MCQ
The quadratic equation whose one root is $2 - \sqrt 3 $ will be
- A${x^2} - 4x - 1 = 0$
- ✓${x^2} - 4x + 1 = 0$
- C${x^2} + 4x - 1 = 0$
- D${x^2} + 4x + 1 = 0$
Hence required equation is ${x^2} - 4x + 1 = 0$.
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Assertion $(A)$ : The circle ${x^2} + {y^2} = 1$ has exactly two tangents parallel to the $x$ - axis
Reason $(R)$ : $\frac{{dy}}{{dx}} = 0$ on the circle exactly at the point $(0, \pm 1)$.
Of these statements