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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If _ x x ≥ 2 and 0 ≤ x ≤ 3π , then the value of x lies in the...

If log⁡cos⁡xsin⁡x≥2\log_{\cos x} \sin x \ge 2 and 0≤x≤3π0 \le x \le 3\pi, then the value of sin⁡x\sin x lies in the interval

Options

  1. A[5−12,1]\left[\dfrac{\sqrt{5} - 1}{2}, 1\right]
  2. B(0,5−12]\left(0, \dfrac{\sqrt{5} - 1}{2}\right]
  3. C[0,12]\left[0, \dfrac{1}{2}\right]
  4. DNone of these

Correct answer

B. (0,5−12]\left(0, \dfrac{\sqrt{5} - 1}{2}\right]

Step-by-step solution

Since, sin⁡x>0,cos⁡x>0,cos⁡x≠1\sin x > 0, \cos x > 0, \cos x \ne 1
So, sin⁡x≤cos⁡2x\sin x \le \cos^{2} x

⇒sin⁡2x+sin⁡x−1≤0⇒ \sin^{2} x + \sin x - 1 \le 0

⇒(sin⁡x+12)2−54≤0⇒ \left(\sin x + \dfrac{1}{2}\right)^{2} - \dfrac{5}{4} \le 0

⇒∣sin⁡x+12∣≤52⇒ | \sin x + \dfrac{1}{2} | \le \dfrac{\sqrt{5}}{2}

∴sin⁡x+12∈[−52,52]∴ \sin x + \dfrac{1}{2} \in [ \dfrac{- \sqrt{5}}{2}, \dfrac{5}{2} ]

(As, sin⁡x>0\sin x > 0)

∴0< sin⁡x≤5−12∴ \text{0< }\sin x \le \dfrac{\sqrt{5} - 1}{2}
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