Let - 6 12 . Suppose 1 and 1 are the roots of the equation x 2 - 2 x sec + 1 = 0 and 2 and 2 are the roots of the equation x 2 + 2 x tan - 1 = 0 . If 1 1 and 2 2 , then 1 + 2 equals [2016]

Question

Let $- \frac{π}{6} < θ < - \frac{π}{12}$ . Suppose $α_{1}$ and $β_{1}$ are the roots of the equation $x^{2} - 2 x \sec θ + 1 = 0$ and $α_{2}$ and $β_{2}$ are the roots of the equation $x^{2} + 2 x \tan θ - 1 = 0 .$ If $α_{1} > β_{1}$ and $α_{2} > β_{2}$ , then $α_{1} + β_{2}$ equals [2016]

Smart Achievers · Accepted Answer

(3)

$x^{2} - 2 x \sec θ + 1 = 0 \Rightarrow x = \sec θ \pm \tan θ$

$and x^{2} + 2 x \tan θ - 1 = 0 \Rightarrow x = - \tan θ \pm \sec θ$

$∵$ $- \frac{π}{6} < θ < - \frac{π}{12} \Rightarrow \sec \frac{π}{6} > \sec θ > \sec \frac{π}{12}$

$and - \tan \frac{π}{6} < \tan θ < - \frac{\tan π}{12}$

$Also \tan \frac{π}{12} < - \tan θ < \tan \frac{π}{6}$

$Since, α_{1}, β_{1} are roots of x^{2} - 2 x \sec θ + 1 = 0 and α_{1} > β_{1}$

$∴$ $α_{1} = \sec θ - \tan θ and β_{1} = \sec θ + \tan θ$

$Since, α_{2}, β_{2} are roots of x^{2} + 2 x \tan θ - 1 = 0 and α_{2} > β_{2}$

$∴$ $α_{2} = - \tan θ + \sec θ, β_{2} = - \tan θ - \sec θ$

$∴$ $α_{1} + β_{2} = \sec θ - \tan θ - \tan θ - \sec θ = - 2 \tan θ$

Solution

1 $2 (\sec θ - \tan θ)$

2 $2 \sec θ$

3 $- 2 \tan θ$

4 $0$

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Solution

Q. Let -π6<θ<-π12. Suppose α1 and β1 are the roots of the equation x2-2xsecθ+1=0 and α2 and β2 are the roots of the equation x2+2xtanθ-1=0. If α1>β1 and α2>β2, then α1+β2 equals [2016]

1 2(secθ-tanθ)

2 2secθ

3 -2tanθ

4 0

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1 $2 (\sec θ - \tan θ)$

2 $2 \sec θ$

3 $- 2 \tan θ$

4 $0$