Let $\lbrack \bullet \rbrack$ denote the greatest integer function, and let $f(x) = min\left\{ \sqrt

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Let $\lbrack \bullet \rbrack$ denote the greatest integer function, and let $f(x) = min\left\{ \sqrt{2}x,x^{2} \right\}$. Let $S = \{ x \in ( - 2,2):$ the function $g(x) = |x|\left\lbrack x^{2} \right\rbrack$ is discontinuous at x$\}$.

Then $\sum_{x \in S}\mspace{2mu} f(x)$ equals :

Accepted Answer

$\ g(x) = |x|\left\lbrack x^{2} ightbrack$points of discontinuity of $g(x)$ in ( $- 2,2$ ) are$${( \pm 1, \pm \sqrt{2}, \pm \sqrt{3}) }{	herefore S = \{ - 1,1, - \sqrt{2},\sqrt{2}, - \sqrt{3},\sqrt{3}\} }{\because f(x) = min\left\{ \sqrt{2}x,x^{2} ight\} }{	herefore\sum_{x \in S}\mspace{2mu} f(x) = - \sqrt{2} + 1 - 2 + 2 - \sqrt{6} + \sqrt{6} }{= 1 - \sqrt{2}}$$

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