IIT JEE Main Maths -Unit 14 -Inverse trigonometric functions, and their properties- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Let \( \sec^{-1}x = a \) where \( a \in [0,\pi] – \{\frac{\pi}{2}\} \) and \( \csc^{-1}x = \frac{\pi}{2} – a \). Then \[ 16\left[ a^2 + \left( \frac{\pi}{2} – a \right)^2 \right] = 16\left[ 2a^2 – \pi a + \frac{\pi^2}{4} \right] \] Maximum at \( a = \pi \) gives \( 20\pi^2 \), minimum at \( a = \frac{\pi}{4} \) gives \( 2\pi^2 \). Sum = \( 22\pi^2 \).
Hence, the correct answer is (4) \( 22\pi^2 \).
▶️ Answer/Explanation
Ans. (2)
Sol.
For \( \frac{\pi}{3} \le x \le \frac{\pi}{2} \),
\( \cos^{-1}\!\left( \frac{12}{13}\cos x + \frac{5}{13}\sin x \right) = \cos^{-1}\!\left( \cos x \cos \alpha + \sin x \sin \alpha \right) \)
\( = \cos^{-1}(\cos(x – \alpha)) = x – \alpha \)
\( \text{where } \alpha = \tan^{-1}\!\frac{5}{12} \)
\( \therefore \text{ Required value } = x – \tan^{-1}\!\frac{5}{12} \)