MCQ
Find the value of $\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$
- A$\frac{\pi}{2}$
- ✓$\frac{\pi}{4}$
- C$\frac{\pi}{6}$
- D$\frac{\pi}{3}$
Then, $\sin x=\frac{1}{2}=\sin \left(\frac{\pi}{6}\right)$
$\therefore \sin ^{-1} \frac{1}{2}=\frac{\pi}{6}$
$\therefore \tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$
$=\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]$
$=\tan ^{-1}\left[2 \cos \frac{\pi}{3}\right]$
$=\tan ^{-1}\left[2 \times \frac{1}{2}\right]$
$=\tan ^{-1} 1=\frac{\pi}{4}$
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Statement $1:$ $f(x)\, \le \,g\,(x)$ for $x$ in $(0,\infty )$
Statement $2:$ $f(x)\, \le \,1$ for $(x)$ in $(0,\infty )$ but $g(x)\,\to \infty$ as $x\,\to \infty$