1. The curve \(\Gamma\) is given by:
\[ y = b e^{-x/a}. \]
At \(x = 0\), the curve crosses the \(y\)-axis at \(y = b\).
2. The straight line \(L\) is given by:
\[ \frac{x}{a} + \frac{y}{b} = 1 \implies y = b \left(1 - \frac{x}{a}\right). \]
3. Check the intersection point of \(\Gamma\) and \(L\):
4. For the \(x\)-axis:
Thus, \(L\) touches \(\Gamma\) at the point where \(\Gamma\) crosses the \(y\)-axis.
\( \text{M} \xrightarrow{\text{CH}_3\text{MgBr}} \text{N} + \text{CH}_4 \uparrow \xrightarrow{\text{H}^+} \text{CH}_3\text{COCH}_2\text{COCH}_3 \)
Identify the ion having 4f\(^6\) electronic configuration.