Dear Ajay Kumar, in my view it is entirely possible that the v(C=O) of an acid would be higher than v(C=O) of the corresponding ester. Such trend can be found in some IR frequencies tables, for example: https://www.sigmaaldrich.com/FI/en/technical-documents/technical-article/analytical-chemistry/photometry-and-reflectometry/ir-spectrum-table

(see here 1760 cm-1 for monomeric acids and 1750-1735 for esters).

The reason for measured v(C=O) of acids being often lower than for esters may be the hydrogen bonding interaction between acid molecules (C=O...H), which weakens the double bond and thus reduces v(C=O), whereas esters do not interact this way if there are no suitable hydrogen bond donors.

So my guess would be that the observed v(C=O) trend in your compounds (where the polysaccharide backbone may provide multiple hydrogen bond donors) is due to hydrogen bonding, which may be either very similar for acids and esters or even stronger for the ester due to some spatial arrangements of the polymers.

Dear Ajay Kumar please have a look at the following lecture which might help you in your analysis:

The C=O stretching frequency

(see atached pdf file)

A nice explanation of the IR bands of esters is provided in the following potentially useful link:

The C=O Bond, Part VI: Esters and the Rule of Three

https://www.spectroscopyonline.com/view/co-bond-part-vi-esters-and-rule-three

I hope this helps.

Dipole moments, Bond length, Resonance, s character (hybridization), and Hydrogen bonding- important factors in IR!