jee-advanced 2016 Q37

jee-advanced · India · paper1 Standard trigonometric equations Evaluate trigonometric expression given a constraint

Let $-\frac{\pi}{6} < \theta < -\frac{\pi}{12}$. 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
(A) $2(\sec\theta - \tan\theta)$
(B) $2\sec\theta$
(C) $-2\tan\theta$
(D) $0$

Let $-\frac{\pi}{6} < \theta < -\frac{\pi}{12}$. 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\\
(A) $2(\sec\theta - \tan\theta)$\\
(B) $2\sec\theta$\\
(C) $-2\tan\theta$\\
(D) $0$

Paper Questions

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