If $f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \\ 37 - x, & 2 < x \le 3 \end{cases}$,then:

Let $R$ be the set of all real numbers and $f(x) = \sin^{10} x (\cos^8 x + \cos^4 x + \cos^2 x + 1)$ for $x \in R$. Let $S = \{\lambda \in R : \text{there exists a point } c \in (0, 2\pi) \text{ with } f'(c) = \lambda f(c)\}$. Then,

Let $k$ and $m$ be positive real numbers such that the function $f(x) = \begin{cases} 3x^2 + k\sqrt{x+1}, & 0 < x < 1 \\ mx^2 + k^2, & x \geq 1 \end{cases}$ is differentiable for all $x > 0$. Then $\frac{8f'(8)}{f'(\frac{1}{8})}$ is equal to $.............$.

If $f(x) = \begin{cases} \frac{8}{x^3} - 6x, & x \le 1 \\ \sqrt{x} + 1, & x > 1 \end{cases}$,then at $x = 1$,$f$ is:

Differentiation of $(x^2-5x+8) \times (x^3+7x+9)$ can be done by

If $f(x) = \begin{cases} ax+b, & \text{if } x \leq 1 \\ ax^2+c, & \text{if } 1 < x \leq 2 \\ \frac{dx^2+1}{x}, & \text{if } x > 2 \end{cases}$ is differentiable on $\mathbb{R}$,then $ad-bc = $