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 :

If the sum of the series $$\frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots+\frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}$$ is equal to 5, then $$50 \mathrm{~d}$$ is equal to :

The coefficient of $$x^{70}$$ in $$x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots+x^{54}(1+x)^{46}$$ is $${ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$$. Then a possible value of $$\mathrm{p}+\mathrm{q}$$ is :