No Author Solutions for Exercise 1: Mathematics

If $f\left(x+y\right)=f\left(x\right)f\left(y\right)$ for all $x, y\in R, f\left(5\right)=2, f^{\text{'}}\left(0\right)=3$, then, $f\left(5\right)$ is equal to

If $y={\cos }^{-1}\left(\cos x\right),$ then $\frac{dy}{dx}$ at $x=\frac{5\pi }{4}$ is equal to

Let $f\left(x\right)=\left\{\begin{array}{l}\left|x\right|,\text{ for }0<\left|x\right|\le 2\\ 1, \text{ for }x=0\end{array}\right.$, then at $x=0$, $f$ has

Statement $\mathrm{I}$: The number of solutions of $\sin x=x^2+x+1$ is $0$

Statement $\mathrm{II}$: While finding a solution, avoid square and denominator function.

Statement $\mathrm{I}$: If $a$ and $b$ are positive integers and $\left[x\right]$ denotes greatest integer $\le x,$ then $\underset{x\to 0^{+}}{\lim }\left(\frac{x}{a}\left[\frac{b}{x}\right]\right)=\frac{b}{a}$
Statement $\mathrm{II}$: $\underset{x\to \infty }{\lim }\left(\frac{\left\{x\right\}}{x}\right)\to 0$, where $\left\{x\right\}$ denotes fractional part of $x$.

Statement $\mathrm{I}$: Matrix ${\left[A\right]}_{3\times 3}$, is defined as $a_{ij}=\left(\frac{i-j}{i+2j}\right),$ which cannot be expressed as a sum of symmetric and skew-symmetric matrix.

Statement $\mathrm{II}$: Matrix ${\left[A\right]}_{3\times 3}, a_{ij}=\left(\frac{i-j}{i+2j}\right)$ is neither symmetric nor skew-symmetric matrix.

Statement $\mathrm{I}$: $\underset{x\to 0}{\lim }\left[\frac{{\tan }^{-1}x}{x}\right]=0$, where, $\left[·\right]$ represents the greatest integer function.

Statement $\mathrm{II}$: $\left(\frac{{\tan }^{-1}x}{x}\right)<1$ in the neighbourhood of $x=0$.

Suppose $f\left(x\right)=x\sin x-\frac{1}{2}{\sin }^{2}x, x\in \left(0,\frac{\pi }{2}\right)$
Statement $\mathrm{I}$ Range of $f\left(x\right)$ is $\left(0,\frac{\pi -1}{2}\right)$.

Statement $\mathrm{II}$ Range of $f\left(x\right)$ is not determined.