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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 value of a for which ax^ 2 + ^ -1 (x^ 2 - 2x + 2) + ^ -1 (x^ 2...

The value of aa for which ax2+sin⁡−1(x2−2x+2)+cos⁡−1(x2−2x+2)=0ax^{2} + \sin^{-1} \left(x^{2} - 2x + 2\right) + \cos^{-1} \left(x^{2} - 2x + 2\right) = 0 has a real solution, is

Options

  1. A−2π- \dfrac{2}{\pi}
  2. B2π\dfrac{2}{\pi}
  3. C−π2- \dfrac{\pi}{2}
  4. Dπ2\dfrac{\pi}{2}

Correct answer

C. −π2- \dfrac{\pi}{2}

Step-by-step solution

Here, x2−2x+2=(x−1)2+1≥1x^{2} - 2x + 2 = {\left(x - 1\right)}^{2} + 1 \ge 1

But, −1≤(x2−2x+2)≤1- 1 \le ( x^{2} - 2x + 2 ) \le 1

Which is possible only when

x2−2x+2=1x^{2} - 2x + 2 = 1

⇒x=1⇒ x = 1

Then, a(1)2+sin⁡−1(1)+cos⁡−1(1)=0a{\left(1\right)}^{2} + \sin^{-1} \left(1\right) + \cos^{-1} \left(1\right) = 0

⇒a+π2+0=0⇒ a + \dfrac{\pi}{2} + 0 = 0

⇒a=−π2⇒ a = - \dfrac{\pi}{2}
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