The derivative of $y = 1 - |x|$ at $x = 0$ is

If $\alpha$ and $\beta$ are such that the function $f(x)$ defined by $f(x) = \begin{cases} \alpha x^2 - \beta, & |x| < 1 \\ \frac{-1}{|x|}, & |x| \ge 1 \end{cases}$ is differentiable everywhere,then the ordered pair $(\alpha, \beta) =$

Let $f(x) = x|x|$ and $g(x) = \sin x$.
Statement-$1$: $gof$ is differentiable at $x=0$ and its derivative is continuous at that point.
Statement-$2$: $gof$ is twice differentiable at $x=0$.

Suppose that $f(x)$ is a differentiable function such that $f^{\prime}(x)$ is continuous,$f^{\prime}(0)=1$ and $f^{\prime \prime}(0)$ does not exist. Let $g(x)=x f^{\prime}(x)$. Then,

If $f(x) = \begin{cases} \frac{x^2 \ln \cos x}{\ln (1+x^2)} & , x \neq 0 \\ 0 & , x=0 \end{cases}$,then $f(x)$ is

If $f(x) = \max(|2-x|, 2-x^3)$ for $x \in R$,then which of the following is true?