- A$|x|>\frac{1}{\sqrt{2}}$
- ✓$|x|<\frac{1}{\sqrt{2}}$
- C$0$$<$$x$$<$$1$
- D$\frac{1}{\sqrt{2}}$$<$$x$$<$$1$
ધારો કે, $si{n^{ - 1}}x = \theta ,\;\theta \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$
$\therefore \ x = sin \theta$
$= sin^{-1} \left(2 x \sqrt{1 - x^2}\right)$
$= sin^{-1} \left(2 sin \theta \sqrt{1 - sin^2 \theta}\right)$
$= sin^{-1} \left(2 sin \theta \sqrt{cos^2 \theta}\right)$
$= sin^{-1} \left(2 sin \theta cos \theta\right)$
$= sin^{-1} \left(sin2 \theta\right)$
$ \left\{ \begin{array}{l l}|x| < \frac{1}{\sqrt{2}} & \quad \\\frac{-1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} & \quad \\\end{array} \right.$
$= 2\theta sin \left(-\frac{\pi}{4}\right) < sin\theta < sin \frac{\pi}{4}$
$= 2sin^{-1} x$
$ \left\{ \begin{array}{l l}\frac{\pi}{2} < 2\theta < \frac{\pi}{2} & \quad \\\end{array} \right.$
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${I_1} = \int_{1 - k}^k {x\,f\left\{ {x(1 - x)} \right\}} \,dx$, ${I_2} = \int_{1 - k}^k {\,f\left\{ {x(1 - x)} \right\}} \,dx$
કે જ્યાં $2k - 1 > 0$ તો ${I_1}/{I_2}$ મેળવો.
