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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 real solutions of the equation ^ -1 x (x + 1) + ^ -1 and x^ 2 +...

The real solutions of the equation tan⁡−1x(x+1)+sin⁡−1 and x2+x+1=π2 and \tan^{-1 } \sqrt{\text{x} \left(\text{x} + 1\right)} + \sin^{-1 \text{ and } } \sqrt{x^{2} + \text{x} + 1} = \dfrac{\pi}{2 \text{ and } } are

Options

  1. A−1,0- 1, 0
  2. B0,10, 1
  3. C−1,1- 1, 1
  4. D−1,2- 1, 2

Correct answer

A. −1,0- 1, 0

Step-by-step solution

tan⁡−1x(x+1)+sin⁡−1x2+x+1=π2\tan^{-1 } \sqrt{\text{x} \left(\text{x} + 1\right)} + \sin^{-1 } \sqrt{x^{2} + \text{x} + 1} = \dfrac{\pi}{2}

⇒tan⁡−1x2+x+sin⁡−1x2+x+1=π2⇒ \tan^{-1 } \sqrt{x^{2} + \text{x}} + \sin^{-1 } \sqrt{x^{2} + \text{x} + 1} = \dfrac{\pi}{2}

This equation holds true, if

x2+x≥0and0≤x2+x+1≤1x^{2} + \text{x} \ge 0 \text{and} 0 \le x^{2} + \text{x} + 1 \le 1

Now, x2+x≥0and0≤x2+x+1≤1x^{2} + \text{x} \ge 0 \text{and} 0 \le x^{2} + \text{x} + 1 \le 1

⇒x2+x≥0andx2+x+1≤1[∵x2+x+1>0for all x]<br/><br/>⇒ x^{2} + \text{x} \ge 0 \text{and} x^{2} + \text{x} + 1 \le 1 \left[∵ x^{2} + \text{x} +1 > 0 \text{for all x}\right]<br /><br />⇒ x^{2} + x≥0and\text{x} \ge0 \text{and} x^{2} + x≤0\text{x} \le0

⇒x2+x=0⇒x=0,−1.<br/><br/>Clearly,thesetwovaluessatisfythegivenequation.Hence,⇒ x^{2} + \text{x} =0 ⇒ \text{x} =0 \text{,} -1 \text{.}<br /><br />Clearly, these two values satisfy the given equation. Hence, x = - 1, 0arethesolutionsofthegivenequation.are the solutions of the given equation.
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