Let $f(x)$ be a polynomial of degree $6$ in $x$,in which the coefficient of $x^{6}$ is unity and it has extrema at $x=-1$ and $x=1$. If $\lim_{x \rightarrow 0} \frac{f(x)}{x^{3}}=1$,then $5 \cdot f(2)$ is equal to .............

Let $f$ be a real-valued function defined on the interval $(0, \infty)$ by $f(x)=\ln x+\int_0^x \sqrt{1+\sin t} \, dt$. Then which of the following statement$(s)$ is (are) true?
$(A)$ $f^{\prime \prime}(x)$ exists for all $x \in(0, \infty)$
$(B)$ $f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and $f^{\prime}$ is continuous on $(0, \infty)$,but not differentiable on $(0, \infty)$
$(C)$ there exists $\alpha>1$ such that $|f^{\prime}(x)|<|f(x)|$ for all $x \in(\alpha, \infty)$
$(D)$ there exists $\beta>0$ such that $|f(x)|+|f^{\prime}(x)| \leq \beta$ for all $x \in(0, \infty)$

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 :

The derivative of $f(x)=\cos ^{-1}\left[\sin \sqrt{\frac{1+x}{2}}\right]+x^x$ with respect to $x$ at $x=1$ is equal to

If $y = \frac{\tan x \cos^{-1} x}{\sqrt{1-x^2}}$, then the value of $\frac{dy}{dx}$ when $x = 0$ is:

Let $f(x) = \begin{cases} 3-x & \text{if } x < -3 \\ 6 & \text{if } -3 \leq x \leq 3 \\ 3+x & \text{if } x > 3 \end{cases}$. Let $\alpha$ be the number of points of discontinuity of $f$ and $\beta$ be the number of points where $f$ is not differentiable. Then $\alpha+\beta=$