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ExamsJEE MainMaths

(i) Solve the inequality (arcsec x)² - 6*(arcsec x) + 8 > 0. (ii) If sin²(x) + sin²(y) < 1 for x, y real, prove that arcsin(tan(x)*tan(y)) lies in (-pi/2, pi/2).

  1. (i) arcsec x < 2 or arcsec x > 4, i.e. x in [1, sec 2)... with the range restriction giving sec x in valid set; (ii) proof
  2. (i) 2 < arcsec x < 4; (ii) proof
  3. (i) arcsec x = 2 or 4; (ii) cannot be proved
  4. (i) all real x; (ii) proof

Correct answer: (i) arcsec x < 2 or arcsec x > 4, i.e. x in [1, sec 2)... with the range restriction giving sec x in valid set; (ii) proof

Solution

(i) Put t = arcsec x in [0,pi/2) U (pi/2,pi]. (t-2)(t-4) > 0 means t < 2 or t > 4. Since the maximum of arcsec is pi (~3.14) < 4, the condition t > 4 is impossible. So we need arcsec x < 2 (and within the allowed range), which translates to a set of x values via the monotonic branches of sec. (ii) Given sin² x + sin² y < 1: sin² x < cos² y. Then tan² x * tan² y = (sin² x/cos² x)(sin² y/cos² y). Using sin² x < cos² y and sin² y < cos² x gives sin² x sin² y < cos² x cos² y, i.e. tan² x tan² y < 1, so |tan x tan y| < 1, ensuring arcsin(tan x tan y) is defined and strictly inside (-pi/2, pi/2).

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