Let -6-1 and -2-2- then -1-2 equalsA- 0B- 2-tan-C- 2 -D- -2 tan-

The correct option is D 2 tan θHere, x^2 2x sec θ +1 =0 has roots α_1 and β_1. ∴α_1, β_1 =2 sec θ±√(4 sec^2 θ 4)/2 × 1 =2 sec θ± 2 |tan θ|/2 Since, θ∈ ( π/ ...

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Byju's Answer

Standard XII

Mathematics

Quadratic Formula for Finding Roots

Let -π/6β1 an...

Question

Let $- \frac{π}{6} < θ < - \frac{π}{12}$ . Suppose $α_{1}$ and $β_{1}$ are the roots of the equation $x^{2} - 2 x s e c θ + 1 = 0$ , $α_{2}$ and $β_{2}$ are the roots of the equation $x^{2} + 2 x t a n θ - 1 = 0$ . If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals

2 (s e c θ - t a n θ)

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2 s e c θ

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- 2 t a n θ

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Solution

The correct option is C $- 2 t a n θ$
Here, $x^{2} - 2 x s e c θ + 1 = 0$ has roots $α_{1}$ and $β_{1} .$
$∴ α_{1}, β_{1} = \frac{2 s e c θ \pm \sqrt{4 s e c^{2} θ - 4}}{2 \times 1} = \frac{2 s e c θ \pm 2 | t a n θ |}{2}$
Since, $θ \in (- \frac{π}{6}, \frac{- π}{12}),$
i.e. $θ \in I V$ quadrant $= \frac{2 s e c θ \mp 2 t a n θ}{2}$
$α_{1} = s e c θ - t a n θ$ and $β_{1} = s e c θ + t a n θ$ [as $α_{1} > β_{1}]$ and $x^{2} + 2 x t a n θ - 1 = 0$ has roots $α_{2}$ and $β_{2} .$
i.e. $α_{2} . β_{2} = \frac{- 2 t a n θ \pm \sqrt{4 t a n^{2} θ + 4}}{2}$
and $∴ α_{2} = - t a n θ + s e c θ$
$β_{2} = - t a n θ - s e c θ [a s α_{2} > β_{2}]$
Thus, $α_{1} + β_{2} = - 2 t a n θ$

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Let −π6<θ<−π12. Suppose α1 and β1 are the roots of the equation x2−2xsecθ+1=0, α2 and β2 are the roots of the equation x2+2xtanθ−1=0. If α1>β1 and α2>β2, then α1+β2 equals

Let $- \frac{π}{6} < θ < - \frac{π}{12}$ . Suppose $α_{1}$ and $β_{1}$ are the roots of the equation $x^{2} - 2 x s e c θ + 1 = 0$ , $α_{2}$ and $β_{2}$ are the roots of the equation $x^{2} + 2 x t a n θ - 1 = 0$ . If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals