JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 6)
Let f, g : R $$\to$$ R be functions defined by
$$f(x) = \left\{ {\matrix{ {[x]} & , & {x < 0} \cr {|1 - x|} & , & {x \ge 0} \cr } } \right.$$ and $$g(x) = \left\{ {\matrix{ {{e^x} - x} & , & {x < 0} \cr {{{(x - 1)}^2} - 1} & , & {x \ge 0} \cr } } \right.$$ where [x] denote the greatest integer less than or equal to x. Then, the function fog is discontinuous at exactly :
one point
two points
three points
four points
Explanation
$f(x)=\left\{\begin{array}{ll}{[x],} & x<0 \\ |1-x|, & x \geq 0\end{array}\right.$ and $g(x)= \begin{cases}e^{x}-x, & x<0 \\ (x-1)^{2}-1, & x \geq 0\end{cases}$
$$ f \circ g(x)= \begin{cases}{[g(x)],} & g(x)<0 \\ |1-g(x)|, & g(x) \geq 0\end{cases} $$
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$$ =\left\{\begin{array}{cc} \left|1+x-e^{x}\right|, & x<0 \\ 1, & x=0 \\ {\left[(x-1)^{2}-1\right],} & 0 < x < 2 \\ \left|2-(x-1)^{2}\right|, & x \geq 2 \end{array}\right. $$
So, $x=0,2$ are the two points where fog is discontinuous.
$$ f \circ g(x)= \begin{cases}{[g(x)],} & g(x)<0 \\ |1-g(x)|, & g(x) \geq 0\end{cases} $$
$$ =\left\{\begin{array}{cc} \left|1+x-e^{x}\right|, & x<0 \\ 1, & x=0 \\ {\left[(x-1)^{2}-1\right],} & 0 < x < 2 \\ \left|2-(x-1)^{2}\right|, & x \geq 2 \end{array}\right. $$
So, $x=0,2$ are the two points where fog is discontinuous.
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