‹ Back to the paper
Choose the correct option.
If θ = sin⁻¹x + cos⁻¹x - tan⁻¹x, x ≥ 0, then the smallest interval in which θ lies is:
If $\theta = \sin^{-1}x + \cos^{-1}x - \tan^{-1}x$, $x \ge 0$, then the smallest interval in which $\theta$ lies is:
- a$\frac{\pi}{2} \le \theta \le \frac{3\pi}{4}$
- b$0 < \theta < \pi$
- c$-\frac{\pi}{4} \le \theta \le 0$
- d$\frac{\pi}{4} \le \theta \le \frac{\pi}{2}$
Answer
Answer
Official answer keyCorrect option: d
Answer: (d) $\frac{\pi}{4} \le \theta \le \frac{\pi}{2}$
For $0 \le x \le 1$, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$, so $\theta = \frac{\pi}{2} - \tan^{-1}x$. Since $0 \le \tan^{-1}x \le \frac{\pi}{4}$, we get $\frac{\pi}{4} \le \theta \le \frac{\pi}{2}$.