Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Quadratic Equation and Inequalities question

2016 · Shift 1 · Q31
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /Quadratic Equation and Inequalities
  5. /2016 · Shift 1 · Q31

Quadratic Equation and Inequalities question

2016 · Shift 1 · Q31

JEE AdvancedMathematicsQuadratic Equation and InequalitiesMCQ+3 / −1
Let −π6<θ<−π12.- {\pi \over 6} \lt \theta \lt - {\pi \over {12}}.−6π​<θ<−12π​. Suppose α1{\alpha _1}α1​ and β1{\beta_1}β1​ are the roots of the equation x2−2xsec⁡θ+1=0{x^2} - 2x\sec \theta + 1 = 0x2−2xsecθ+1=0 and α2{\alpha _2}α2​ and β2{\beta _2}β2​ are the roots of the equation x2+2x tan⁡θ−1=0.If α1>β1{x^2} + 2x\,\tan \theta - 1 = 0.If\,{\alpha _1} \gt {\beta _1}x2+2xtanθ−1=0.Ifα1​>β1​ and α2>β2,{\alpha _2} \gt {\beta _2},α2​>β2​, then α1+β2{\alpha _1} + {\beta _2}α1​+β2​ equals
  1. A
    2(sec⁡θ−tan⁡θ)2\left( {\sec \theta - \tan \theta } \right)2(secθ−tanθ)
  2. B
    2 sec⁡ θ2\,\sec \,\theta2secθ
  3. C
    −2tan⁡θ- 2\tan \theta−2tanθ
  4. D
    000
View written solutionFree

Correct answer: C

  1. Solve the first quadratic

Given x2−2xsec⁡θ+1=0x^2-2x\sec\theta+1=0x2−2xsecθ+1=0 Its roots are x=2sec⁡θ±4sec⁡2θ−42=sec⁡θ±sec⁡2θ−1x=\frac{2\sec\theta\pm\sqrt{4\sec^2\theta-4}}{2}=\sec\theta\pm\sqrt{\sec^2\theta-1}x=22secθ±4sec2θ−4​​=secθ±sec2θ−1​ Since sec⁡2θ−1=tan⁡2θ=∣tan⁡θ∣\sqrt{\sec^2\theta-1}=\sqrt{\tan^2\theta}=|\tan\theta|sec2θ−1​=tan2θ​=∣tanθ∣ we need the sign of tan⁡θ\tan\thetatanθ.

Because −π6<θ<−π12-\frac{\pi}{6}<\theta<-\frac{\pi}{12}−6π​<θ<−12π​ θ\thetaθ lies in the fourth quadrant, so tan⁡θ<0,sec⁡θ>0\tan\theta<0,\qquad \sec\theta>0tanθ<0,secθ>0 Hence ∣tan⁡θ∣=−tan⁡θ|\tan\theta|=-\tan\theta∣tanθ∣=−tanθ Therefore the roots are sec⁡θ±(−tan⁡θ)=sec⁡θ∓tan⁡θ\sec\theta\pm(-\tan\theta)=\sec\theta\mp\tan\thetasecθ±(−tanθ)=secθ∓tanθ So the two roots are sec⁡θ−tan⁡θandsec⁡θ+tan⁡θ\sec\theta-\tan\theta \quad \text{and} \quad \sec\theta+\tan\thetasecθ−tanθandsecθ+tanθ Now since tan⁡θ<0\tan\theta<0tanθ<0, sec⁡θ−tan⁡θ>sec⁡θ+tan⁡θ\sec\theta-\tan\theta>\sec\theta+\tan\thetasecθ−tanθ>secθ+tanθ Thus α1=sec⁡θ−tan⁡θ,β1=sec⁡θ+tan⁡θ\alpha_1=\sec\theta-\tan\theta,\qquad \beta_1=\sec\theta+\tan\thetaα1​=secθ−tanθ,β1​=secθ+tanθ

  1. Solve the second quadratic

Given x2+2xtan⁡θ−1=0x^2+2x\tan\theta-1=0x2+2xtanθ−1=0 Its roots are

=-\tan\theta\pm\sqrt{\tan^2\theta+1}$$ Using $$\tan^2\theta+1=\sec^2\theta$$ and since $\sec\theta>0$ in the given interval, $$\sqrt{\tan^2\theta+1}=\sec\theta$$ So the roots are $$-\tan\theta\pm\sec\theta$$ that is, $$\sec\theta-\tan\theta \quad \text{and} \quad -\sec\theta-\tan\theta$$ Clearly, $$\sec\theta-\tan\theta>-\sec\theta-\tan\theta$$ Hence $$\alpha_2=\sec\theta-\tan\theta,\qquad \beta_2=-\sec\theta-\tan\theta$$ 3. **Compute $\alpha_1+\beta_2$** $$\alpha_1+\beta_2=(\sec\theta-\tan\theta)+(-\sec\theta-\tan\theta)$$ $$= -2\tan\theta$$ 4. **Match with the options** $$\alpha_1+\beta_2=-2\tan\theta$$ So the correct option is: **C. $-2\tan\theta$**
PreviousNext

More from Quadratic Equation and Inequalities

  • Let S be the set of all non-zero real numbers α such that the quadratic equation αx2−x+α=0 has two distinct real roots x1​ and x2​ satisfying the inequality ∣x1​−x2​∣<1.…2015 · Multiple correct
  • The quadratic equation p(x)=0 with real coefficients has purely imaginary roots. Then the equation p(p(x))=0 has2014 · MCQ
  • If 3x=4x−1, then x=2013 · Multiple correct
  • The value of 6+log3/2​(32​1​4−32​1​4−32​1​4−32​1​...​​​) is ​.2012 · Numerical
  • Let α(a) and β(a) be the roots of the equation (31+a​−1)x2+(1+a​−1)x+(61+a​−1)=0 where a>−1. Then a→0+lim​α(a) and a→0+lim​β(a)…2012 · MCQ
  • The minimum value of the sum of real numbers a−5,a−4,3a−3,1,a8 and a10 where a>0 is2011 · Numerical
  • Let α and β be the roots of x2−6x−2=0, with α>β. If an​=αn−βn for n≥1 then the value of 2a9​a10​−2a8​​ is2011 · MCQ
  • Let (x0​,y0​) be the solution of the following equations (2x)ℓn2=(3y)ℓn33ℓnx=2ℓny​ Then x0​…2011 · MCQ