The value of $k$ for which the point $(\alpha,\arcsin \alpha)(\alpha>0)$ lies inside the triangle formed by $x+y=k$ with the coordinate axes is-

Question

The value of $k$ for which the point $(\alpha,\arcsin \alpha)(\alpha>0)$ lies inside the triangle formed by $x+y=k$ with the coordinate axes is-

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As the point $(\alpha,\arcsin \alpha)(\alpha>0)$ lies inside the triangle formed by $x+y=k$ with the coordinate axes.So $\alpha+\arcsin \alpha<k,\alpha>0$

Now i cannot solve this.The answer given is $k\in(1+\frac{\pi}{2},\infty)$

Answer 1

Function $\arcsin\alpha$ is defined only for $-1\le\alpha\le1$, but in this case we are told that $\alpha>0$, hence we conclude that $0<\alpha\le1$. In this range $\arcsin\alpha$ is increasing and varies from $\arcsin0=0$ and $\arcsin1=\pi/2$: $$ 0<\arcsin\alpha\le{\pi\over2}. $$ You can see below, in red, the graph of $y=\arcsin x$ for $0<x\le1$. Point $P$ has coordinates $(1,\pi/2)$ and the line passing through it has then equation $x+y=1+\pi/2$. The value $k=1+\pi/2$ is then the lowest allowing all the graph to be contained inside the triangle formed by line $x+y=k$ with the coordinate axes.