Suppose $a$ is a positive real number such that $a^5-a^3+a=2$. Then,

Let $f$ and $g$ be twice differentiable functions on $R$ such that
$f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x$
$f^{\prime}(1)=4, g^{\prime}(1)=3$
$f(2)=12, g(2)=4$
Then which of the following is $NOT$ true?

Let $f: R \rightarrow R$ be defined as $f(x)=\begin{cases} \frac{a-b \cos 2 x}{x^2} & ; x<0 \\ x^2+c x+2 & ; 0 \leq x \leq 1 \\ 2 x+1 & ; x>1 \end{cases}$. If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is $NOT$ differentiable,then $m+a+b+c$ equals:

Let $f:\left[-\frac{1}{2}, 2\right] \rightarrow R$ and $g:\left[-\frac{1}{2}, 2\right] \rightarrow R$ be functions defined by $f(x)=\left[x^2-3\right]$ and $g(x)=|x| f(x)+|4 x-7| f(x)$,where $[y]$ denotes the greatest integer less than or equal to $y$ for $y \in R$. Then
$(A)$ $f$ is discontinuous exactly at three points in $\left[-\frac{1}{2}, 2\right]$
$(B)$ $f$ is discontinuous exactly at four points in $\left[-\frac{1}{2}, 2\right]$
$(C)$ $g$ is $NOT$ differentiable exactly at four points in $\left(-\frac{1}{2}, 2\right)$
$(D)$ $g$ is $NOT$ differentiable exactly at five points in $\left(-\frac{1}{2}, 2\right)$

Let $f : (0, \pi) \rightarrow \mathbb{R}$ be a twice differentiable function such that $\lim _{t \rightarrow x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x$ for all $x \in (0, \pi)$. If $f \left(\frac{\pi}{6}\right) = -\frac{\pi}{12}$,then which of the following statement$(s)$ is (are) $TRUE$?
$(A) f \left(\frac{\pi}{4}\right) = \frac{\pi}{4 \sqrt{2}}$
$(B) f(x) < \frac{x^4}{6} - x^2$ for all $x \in (0, \pi)$
$(C)$ There exists $\alpha \in (0, \pi)$ such that $f^{\prime}(\alpha) = 0$
$(D) f^{\prime \prime}\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = 0$

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 = $