JEE Advanced · Mathematics · 2. Quadratic Equations

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

A $2 (\sec θ - \tan θ)$
B $2 \sec θ$
C $- 2 \tan θ$
D $0$

Verified Solution

Answer & Solution

Correct Answer

(C) $- 2 \tan θ$

Step-by-step Solution

Detailed explanation

α_{1} > β_{1}

' +'

sign will be used for

α_{1}

β_{2} < α_{2}

' -'

sign for

β_{2}

α_{1} = \frac{2 \sec θ \pm \sqrt{4 \sec^{2} θ - 4}}{2},

β_{2} = \frac{- 2 \tan θ - \sqrt{4 \tan^{2} θ + 4}}{2}

α_{1} = \sec θ + |\tan θ|

β_{2} = - \tan θ - \sec θ

α_{1} = \sec θ - \tan θ

\{∵ θ \in (- \frac{π}{6}, - \frac{π}{12})\}

So,

\tan θ

- v e

&

\sec θ

+ v e

β_{2} = - t a n θ - s e c θ

α_{1} + β_{2} = - 2 \tan θ

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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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

Step-by-step Solution

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