Given the following properties of a function $f(x)$:
$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given by the following number line:
$f'(x)$ is positive for $x < -5$,negative for $-5 < x < 2$,positive for $2 < x < 4$,and negative for $x > 4$.
On the possible graph of $y = f(x)$,we have:

Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by $f(x)=\frac{x^2-a x+1}{x^2+a x+1}, 0 < a < 2 .$
$1.$ Which of the following is true?
$(A)$ $(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0$
$(B)$ $(2-a)^2 f^{\prime}(1)-(2+a)^2 f^{\prime \prime}(-1)=0$
$(C)$ $f^{\prime}(1) f^{\prime}(-1)=(2-a)^2$
$(D)$ $f^{\prime}(1) f^{\prime}(-1)=-(2+a)^2$
$2.$ Which of the following is true?
$(A)$ $f(x)$ is decreasing on $(-1,1)$ and has a local minimum at $x=1$
$(B)$ $f(x)$ is increasing on $(-1,1)$ and has a local maximum at $x=1$
$(C)$ $f(x)$ is increasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$
$(D)$ $f(x)$ is decreasing on $(-1,1)$ but has neither a local maximum nor a local minimum at $x=1$
$3.$ Let $g(x)=\int_0^{e^x} \frac{f^{\prime}(t)}{1+t^2} d t$. Which of the following is true?
$(A)$ $g^{\prime}(x)$ is positive on $(-\infty, 0)$ and negative on $(0, \infty)$
$(B)$ $g^{\prime}(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$
$(C)$ $g^{\prime}(x)$ changes sign on both $(-\infty, 0)$ and $(0, \infty)$
$(D)$ $g^{\prime}(x)$ does not change sign on $(-\infty, \infty)$
Give the answer for questions $1, 2$ and $3.$

If $y=\frac{(\sqrt{x}+1)(x^2-\sqrt{x})}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}(3 \cos^2 x-5) \cos^3 x$,then $96 y'(\frac{\pi}{6})$ is equal to :

If $y = \log(\tan(x/2)) + \sin^{-1}(\cos x)$,then $dy/dx$ is

If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then $f(x)$ at $x = 0$ is

The function $f(x) = \begin{cases} \frac{\pi}{4} + \tan^{-1} x, & |x| \leq 1 \\ \frac{1}{2}(|x|-1), & |x| > 1 \end{cases}$ is: