Difficult
If $\tan A$ and $\tan B$ are the roots of the equation $5x^2 - 4x + 1 = 0$, then the value of $A + B$ is...
- A$\frac{\pi}{6}$
- B$\frac{\pi}{4}$
- C$\frac{\pi}{3}$
- D$\frac{\pi}{2}$
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View SolutionLet $x, y$ be real numbers such that $x \neq y$ and $xy \neq 1$. If $ax + b \sec(\tan^{-1} x) = c$ and $ay + b \sec(\tan^{-1} y) = c$,then $\frac{x+y}{1-xy} =$
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Let $\alpha$ and $\beta$ be the roots of the quadratic equation $a x^2+b x+c=0$. Match the conditions in List-$I$ with the corresponding relations in List-$II$.
List-$I$ List-$II$ $(i) \alpha = \beta$ $(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$ $(ii) \alpha = 2\beta$ $(B) 2b^2 = 9ac$ $(iii) \alpha = 3\beta$ $(C) b^2 = 6ac$ $(iv) \alpha = \beta^2$ $(D) 3b^2 = 16ac$ $(E) b^2 = 4ac$ $(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$
| List-$I$ | List-$II$ |
| $(i) \alpha = \beta$ | $(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$ |
| $(ii) \alpha = 2\beta$ | $(B) 2b^2 = 9ac$ |
| $(iii) \alpha = 3\beta$ | $(C) b^2 = 6ac$ |
| $(iv) \alpha = \beta^2$ | $(D) 3b^2 = 16ac$ |
| $(E) b^2 = 4ac$ | |
| $(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$ |
DifficultTS EAMCET 2008
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