Class 12th

The problem is to evaluate the integral:
I = ∫? ¹? [x] * e^ [x] / e^ (x-1) dx, where [x] denotes the greatest integer function.

The solution breaks the integral into a sum of integrals over unit intervals:
I = ∑? ∫? ¹ n * e? / e^ (x-1) dx
= ∑? n * e? ∫? ¹ e^ (1-x) dx
= ∑? n * e? [-e^ (1-x)] from n to n+1
= ∑? n * e? [-e? - (-e¹? )]
= ∑? n * e? (e¹? - e? )
= ∑? n * e? * e? (e - 1)
= (e - 1) ∑? n
= (e - 1) * (0 + 1 + 2 + . + 9)
= (e - 1) * (9 * 10 / 2)
= 45 (e - 1)

Yes, candidate who have secured 60% in Class 12 board exams can secure an admission into BSc courses offered by the NIU. The candidates must clear Class 12 in the PCB stream to be eligible for most BSc courses and then appear for the CUET entrance exam to qualify for admissions. The NIU BSc admissions are decided through the CUET entrance exam.

Fat-soluble vitamins are stored in our body for a relatively longer duration as compared to water-soluble vitamins.

• Vitamin B and C are water-soluble. (Thiamine is vitamin B1, while ascorbic acid is vitamin C).
• Vitamin A and vitamin D are fat-soluble, so they are stored in our body for a relatively longer duration.

The E° value for Ce? /Ce³? is +1.74 V, which suggests that Ce? is a strong oxidant, reverting to its common +3 oxidation state. So, Ce³? is more stable than Ce?

The size of the Bk³? ion is less than the Np³? ion because Berkelium (Bk) lies beyond Neptunium (Np) in the actinoid series, and the size variation here is because of the actinoid contraction.

The enol form of acetone exists in less than 0.1% quantity, since its keto form is highly stable. But in the case of acetylacetone, the enol form is stabilized by intramolecular H-bonding, so its quantity increases to approximately 15%.
The intramolecular H-bond in the enol form of acetylacetone is shown.

DIBAL-H at low temperature in a non-polar solvent, followed by hydrolysis, reduces esters to an aldehyde and an alcohol as a byproduct. The reaction shown is:

Kindly consider the following Image