JEE Main 2013 — Mathematics Calculus

$$\text { } \begin{aligned} & f \circ g=f(g(x))=f(1-|x|) \\ & =-1+|1-| x|-2| \\ & =-1+|-| x|-1|=-1+|| x|+1| \end{aligned}$$ ​f∘g=f(g(x))=f(1−∣x∣)=−1+∣1−∣x∣−2∣=−1+∣−∣x∣−1∣=−1+∣∣x∣+1∣​ Let $f \circ g=y$f∘g=y $$\begin{aligned} & \therefore y=-1+|| x|+1| \\ & \Rightarrow y= \begin{cases}-1+x+1, & x \geq 0 \\ -1-x+1, & x < 0\end{cases} \\ & \Rightarrow y=\left\{\begin{array}{cc} x, & x \geq 0 \\ -x, & x < 0 \end{array}\right. \end{aligned}$$​∴y=−1+∣∣x∣+1∣⇒y={−1+x+1,−1−x+1,​x≥0x<0​⇒y={x,−x,​x≥0x<0​​ LHL at $\displaystyle (x=0)=\lim _{x \rightarrow 0}(-x)=0$(x=0)=x→0lim​(−x)=0 RHL at $\displaystyle (x=0)=\lim _{x \rightarrow 0}(x)=0$(x=0)=x→0lim​(x)=0 When $x=0$x=0, then $y=0$y=0 Hence, LHL at $(x=0)=$(x=0)= RHL at $(x=0)=$(x=0)= value of $y$y at $(x=0)$(x=0) Hence $y$y is continuous at $x=0$x=0. Clearly at all other point $y$y continuous. Therefore, the set of all points where fog is discontinuous is an empty set.