$\frac{1}{e^{3x}}(e^x + e^{5x}) = a_0 + a_1x + a_2x^2 + \ldots$
$\Rightarrow 2a_1 + 2^3a_3 + 2^5a_5 + \ldots$ is equal to

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 $\alpha+\beta=$

If $f:R \to R$ and $f(x)$ is a polynomial function of degree $10$ such that $f(x)=0$ has all real and distinct roots,then the equation $(f'(x))^2 - f(x)f''(x) = 0$ has:

If $y = \frac{1}{1 + x^{n-m} + x^{p-m}} + \frac{1}{1 + x^{m-n} + x^{p-n}} + \frac{1}{1 + x^{m-p} + x^{n-p}}$,then $\frac{dy}{dx}$ at $x = e^{m^{n^p}}$ is equal to:

The function $f(x) = |x|$ at $x = 0$ is

Let $f(x) = \begin{cases} x^{3/5} & \text{if } x \le 1 \\ -(x - 2)^3 & \text{if } x > 1 \end{cases}$. Then the number of critical points on the graph of the function is: