The largest value of $n$, for which $40^n$ divides $60!$, is

Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0, x=1, y^2=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^2$. Then $(f(0)+f(1))$ is equal to

Let the image of parabola $x^2=4 y$, in the line $x-y=1$ be $(y+a)^2=b(x-c)$, $a, b, c \in \mathrm{~N}$. Then $a+b+c$ is equal to

Let the angles made with the positive $x$-axis by two straight lines drawn from the point $\mathrm{P}(2,3)$ and meeting the line $x+y=6$ at a distance $\sqrt{\frac{2}{3}}$ from the point P be $\theta_1$ and $\theta_2$. Then the value of $\left(\theta_1+\theta_2\right)$ is: