Let $y(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$. Then $y'(x) - y''(x)$ at $x = -1$ is equal to:

Let $f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geqslant 1 \\ ax^2 + b, & |x| < 1 \end{cases}$ be continuous and differentiable everywhere. Then $a$ and $b$ are

$f(x)$ is a differentiable function and given $f^{\prime}(2)=6$ and $f^{\prime}(1)=4$, then $L=\lim _{h \rightarrow 0} \frac{f\left(2+2 h+h^2\right)-f(2)}{f\left(1+h-h^2\right)-f(1)}$

The number of polynomials $p: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $p(0)=0$,$p(x) > x^2$ for all $x \neq 0$,and $p^{\prime \prime}(0) = \frac{1}{2}$ is

Match the following:
In the following,$[x]$ denotes the greatest integer less than or equal to $x$.

$(a)$ $x|x|$	$(i)$ continuous in $(-1, 1)$
$(b)$ $\sqrt{|x|}$	$(ii)$ differentiable in $(-1, 1)$
$(c)$ $x+[x]$	$(iii)$ strictly increasing in $(-1, 1)$
$(d)$ $|x-1|+|x+1|$	$(iv)$ not differentiable at,at least one point in $(-1, 1)$

Let $f:R \to R$ be a continuous function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$.
Statement-$1$: $f(c) = \frac{1}{3}$ for some $c \in R$.
Statement-$2$: $0 < f(x) < \frac{1}{2\sqrt{2}}$ for all $x \in R$.