Class 12th

R-CONH? + Br? + 4NaOH → R-NH? + 2NaBr + Na? CO? + 2H? O
This reaction is the Hoffmann bromamide degradation, in which an amide is converted to a 1° amine.

The shortest distance D between two skew lines is given by the formula:
D = | (a? - a? ) ⋅ (b? x b? )| / |b? x b? |
Line L? : (x-1)/2 = (y-2)/3 = (z-4)/4
Line L? : (x-2)/3 = (y-4)/4 = (z-5)/5

Here, a? = I + 2j + 4k, b? = 2i + 3j + 4k
a? = 2i + 4j + 5k, b? = 3i + 4j + 5k

a? - a? = I + 2j + k
b? x b? = | I j k |
| 2 3 4 |
| 3 4 5 |
= I (15-16) - j (10-12) + k (8-9) = -i + 2j - k

D = | (i + 2j + k) ⋅ (-i + 2j - k)| / √ (-1)² + 2² + (-1)²)
= |-1 + 4 - 1| / √ (1 + 4 + 1)
= 2 / √6

Kindly consider the following Image 

The equation of the plane is given as x + y + z = 42. It is also mentioned that x³ + y³ + z³ = 3xyz.
From the identity, if x³ + y³ + z³ - 3xyz = 0, then x + y + z = 0 or x = y = z.
Given the expression:
3 + (x³ + y³ + z³ - 3xyz) / (xyz)²

Since x³ + y³ + z³ = 3xyz, the expression simplifies to:
3 + 0 = 3

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.