Amit M Agarwal Solutions for Chapter: Continuity and Differentiability, Exercise 4: Target Exercise 6.4

If $f\left(x\right)=\left\{\begin{array}{ll}\left|1-4x^2\right|& 0\le x\le 1\\ \left[x^2-2x\right]& 1\le x<2\end{array}\right.$(where $\left[.\right]$) denotes the greatest integer function). Then:

Let $f\left(x\right)={\int }_x^1\left|x-t\right|t dt$, then

Let $f\left(x\right)$ be a function such that $f(x+y)=f\left(x\right)+f\left(y\right)$ for all $x$ and $y$ and $f\left(x\right)=\left(2x^2+3x\right).g\left(x\right)$ for all $x$, where $g\left(x\right)$ is continuous and $g\left(0\right)=3$. Then, $f^{\text{'}}\left(x\right)$ is equal to

Given a function $g\left(x\right)$ which has derivates $g^{\text{'}}\left(x\right)$ for every real $x$ and which satisfies the equation $g\left(x+y\right)=e^{y}g\left(x\right)+e^{x}g\left(y\right)$ for all $x$ and $y$ and $g^{\text{'}}\left(0\right)=2$, then the value of $\left\{g^{\text{'}}\left(x\right)-g\left(x\right)\right\}$ is equal to

Let $f:R\to R$ be a function satisfying $f\left(\frac{xy}{2}\right)=\frac{f\left(x\right)f\left(y\right)}{2},\forall x,y\in R$ and  $f\left(1\right)=f^{\text{'}}\left(1\right)\ne 0$ Then, $f\left(x\right)+f\left(1-x\right)$ is (for all non zero real values of $x$):

Let, $f:R\to \left(-\mathrm{\pi },\mathrm{\pi }\right)$, be a differentiable function such that $f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1-xy}\right)$, when $xy\ne 1$. If $f\left(1\right)=\frac{\mathrm{\pi }}{2}$ and $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=2$. Then, $f\left(x\right)$ is equal to

Let, $f\left(x\right)$ be a derivable function at $x=0$  and  $f\left(\frac{x+y}{K}\right)=\frac{f\left(x\right)+f\left(y\right)}{K}$ $\left(K\in R, K\ne 0,2\right)$. Then, $f\left(x\right)$ is

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