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QJEEAdv16Maa32a: JEE Advanced 2016 Mathematics - Class 11 JEEAdv16Maa32a Solution: Let and then equals

Let - {\pi \over 6} < \theta < - {\pi \over {12}}. Suppose α1{\alpha _1} and β1{\beta_1} are the roots of the equation x2−2xsec⁡θ+1=0{x^2} - 2x\sec \theta + 1 = 0 and α2{\alpha _2} and β2{\beta _2} are the roots of the equation x2+2x tan⁡θ−1=0.{x^2} + 2x\,\tan \theta - 1 = 0. If\,{\alpha _1} > {\beta _1} and {\alpha _2} > {\beta _2}, then α1+β2{\alpha _1} + {\beta _2} equals

Multiple Choice Options

  • A 2(sec⁡θ−tan⁡θ)2\left( {\sec \theta - \tan \theta } \right)
  • B 2 sec⁡ θ2\,\sec \,\theta
  • C −2tan⁡θ- 2\tan \theta
  • D 00

Step-by-Step Verified Solution

Given, first equation x2−2xsec⁡θ+1=0x^2-2 x \sec \theta+1=0

Using quadratic equation formula we get,

xamp;=−(−2sec⁡θ)±(−2sec⁡θ)2−42⇒amp;x=2sec⁡θ±4sec⁡2θ−42⇒amp;x=2sec⁡θ±2tan⁡θ2⇒amp;x=sec⁡θ±tan⁡θ as θ∈(−π6,−π2)⇒amp;α1=sec⁡θ−tan⁡θ and β1=sec⁡θ+tan⁡θ\begin{aligned}x & =\frac{-(-2 \sec \theta) \pm \sqrt{(-2 \sec \theta)^2-4}}{2} \\\\\Rightarrow & x=\frac{2 \sec \theta \pm \sqrt{4 \sec ^2 \theta-4}}{2} \\\\\Rightarrow & x=\frac{2 \sec \theta \pm 2 \tan \theta}{2} \\\\\Rightarrow & x=\sec \theta \pm \tan \theta \text { as } \theta \in\left(\frac{-\pi}{6}, \frac{-\pi}{2}\right) \\\\\Rightarrow & \alpha_1=\sec \theta-\tan \theta \text { and } \beta_1=\sec \theta+\tan \theta\end{aligned}

Now, for the equation x2+2xtan⁡θ−1=0x^2+2 x \tan \theta-1=0

amp;x=−2tan⁡θ±4tan⁡2θ+42⇒amp;x=−2tan⁡θ±2sec⁡θ2⇒amp;x=−tan⁡θ±sec⁡θ⇒amp;x=(sec⁡θ−tan⁡θ),−(sec⁡θ+tan⁡θ)\begin{aligned}& x=\frac{-2 \tan \theta \pm \sqrt{4 \tan ^2 \theta+4}}{2} \\\\\Rightarrow & x=\frac{-2 \tan \theta \pm 2 \sec \theta}{2} \\\\\Rightarrow & x=-\tan \theta \pm \sec \theta \\\\\Rightarrow & x=(\sec \theta-\tan \theta),-(\sec \theta+\tan \theta)\end{aligned}

Given, that α2\alpha_2 and β2\beta_2 are the roots of equation and \alpha_2>\beta_2

amp;⇒α2=sec⁡θ−tan⁡θ,β2=−(sec⁡θ+tan⁡θ)amp;⇒α1+β2=(sec⁡θ−tan⁡θ)−(sec⁡θ+tan⁡θ)amp;=sec⁡θ−tan⁡θ−sec⁡θ−tan⁡θamp;⇒α1+β2=−2tan⁡θ\begin{aligned}&\Rightarrow \alpha_2 =\sec \theta-\tan \theta, \beta_2=-(\sec \theta+\tan \theta) \\\\&\Rightarrow \alpha_1+\beta_2 =(\sec \theta-\tan \theta)-(\sec \theta+\tan \theta) \\\\& =\sec \theta-\tan \theta-\sec \theta-\tan \theta \\\\&\Rightarrow \alpha_1+\beta_2 =-2 \tan \theta\end{aligned}

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