SATHEE: Theory of Equations 1 Question 16

Theory of Equations 1 Question 16

17. 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 Adv.)

(a) $2 (\sec θ - \tan θ)$

(b) $2 \sec θ$

(d) 0

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Answer:

Correct Answer: 17. (c)

Solution:

Here, $x^{2} - 2 x \sec θ + 1 = 0$ has roots $α_{1}$ and $β_{1}$ .

$\begin{aligned} ∴ α_{1}, β_{1} & = \frac{2 \sec θ \pm \sqrt{4 \sec^{2} θ - 4}}{2 \times 1} \\ = \frac{2 \sec θ \pm 2 | \tan θ |}{2} \end{aligned}$

Since, $θ \in - \frac{π}{6}, - \frac{π}{12}$ ,

i.e. $θ \in I V$ quadrant $= \frac{2 \sec θ \mp 2 \tan θ}{2}$

$∴ α_{1} = \sec θ - \tan θ and β_{1} = \sec θ + \tan θ [as α_{1} > β_{1}]$ and $x^{2} + 2 x \tan θ - 1 = 0$ has roots $α_{2}$ and $β_{2}$ .

$\begin{array}{lll} i.e. & α_{2}, β_{2} & = \frac{- 2 \tan θ \pm \sqrt{4 \tan^{2} θ + 4}}{2} \\ ∴ & α_{2} & = - \tan θ + \sec θ \\ and & β_{2} & = - \tan θ - \sec θ [as α_{2} > β_{2}] \\ Thus, & α_{1} + β_{2} & = - 2 \tan θ \end{array}$

Theory of Equations 1 Question 16

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Theory of Equations 1 Question 16

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

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

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Sample Mock Test

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