The compound shown is a monosaccharide with an aldehyde group (-CHO) and multiple hydroxyl (-OH) groups. When treated with Br$_2$ water, it is not oxidized to a dicarboxylic acid but rather forms a monocarboxylic acid, making option (1) incorrect.
Statement (2) is correct: Despite having a –CHO group, certain sugars like glucose do not respond to Schiff’s test.
Statement (3) is correct: There are 4 chiral (asymmetric) carbon atoms present in the structure.
Statement (4) is correct: The compound can exist in equilibrium with two other cyclic forms, $\alpha$-D-glucose and $\beta$-D-glucose.
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: