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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

Statement \(I:\) The equation \({({\sin ^{ - 1}}\,x)^3} + {({\cos ^{ - 1}}\,x)^3} - a{\pi ^3} = 0\) has a solution for all \(a \ge \frac{1}{{32}}.\) Statement \(II:\)  For any \(x \in R ,\) \({\sin ^{ - 1}}\,x + {\cos ^{ - 1}}\,x = \frac{\pi }{2}\) and \(0 \le {\left( {{{\sin }^{ - 1}}\,x - \frac{\pi }{4}} \right)^2} \le \frac{{9{\pi ^2}}}{{16}}\)

  1. A Both statements \(I\) and \(II\) are true.
  2. B Both statements \(I\) and \(II\) are false.
  3. C Statement \(I\) is true and statement \(II\) is false
  4. D Statement \(I\) is false and statement \(II\) is true.
Verified Solution

Answer & Solution

Correct Answer

(A) Both statements \(I\) and \(II\) are true.

Step-by-step Solution

Detailed explanation

\({\sin ^{ - 1}}x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\) \( \Rightarrow - \frac{{3\pi }}{4} \le \left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right) \le \frac{\pi }{4}\)…