JEE MAIN - Mathematics (2023 - 15th April Morning Shift - No. 6)
Let $[x]$ denote the greatest integer function and
$f(x)=\max \{1+x+[x], 2+x, x+2[x]\}, 0 \leq x \leq 2$. Let $m$ be the number of
points in $[0,2]$, where $f$ is not continuous and $n$ be the number of points in
$(0,2)$, where $f$ is not differentiable. Then $(m+n)^{2}+2$ is equal to :
$f(x)=\max \{1+x+[x], 2+x, x+2[x]\}, 0 \leq x \leq 2$. Let $m$ be the number of
points in $[0,2]$, where $f$ is not continuous and $n$ be the number of points in
$(0,2)$, where $f$ is not differentiable. Then $(m+n)^{2}+2$ is equal to :
3
6
2
11
Explanation
$$
\begin{aligned}
& \text { Let } g(x)=1+x+[x]=\left\{\begin{array}{cc}
1+x ; & x \in[0,1) \\\
2+x ; & x \in[1,2) \\
5 ; & x=2
\end{array}\right. \\\\
& \lambda(x)=x+2[x]=\left\{\begin{array}{cc}
x ; & x \in[0,1) \\
x+2 ; & x \in[1,2) \\
6 ; & x=2
\end{array}\right. \\\\
& r(x)=2+x \\\\
& f(x)=\left\{\begin{array}{cc}
2+x ; & x \in[0,2) \\
6 ; & x=2
\end{array}\right.
\end{aligned}
$$
$\mathrm{f}(\mathrm{x})$ is discontinuous only at $x=2 \Rightarrow \mathrm{m}=1$
$\mathrm{f}(\mathrm{x})$ is differentiable in $(0,2) \Rightarrow \mathrm{n}=0$
$$ (m+n)^2+2=3 $$
$\mathrm{f}(\mathrm{x})$ is discontinuous only at $x=2 \Rightarrow \mathrm{m}=1$
$\mathrm{f}(\mathrm{x})$ is differentiable in $(0,2) \Rightarrow \mathrm{n}=0$
$$ (m+n)^2+2=3 $$
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