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Expression = (49)¹²? - 1) / 48
This uses the sum of a geometric series or a? - b? factorization.
(x? - 1) / (x - 1) = 1 + x + x² + . + x? ¹.
Let x = 49. (49¹²? - 1)/48 is an integer.
The solution shows (49? ³-1) (49? ³+1) / 48. This is correct factorization. Since 49 is odd, 49? ³ is odd. So 49? ³-1 and 49? ³+1 are consecutive even numbers. One is divisible by 2, the other by 4, so their product is divisible by 8. Also, 49 ≡ 1 (mod 3), so 49? ³-1 is divisible by 3. Hence the numerator is divisible by 24. It is also divisible by 48.

Uttar Pradesh University of Medical Sciences BSc admission is entrance-based. Candidates are selected based on scores of Combined Pharmacy and Nursing Entrance Test (CPNET). The candidates who are Class 12 or equivalent pass can apply for the course admission. Additionally, the forms can be submitted in online mode. To fill out the offline form, candidates should contact the admission office. Once the seat allotment is done, the fee payment can be done for admission confirmation.

If any student fail in maximum two subjects, they can apply to appear for the TBSE 12th compartment exams. The board will announce a date to submit the application form for TBSE 12th compartment exams. Thereafter, an official TBSE HS supplementary exam date will be announced.

IIIT Bhopal CSE cutoff 2025 ranged between 24300 and 35569 for the students belonging to the General AI quota. Therefore, the chances of you getting CSE with a JEE Main cutoff rank of 20000 at IIIT Bhopal are high if you belong to the General category. Please note that the cutoff ranks vary every year for each round. So, it is advisable to target a lower rank to improve your chances of admission. Tabulated below are the round-wise CSE cutoff ranks of 2025 for the students belonging to the General AI quota.

Yes, the students who are notsatisfied with their marks in TBSE 12th exams can apply for re-evaluation or re-tataling. The board annouce the date the accept the scrutniy application form after the declaration of result. To apply for the scrutiny, the students need to pay the required fee. Thereafter, the board will re-valuation the asnwer sheet and prepare an updated result for the students. 

A = [i, -i], [-i, i]
A² = [-2, 2], [2, -2]
A? = [8, -8], [-8, 8]
A? = [-128, 128], [128, -128]
A? [x, y]? =?
-128x + 128y = 8 ⇒ -16x + 16y = 1 ⇒ x - y = -1/16 (I)
128x - 128y = 64 ⇒ 16x - 16y = 8 ⇒ x - y = 1/2 (II)
System is inconsistent hence No solution

The registrations for Uttar Pradesh University of Medical Sciences BSc registrations commence with the CPNET form release. As per the past schedule, the university releases in Jun. The entrance examination is further conducted in Jul. Candidates seeking admission must apply on or before the prescribed deadline.

Equation of chord of x² + y² = 25 with mid point (h, k) is xh + yk = h² + k².
Or, y = (-h/k)x + (h² + k²)/k.
If this touches the ellipse x²/9 + y²/16 = 1, then the condition for tangency c² = a²m² + b² must be satisfied.
Here, m = -h/k, c = (h²+k²)/k, a²=9, b²=16.
(h² + k²)/k)² = 9 (-h/k)² + 16
(h² + k²)²/k² = 9h²/k² + 16
⇒ (h² + k²)² = 9h² + 16k²
∴ Required locus (x² + y²)² = 9x² + 16y².

Yes, candidates can challenge multiple questions in AP EAMCET preliminary answer key. For each objection, candidates must submit justification with relevant proof through official online portal within given deadline.

Kindly consider the following Image 

Tripura Board releases the TBSE 12th result online on its website. To check result, all the students are required to use their roll number. Without it, no students can check the TBSE 12th result. However, some schools publishes the TBSE 12th result on notice board. To check result quickly online, the students must keep their TBSE 12th admit card handy. 

The commencement dates for the M.Tech course at NIELIT Calicut are usually announced after the CCMT counselling process. The orientation of the course was held at the beginning of August. Moreover, the classes of the first semester commence either in the last week of August or the initial week of September. 

Yes, Uttar Pradesh University of Medical Sciences offer a four-year BSc in Nursing programme. The mode of study is full-time for the undergraduate course. It is being offered through the Faculty of Nursing. Candidates seeking admission have to take the UPUMS Combined Pharmacy and Nursing Entrance Test (CPNET). The eligible students can go to the official website to apply.

The TBSE 12th result include the following details.

• Student's name
• Registration number
• Roll number
• Enrollment number
• Grade
• Total marks & more

Students must check these details carefully. In case any mistake/error is found on scorecard, they should immediately contact the school or board to report the same. 

Area A = 2π - ∫? ¹ (√x - x) dx is incorrect. The area is likely between two curves.
The calculation shown is:
A = 2π - [2/3 x^ (3/2) - x²/2] from 0 to 1.
A = 2π - (2/3 - 1/2) = 2π - (4/6 - 3/6) = 2π - 1/6 = (12π - 1)/6.

Two answer keys are released for Andhra Pradesh EAMCET:

Preliminary Answer Key – Published soon after the exam. Candidates can check their responses and raise objections if they find discrepancies.

Final Answer Key – Released after evaluating all objections. This key is final and is used for preparing the result.