First slide

Properties of ITF

Question

$\sin^{- 1} \{\sin (\frac{2 x^{2} + 4}{x^{2} + 1})\} < π - 3,$ if

Moderate

A

$x \in [- 1, 0]$
B

$x \in [0, 1]$
C

$x \in (- 1, 1)$
D

$x \in (1, \infty)$

Solution

We have,
$\frac{2 x^{2} + 4}{x^{2} + 1} = \frac{2 (x^{2} + 1) + 2}{x^{2} + 1} = 2 + \frac{2}{x^{2} + 1}$
$∴ 2 < \frac{2 x^{2} + 4}{x^{2} + 1} \leq 4$ for all $x \in R$
$\Rightarrow π - 4 < π - \frac{2 x^{2} + 4}{x^{2} + 1} < π - 2$ for all $x \in R$
$\Rightarrow - \frac{π}{2} < π - \frac{2 x^{2} + 4}{x^{2} + 1} < \frac{π}{2}$ for all $x \in R$
$∴ \sin^{- 1} \{\sin (\frac{2 x^{2} + 4}{x^{2} + 1})\} = \sin^{- 1} \{\sin (π - \frac{2 x^{2} + 4}{x^{2} + 1})\}$
$= π - \frac{2 x^{2} + 4}{x^{2} + 1}$
Hence, $\sin^{- 1} \{\sin (\frac{2 x^{2} + 4}{x^{2} + 1})\} < π - 3$
$\begin{array}{l} \Rightarrow π - \frac{2 x^{2} + 4}{x^{2} + 1} < π - 3 \\ \Rightarrow \frac{2 x^{2} + 4}{x^{2} + 1} > 3 \\ \Rightarrow 2 + \frac{2}{x^{2} + 1} \Rightarrow 2 > 3 > x^{2} + 1 \Rightarrow x^{2} - 1 < 0 \Rightarrow x \in (- 1, 1) \end{array}$

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