Let the functions $f: [-1,1] \rightarrow \mathbb{R}$ and $g: [-1,1] \rightarrow [-1,1]$ be defined by $f(x)…

Let the functions $f: [-1,1] \rightarrow \mathbb{R}$ and $g: [-1,1] \rightarrow [-1,1]$ be defined by $f(x) = |2x-1| + |2x+1|$ and $g(x) = x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. Let $f \circ g: [-1,1] \rightarrow \mathbb{R}$ be the composite function defined by $(f \circ g)(x) = f(g(x))$. Suppose $c$ is the number of points in the interval $[-1,1]$ at which $f \circ g$ is not continuous, and suppose $d$ is the number of points in the interval $[-1,1]$ at which $f \circ g$ is not differentiable. Then the value of $c+d$ is _____________

Solution

Graph of $g(x)$ is $f(x) = |2x - 1| + |2x + 1| = \begin{cases} -4x, & x < -\frac{1}{2} \\ 2, & -\frac{1}{2} \leq x \leq \frac{1}{2} \\ 4x, & x > \frac{1}{2} \end{cases}$ $f(g(x)) = \begin{cases} -4g(x), & g(x) < -\frac{1}{2} \\ 2, & -\frac{1}{2} \leq g(x) \leq \frac{1}{2} \\ 4g(x), & g(x) > \frac{1}{2} \end{cases}$

fgx=2,x-1,-120,124x+1x-12,04xx12,1

c=1 at x=0

d=3 at x=-12, 0, 12,

c+d=4.

$g(x) = \{x\}$

Asked in: JEE Advanced 2020 (Paper 2)

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