Let $$f(x)=a x^3+b x^2+c x+41$$ be such that $$f(1)=40, f^{\prime}(1)=2$$ and $$f^{\prime \prime}(1)=4$$. Then $$a^2+b^2+c^2$$ is equal to:

The parabola $$y^2=4 x$$ divides the area of the circle $$x^2+y^2=5$$ in two parts. The area of the smaller part is equal to :

The solution of the differential equation $$(x^2+y^2) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0$$, is :

Let $$|\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8}, \theta \epsilon[0,2 \pi]$$. Then, the sum of all $$\theta \in[0,2 \pi]$$, where $$\cos 3 \theta$$ attains its maximum value, is :