If $f(x) = \begin{cases} ax - 1 & {if } x \ > 1\\ \ 2x + 1 & {if } x < 1 \end{cases}$ is continuous at $x = 1$, then $a$ equals

If $x = e^t$ and $y = e^{2t}$ then $\frac{d^2y}{dx^2} =$

If $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ 5, & x = 3 \end{cases}$ then $f(x)$ :

Let $e^y(x+1) = 1$. Then which of the following are TRUE?

(A) $\frac{d^2y}{dx^2} = -\frac{1}{(x+1)^2}$

(B) $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$

(C) $\left.\frac{d^2y}{dx^2}\right|_{x=0} = -1$

(D) $\left.\frac{d^2y}{dx^2}\right|_{x=0} = 1$

(E) $\left.\frac{d^2y}{dx^2}\right|_{x=1} = \frac{1}{4}$

Choose the correct answer from the options given below:

The equation of the tangent to the curve $\mathrm{x}^{\frac{5}{2}}+\mathrm{y}^{\frac{5}{2}}=33$ at the point $(1,4)$ is :

If $y = \frac{1}{\sqrt[3]{1-x^3}}$ then $\frac{dy}{dx}$ is equal to

If $y = \frac{\sqrt{x+1} + \sqrt{x-1}}{\sqrt{x+1} - \sqrt{x-1}}$, then $(x^2-1)^{3/2} \frac{d^2y}{dx^2} =$