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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

The complete solution set of the inequality ^ -1 ( 4 ) > 3x^ 2 - 4x...

The complete solution set of the inequality cos⁡−1(cos⁡4)>3x2−4x\cos^{-1} ( \cos 4 ) > 3x^{2} - 4x is

Options

  1. A(0,2+6π−83)\left(0, \dfrac{2 + \sqrt{6\pi - 8}}{3}\right)
  2. B(2−6π−83,0)\left(\dfrac{2 - \sqrt{6\pi - 8}}{3}, 0\right)
  3. C(−2,2)( - 2, 2 )
  4. D(2−6π−83,2+6π−83)\left(\dfrac{2 - \sqrt{6\pi - 8}}{3}, \dfrac{2 + \sqrt{6\pi - 8}}{3}\right)

Correct answer

D. (2−6π−83,2+6π−83)\left(\dfrac{2 - \sqrt{6\pi - 8}}{3}, \dfrac{2 + \sqrt{6\pi - 8}}{3}\right)

Step-by-step solution

As, cos⁡−1(cos⁡4)=cos⁡−1{cos⁡(2π−4)}=2π−4\cos^{-1} ( \cos 4 ) = \cos^{-1} \left\{\cos \left(2\pi - 4\right)\right\} = 2\pi - 4

⇒2π−4>3x2−4x⇒ 2\pi - 4 > 3x^{2} - 4x

⇒3x2−4x−(2π−4)<0⇒ 3x^{2} - 4x - \left(2\pi - 4\right) < 0

⇒2−6π−83<x<2+6π−83⇒ \dfrac{2 - \sqrt{6\pi - 8}}{3} < x < \dfrac{2 + \sqrt{6\pi - 8}}{3}
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