Resonance in $\mathrm{X}_2 \mathrm{Y}$ can be represented as

The enthalpy of formation of $X_2Y$ $ \left(X = X(g) + \frac{1}{2} Y = Y(g) \rightarrow X_2Y(g) \right) $ is 80 kJ mol$^{-1}$. The magnitude of resonance energy of $X_2Y$ is __ kJ mol$^{-1}$ (nearest integer value).

Given: Bond energies of $X \equiv X$, $X = X$, $Y = Y$ and $X = Y$ are 940, 410, 500, and 602 kJ mol$^{-1}$ respectively.
valence $X$: 3, $Y$: 2

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$
and $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the
points $(0, 0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

Let $f(x) = x - 1$ and $g(x) = e^x$ for $x \in \mathbb{R}$. If $\frac{dy}{dx} = \left( e^{-2\sqrt{x}} g\left(f(f(x))\right) - \frac{y}{\sqrt{x}} \right)$, $y(0) = 0$, then $y(1)$ is

If $ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots \infty= \frac{\pi^4}{90} $,

$\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots \infty= \alpha $,

$ \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots \infty= \beta $,

then $ \frac{\alpha}{\beta} $ is equal to :