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JEE Advanced 2016 · Paper 1, Mathematics question 37

Let −π6<θ<−π12-\dfrac{\pi}{6}<\theta<-\dfrac{\pi}{12}. Suppose α1\alpha_1 and β1\beta_1 are the roots of the equation x2−2xsec⁡θ+1=0x^2-2x\sec\theta+1=0 and α2\alpha_2 and β2\beta_2 are the roots of the equation x2+2xtan⁡θ−1=0x^2+2x\tan\theta-1=0. If α1>β1\alpha_1>\beta_1 and α2>β2\alpha_2>\beta_2, then α1+β2\alpha_1+\beta_2 equals

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Answer and worked solution

Official answer: C

Roots.

  • x2−2xsec⁡θ+1=0x^2-2x\sec\theta+1=0 has roots sec⁡θ±tan⁡θ\sec\theta\pm\tan\theta.
  • x2+2xtan⁡θ−1=0x^2+2x\tan\theta-1=0 has roots −tan⁡θ±sec⁡θ-\tan\theta\pm\sec\theta.

Signs. For −π6<θ<−π12-\frac\pi6<\theta<-\frac\pi{12}, sec⁡θ>0\sec\theta>0 and tan⁡θ<0\tan\theta<0.

  • The larger root of the first is α1=sec⁡θ−tan⁡θ\alpha_1=\sec\theta-\tan\theta.
  • The smaller root of the second is β2=−tan⁡θ−sec⁡θ\beta_2=-\tan\theta-\sec\theta.

Sum. α1+β2=−2tan⁡θ\alpha_1+\beta_2=-2\tan\theta.

(C)

Topics:Roots and coefficientsTrigonometric ratios and identities
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