Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 3: Level 3

Let, $f\left(x\right)=\sin x$ and $g\left(x\right)=\left\{\begin{array}{ll}\max \left\{f\right(t),& 0\le t\le x , 0\le \mathrm{x}\le \mathrm{\pi }\}\\ \frac{1-\cos x}{2},& x>\mathrm{\pi }\end{array}\right.$. Then $g\left(x\right)$ is

Let $f:R\to R$ be a function defined as $f\left(x\right)=\left\{\begin{array}{ccc}3\left(1-\frac{\left|x\right|}{2}\right)& \text{if}& \left|x\right|\le 2\\ 0& \text{if}& \left|x\right|>2\end{array}\right..$

Let $g:R\to R$ be given by $g\left(x\right)=f(x+2)-f(x-2).$ If $n$ and $m$ denote the number of points in $R$ where $g$ is not continuous and not differentiable, respectively, then $n+m$ is equal to ________.

A function $f$ is defined on $[-3,3]$ as

$f\left(x\right)=\left\{\begin{array}{c}\min \left\{\left|x\right|,2-x^2\right\},-2\le x\le 2\\ \left[\right|x\left|\right] , 2<\left|x\right|\le 3\end{array}\right.$

where $\left[x\right]$ denotes the greatest integer $\le x$. The number of points, where $f$ is not differentiable in $\left(-3,3\right)$ is ___ .

If $f\left(x\right)=\sqrt{1-\sqrt{1-x^2}},$ then $f\left(x\right)$ is

Let $g\left(x\right)=\frac{{\left(x-1\right)}^n}{{\mathrm{logcos}}^m\left(x-1\right)}; 0<x<2, m$ and $n$ are integers, $m\ne 0, n>0$ and let $p$ be the left-hand derivative of $|x-1|$ at $x=1$. If$\underset{x\to 1^{+}}{\lim }g\left(x\right)=p$, then

Let $S=t\in R:f\left(x\right)=|x-\pi |·\left(e^{\left|x\right|}-1\right)\sin \left|x\right|$ is not differentiable at $t$. Then the set $S$ is equal to

Let $f\left(x\right)$ be defined in$[-2,2]$ by $f\left(x\right)=\left\{\begin{array}{ll}\max (\sqrt{4-x^2},& \sqrt{1+x^2}), -2\le x\le 0\\ \min (\sqrt{4-x^2},& \sqrt{1+x^2}), 0<x\le 2\end{array}\right.$, then $f\left(x\right)$

$$ Given $f\left(x\right)=\left\{\begin{array}{ccc}{\log }_a{\left(a\left|\left[x\right]+[-x]\right|\right)}^x\left(\frac{a^{{\displaystyle \frac{2}{\left(\frac{\left[x\right]+\left[-x\right]}{\left|x\right|}\right)}-5}}}{3+a^{\frac{1}{\left|x\right|}}}\right)& \text{for}& x\ne 0\\ 0& \text{for}& x=0\end{array}\right.$; $a>1$ where [.] represents the integral part function, then