The reaction is the Gattermann–Koch reaction, which involves the formylation of benzene to produce benzaldehyde (\(C_6H_5CHO\)). In the presence of carbon monoxide (CO), hydrogen chloride (HCl), and anhydrous AlCl\(_3\)/CuCl, the formyl group (CHO) is introduced into the benzene ring.
\(C_6H_6 + CO + HCl \xrightarrow{\text{AlCl}_3/\text{CuCl}} C_6H_5CHO\)
Thus, the major product \(X\) is benzaldehyde (3).
Thus the correct answer is Option 3.
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: