Let's analyze each statement:
A) It is an aldohexose:
This is true. Glucose is a six-carbon sugar (hexose) with an aldehyde functional group (aldo-).
B) On heating with HI it forms n-hexane:
This is also true. HI is a strong reducing agent that removes all oxygen atoms from glucose, reducing it to n-hexane.
C) It exists in furanose form:
This is partially true but misleading. While glucose can exist in both pyranose (six-membered ring) and furanose (five-membered ring) forms, the pyranose form is overwhelmingly predominant in solution. The furanose form exists only in minor amounts.
D) It does not give Schiff's test:
This is false. Glucose does give Schiff's test because its open-chain form contains an aldehyde group (in equilibrium with cyclic forms). Schiff's reagent reacts with aldehydes to produce a pink-magenta color.
Conclusion:
While statements A, B, and C have elements of truth, statement D is completely incorrect. The statement that is not true is (D).
Final Answer:
The incorrect statement is (D).
(i) Differentiate between globular and fibrous proteins.
(ii) What is meant by denaturation of protein?
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).