Let $y=y(x)$ be the solution of the differential equation

$\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1$.

Then, $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ equals :

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as :

$$ f(x)= \begin{cases}\frac{a-b \cos 2 x}{x^2} ; & x<0 \\\\ x^2+c x+2 ; & 0 \leq x \leq 1 \\\\ 2 x+1 ; & x>1\end{cases} $$

If $f$ is continuous everywhere in $\mathbf{R}$ and $m$ is the number of points where $f$ is NOT differential then $\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}$ equals :

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a}>\mathrm{b}$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latusrectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :

Let $3, a, b, c$ be in A.P. and $3, a-1, b+1, c+9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :