If the function f(x) = x^2[sin^{-1}x] is disc | vedclass

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(B) The function $f(x) = x^2[\sin^{-1}x]$ is discontinuous where the greatest integer function $[\sin^{-1}x]$ is discontinuous.For the function $g(x)

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(B) The function $f(x) = x^2[\sin^{-1}x]$ is discontinuous where the greatest integer function $[\sin^{-1}x]$ is discontinuous.For the function $g(x) = [\sin^{-1}x]$,the points of discontinuity occur when $\sin^{-1}x$ is an integer.The domain of $\sin^{-1}x$ is $[-1, 1]$. The possible integer values for $\sin^{-1}x$ are $\{-1, 0, 1\}$.$1$. $\sin^{-1}x = -1 \implies x = \sin(-1) = -\sin 1$.$2$. $\sin^{-1}x = 0 \implies x = \sin(0) = 0$.$3$. $\sin^{-1}x = 1 \implies x = \sin(1) = \sin 1$.However,we are given that $\alpha, \beta \in R - \{0\}$. Thus,we exclude $x = 0$.The points of discontinuity in the given domain are $\alpha = \sin 1$ and $\beta = -\sin 1$.Therefore,the value of $\alpha + \beta = \sin 1 + (-\sin 1) = 0$.

If the function $f(x) = x^2[\sin^{-1}x]$ is discontinuous at $x = \alpha$ and $x = \beta$,where $\alpha, \beta \in R - \{0\}$ and $[.]$ denotes the greatest integer function,then the value of $\alpha + \beta$ is:

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