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Let $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$, where $[.]$ is the greatest integer function and $\mathrm{g}(\mathrm{x})=\sin \mathrm{x}$ be two real valued functions over R.
Which one of the following statements is correct?
Options:
Which one of the following statements is correct?
Solution:
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Verified Answer
The correct answer is:
$\lim _{x \rightarrow 0+}($ fog $)(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0^{+}}(\mathrm{gof})(\mathrm{x})$
At $x=0$
For LHL: $g(x)=-2+|-2|=0$
For RHL: $g(x)=|x|-2+|x-2|$
$\begin{aligned} g(x) &=x-2-(x-2)=0 \\ \mathrm{~g}(\mathrm{x}) &=0 \\ \text { For }(x=0): g(x) &=-2+|-2|=0 \end{aligned}$
LHL $=\lim _{x \rightarrow 0^{-}} g(x)=0$
$\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} g(x)=0$
$g(0)=0$
$\Rightarrow g(x)$ is continuous at $x=0$ At $x=2$
For LHL: $g(x)=|x|-2+|x-2|$
$g(x)=x-2-(x-2)$
$g(x)=0$
For RHL: $g(x)=|x|-2+|x-2|$
$g(x)=x-2+x-2$
$2(x)=2 x-4$
For $(x=2): g(x)=|x|-2+|x-2|$
$\mathrm{LHL}=\lim _{x \rightarrow 2^{-}} g(x)=0$
$\mathrm{RHL}=\lim _{x \rightarrow 2^{+}} g(x)=\lim _{x \rightarrow 2} 2 x-4=2(2)-4=0$
$g(2)=|2|-2+|2-2|$
$g(2)=2-2+2-2=0$
$\Rightarrow g(x)$ is continuous at $x=2$. At $x=-1$
For LHL: $g(x)=-2+|-2|=0$
For RHL: $g(x)=-2+|-2|=0$
For $(x=-1): g(x)=-2+|-2|=0$
$\therefore \mathrm{LHL}=\lim _{x \rightarrow-1^{-}} g(x)=0$
$\mathrm{RHL}=\lim _{x \rightarrow-1^{+}} g(x)=0$
$g(-1)=0$
$\Rightarrow g(\mathrm{x})$ is differentiable at $x=-1$.
For LHL: $g(x)=-2+|-2|=0$
For RHL: $g(x)=|x|-2+|x-2|$
$\begin{aligned} g(x) &=x-2-(x-2)=0 \\ \mathrm{~g}(\mathrm{x}) &=0 \\ \text { For }(x=0): g(x) &=-2+|-2|=0 \end{aligned}$
LHL $=\lim _{x \rightarrow 0^{-}} g(x)=0$
$\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} g(x)=0$
$g(0)=0$
$\Rightarrow g(x)$ is continuous at $x=0$ At $x=2$
For LHL: $g(x)=|x|-2+|x-2|$
$g(x)=x-2-(x-2)$
$g(x)=0$
For RHL: $g(x)=|x|-2+|x-2|$
$g(x)=x-2+x-2$
$2(x)=2 x-4$
For $(x=2): g(x)=|x|-2+|x-2|$
$\mathrm{LHL}=\lim _{x \rightarrow 2^{-}} g(x)=0$
$\mathrm{RHL}=\lim _{x \rightarrow 2^{+}} g(x)=\lim _{x \rightarrow 2} 2 x-4=2(2)-4=0$
$g(2)=|2|-2+|2-2|$
$g(2)=2-2+2-2=0$
$\Rightarrow g(x)$ is continuous at $x=2$. At $x=-1$
For LHL: $g(x)=-2+|-2|=0$
For RHL: $g(x)=-2+|-2|=0$
For $(x=-1): g(x)=-2+|-2|=0$
$\therefore \mathrm{LHL}=\lim _{x \rightarrow-1^{-}} g(x)=0$
$\mathrm{RHL}=\lim _{x \rightarrow-1^{+}} g(x)=0$
$g(-1)=0$
$\Rightarrow g(\mathrm{x})$ is differentiable at $x=-1$.
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