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Question-63430




Question Number 63430 by rajesh4661kumar@gamil.com last updated on 04/Jul/19
Answered by MJS last updated on 04/Jul/19
y=sin^(−1)  (√(x−1))  1≤x≤2 ⇒ 0≤(√(x−1))≤1 ⇒  ⇒ 0≤y≤(π/2)
$${y}=\mathrm{sin}^{−\mathrm{1}} \:\sqrt{{x}−\mathrm{1}} \\ $$$$\mathrm{1}\leqslant{x}\leqslant\mathrm{2}\:\Rightarrow\:\mathrm{0}\leqslant\sqrt{{x}−\mathrm{1}}\leqslant\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{0}\leqslant{y}\leqslant\frac{\pi}{\mathrm{2}} \\ $$

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