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(81)^sin²x + (81)^cos²x = 30.
(81)^sin²x + (81)^ (1-sin²x) = 30.
Let y = 81^sin²x.
y + 81/y = 30
y² - 30y + 81 = 0
(y - 3) (y - 27) = 0
⇒ y = 3 or y = 27.
Either 81^sin²x = 3 ⇒ 3^ (4sin²x) = 3¹ ⇒ sin²x = 1/4 ⇒ sin x = ±1/2. x = π/6, 5π/6.
OR, 81^sin²x = 27 ⇒ 3^ (4sin²x) = 3³ ⇒ sin²x = 3/4 ⇒ sin x = ±√3/2. x = π/3, 2π/3.
(as 0 ≤ x ≤ π)
Total possible solutions = 4.

No, you cannot use the score of XAT 2025 for admission to 2026 batch at XLRI as the validity of exam score is one year only. Thus, you will have to appear for XAT 2026 exam if you plan to pursue MBA from XLRI in 2026. The registration process for XAT 2026 is open and the last date to apply is December 5, 2025.

The total course fee of full time MBA at XLRI is INR 28 Lacs according to Shiksha. The fee is to be paid semester wise. The first installment is to be paid at the time of admission. The XLRI course fee includes tuition fees, hostel and mess charges, caution money, library chargers and other miscellaneous services.

There is no process of applying for multiple XAT accepting colleges through one form because there is no common counselling for XAT. Candidates have to apply to the colleges of their choice individually. They must keep in mind the application last date so they don't miss out on it. Also, they must check the admission cut offs and selection criteria.

After completing BEd through UP BEd JEE 2025, candidates can pursue teaching jobs in government and private schools. They may also appear for TET, CTET, and other teaching eligibility tests. Opportunities include school teacher, lecturer, educational consultant, or pursuing higher studies like MEd and PhD.

Yes, final-year graduation or post-graduation students can apply for UP BEd JEE 2025, provided they meet minimum eligibility criteria by the time of counseling. They must produce proof of qualifying their degree during document verification to secure admission into the BEd program.

Yes, all the students can download the TBSE HS result marksheet immidietely after the declaration of result. The board releases the Tripura Board 12th result online on its website at tbresults.tripura.gov.in and tbse.tripura.gov.in. The students can access the TBSE class 12 result scorecard using their roll number and enrollment number. However, original marksheet will be made available in a week or two after the result declaration.

Candidates can clear the UP BEd JEE exam by preparing effectively in the listed subjects. Arts, Science, Commerce, Agriculture, or any other professional stream are pursued by the candidates in their graduation. Candidates can prepare accordingly, as per their chosen subject and what they wish to pursue in BEd.

Given kx^ (k-1) + k * y^ (k-1) * dy/dx = 0.
dy/dx = - (kx^ (k-1) / (ky^ (k-1) = - (x/y)^ (k-1).
The provided solution has dy/dx + (x/y)^ (k-1) = 0.
It seems to relate to k-1 = -1/3, which implies k = 1 - 1/3 = 2/3.

UP BEd JEE 2025 scores are accepted by all state universities and affiliated colleges across Uttar Pradesh offering BEd programs. Top universities such as Lucknow University, CCSU Meerut, MJPRU Bareilly, and BHU Varanasi are among the prominent institutions admitting students through this entrance exam.

Truth table for (p → q) ∧ (q → ~p).
| p | q | p → q | ~p | q → ~p | (p → q) ∧ (q → ~p) |
|-|-|-|-|-|-|
| T | T | T | F | F | F |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | T | T | T |
The final column is F, T, which is the truth table for ~p.
Therefore, (p → q) ∧ (q → ~p) is equivalent to ~p.

Yes, Nursing courses are offered at various levels and in many specializations. You can pursue a Nursing at the UG-level (BSc Nursing), PG-level (MSc Nursing), Diploma-level (GNM and ANM), Post-Basic BSc Nursing, etc. Also, at the PG-level, you'll even get to choose the specialised areas like Oncology, Gastroenterology, Psychiatry, Pediatrics, Ophthalmology, and so on.

The tangent to the parabola y² = 4ax is y = mx + a/m.

For y² = 4x, a=1. So, the tangent is y = mx + 1/m.
The given line is y = mx + 4.
Comparing the two, 1/m = 4 ⇒ m = 1/4.
The line is y = (1/4)x + 4.
This line is also tangent to x² = 2by.
Substitute y into the parabola equation:
x² = 2b (1/4)x + 4)
x² = ( b/2 )x + 8b
x² - ( b/2 )x - 8b = 0.
For tangency, the discriminant (D) is zero.
D = (-b/2)² - 4 (1) (-8b) = 0.
b²/4 + 32b = 0.
b ( b/4 + 32) = 0.
b = 0 (not possible) or b/4 = -32 ⇒ b = -128.

No, GATE scores are not compulsory for admissions to the M.Tech course at NIELIT Calicut. The institute also offers admissions to those students who have not appeared in GATE. The unfilled GATE seats will first be filled with NON-GATE candidates from Kerala state based on the nativity certificate. If seats are still remaining vacant,  candidates from other states will be considered for admission.

Moreover, the institute also allocates 5 seats to each M.Tech branch for sponsored category students. However, apart from all these aspects, applicants with GATE scores are considered more for admission. 

(3¹/? + 5¹/? )?
General term =? C? (3¹/? )? (5¹/? )? =? C? 3^ (60-r)/4) 5^ (r/8)
Terms are rational for r being a multiple of 8 and (60-r) being a multiple of 4.
If r is a multiple of 8, then 60-r is 60 - 8k. Since 60 is a multiple of 4, 60-8k is also a multiple of 4.
So, we just need r to be a multiple of 8.
r = 0, 8, 16, 24, 32, 40, 48, 56. (Total 8 rational terms)
Total terms are 61.
Number of irrational terms = 61 - 8 = 53 = n.
∴ n - 1 = 52.

g (f (x) = f² (x) + f (x) - 1.
g (f (5/4) = f² (5/4) + f (5/4) - 1.
Given g (f (5/4) = 5/4, let f (5/4) = y.
-5/4 = y² + y - 1 (There appears to be a typo in the image's solution)
y² + y - 1 + 5/4 = 0
y² + y + 1/4 = 0
(y + 1/2)² = 0
y = -1/2.
So, f (5/4) = -1/2.