta \pm 2 \tan \theta}{2} \\\\ \Rightarrow & x=\sec \theta \pm \tan \theta \text { as } \theta \in\left(\frac{-\pi}{6}, \frac{-\pi}{2}\right) \\\\ \Rightarrow & \alpha_1=\sec \theta-\tan \theta \text { and } \beta_1=\sec \theta+\tan \theta \end{aligned} $$
Now, for the equation $x^2+2 x \tan \theta-1=0$
$$ \begin{aligned} & x=\frac{-2 \tan \theta \pm \sqrt{4 \tan ^2 \theta+4}}{2} \\\\ \Rightarrow & x=\frac{-2 \tan \theta \pm 2 \sec \theta}{2} \\\\ \Rightarrow & x=-\tan \theta \pm \sec \theta \\\\ \Rightarrow & x=(\sec \theta-\tan \theta),-(\sec \theta+\tan \theta) \end{aligned} $$
Given, that $\alpha_2$ and $\beta_2$ are the roots of equation and $\alpha_2>\beta_2$
$$ \begin{aligned} &\Rightarrow \alpha_2 =\sec \theta-\tan \theta, \beta_2=-(\sec \theta+\tan \theta) \\\\ &\Rightarrow \alpha_1+\beta_2 =(\sec \theta-\tan \theta)-(\sec \theta+\tan \theta) \\\\ & =\sec \theta-\tan \theta-\sec \theta-\tan \theta \\\\ &\Rightarrow \alpha_1+\beta_2 =-2 \tan \theta \end{aligned} $$
Comments (0)
