First slide

Properties of ITF

Question

$\tan^{- 1} (\tan \sqrt{1 - θ}) = \sqrt{1 - θ}$ when

Moderate

A

$- \frac{π}{2} < θ < \frac{π}{2}$
B

$θ > \frac{4 - π}{4}$
C

$θ < \frac{4 - π}{4}$
D

$\frac{4 - π^{2}}{4} < θ \leq 1$

Solution

Clearly, $\sqrt{1 - θ}$ is real, If $θ \leq 1 .$
Now,
$\tan^{- 1} (\tan \sqrt{1 - θ}) = \sqrt{1 - θ}$
$\Rightarrow 0 \leq \sqrt{1 - θ} < \frac{π}{2}$ and $θ \leq 1$
$\Rightarrow 0 \leq 1 - θ < \frac{π^{2}}{4}$ and $θ \leq 1$
$\Rightarrow - 1 \leq - θ \leq \frac{π^{2}}{4} - 1$ and $θ \leq 1$
$\Rightarrow 1 - \frac{π^{2}}{4} < θ \leq 1$ and $θ \leq 1 \Rightarrow \frac{4 - π^{2}}{4} < θ \leq 1$

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