If $y=|\cos x-\sin x|+|\tan x-\cot x|$,then $\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{3}}+\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{6}}=$

Let $f(x)=x^2+a x+b$,where $a, b \in R$. If $f(x)=0$ has all its roots imaginary,then the roots of $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$ are

Let $f: R \rightarrow R$ be given by
$f(x) = \begin{cases} x^5+5x^4+10x^3+10x^2+3x+1, & x < 0 \\ x^2-x+1, & 0 \leq x < 1 \\ \frac{2}{3}x^3-4x^2+7x-\frac{8}{3}, & 1 \leq x < 3 \\ (x-2)\log_e(x-2)-x+\frac{10}{3}, & x \geq 3 \end{cases}$
Then which of the following options is/are correct?
$(1)$ $f^{\prime}$ has a local maximum at $x = 1$
$(2)$ $f$ is onto
$(3)$ $f$ is increasing on $(-\infty, 0)$
$(4)$ $f^{\prime}$ is $NOT$ differentiable at $x = 1$

Let $h$ be a twice continuously differentiable positive function on an open interval $J.$ Let $g(x) = \ln(h(x))$ for each $x \in J$. Suppose $(h'(x))^2 > h''(x) h(x)$ for each $x \in J$. Then

If $f(x) = \begin{cases} ax^2 - bx + 2, & x < 3 \\ bx^2 - 3, & x \geq 3 \end{cases}$ is differentiable at every $x \in R$,then the area (in sq units) of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is

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: