MCQ
If $\pi \le x \le 2\pi $, then ${\cos ^{ - 1}}(\cos x)$ is equal to
- A$x$
- B$ - x$
- C$2\pi + x$
- ✓$2\pi - x$
$ \Rightarrow \,\, - \pi \ge - x \ge - 2\pi \,\,\, \Rightarrow \,\,\pi \ge 2\pi - x \ge 0$
$ \Rightarrow \,\,{\cos ^{ - 1}}\,\left\{ {\cos \,(2\pi - x)} \right\} = 2\pi - x$.
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$f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2} & , x<0 \\ \alpha & , x=0, \text { where } \alpha, \beta \in R \text {. If } \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x>0\end{cases}$
$f$ is continuous at $\mathrm{x}=0$, then $\alpha^2+\beta^2$ is equal to :