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As the TS EAMCET is an entrance exam, candidates will have to enquire about the course fees as it may vary by college, branch, and category.

No, as the TS EAMCET is just an entrance exam, there are no scholarships, however, after getting selected candidates can enquire in their college about the scholarships or fee waivers based on merit/category.

The minimum age limit to apply for TS EAMCET is 16 years. There is no upper age limit for the same.

Students who meet the minimum eligibility criteria like the academic qualification, age limit, etc. can apply for the TS EAMCET exam.

To be eligible for the TS EAMCET exam, candidates need to meet the minimum eligibility criteria. As per the education qualification, they must have qualified in the 10+2 with PCM for Engineering or PCB for Agri/Pharmacy. Min 45% for General, 40% for reserved categories.

For candidates who qualify in the TS EAMCET entrance exam, they will get selected for the Engineering (B.Tech/BE in CSE, ECE, Mechanical, Civil, etc.) course.

Organic compound → AgBr (s)

  0.5 g                    0.40 g

Here; Moles of Br in organic compound = Moles of Br in Ag Br

= Moles of AgBr

=  $\frac{0.40}{188}mol$

Mass of Br in organic compound =  $\frac{0.4}{188}\times 80g$  

$\begin{array}{l}\mathrm{\%}ofBr=\frac{0.4\times 80}{188\times 0.5}\times 100\mathrm{\%}\\ =34\mathrm{\%}\end{array}$

The TS EAMCET application form is expected to be out along with the notification in the month of January 2026.

As the TS EAMCET is an entrance exam, there is not any campus for it. However the campuses of the participating colleges include JNTU Hyderabad, Osmania University, Kakatiya Institute of Technology, etc.

The TS EAMCET question papers consist of Multiple Choice Questions (MCQs) in Physics, Chemistry, Mathematics or Biology.

See, solve as many previous question papers and it is advised to allocate time to yourself while solving the paper.

$f\left(x\right)=a{x}^2+bx+c$

g (x) = px + q

$f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$  

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$  

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$  

->4x² + 6x + 1 = apx² + bpx + cp + q

->ap = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

->b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

It is advisable to start solving CUET PG question papers as part of your exam preparation at least 4-5 months before the exam date. 

The stages in the TS EAMCET exam are as follows - 

Application form filling - Downloading the Admit Card - Appearing in the TS EAMCET Exam - Results declaration and Rank card on display - Counselling Process - Seat Allotment and reporting to college.

The TS EAMCET admission process consists of the following steps - 

Step 1. Notification & application for m release
Step 2. Admit card release
Step 3. Appearing in the exam
Step 4. Result declaration
Step 5. Counselling process  (document verification, choice filling, seat allotment)
Step 6. Admission to allotted college

As of now the 2025 counselling has commenced. However, based on past year events we can expect that the TS EAMCET exam dates will be released in January 2026. It will tentatively be conducted in April or May for the engineering entrance exam.

The TS EAMCET official website is - eapcet.tgche.ac.in. To access all the information related to the exam, like the dates, notification details, direct link to apply, you can check this official link.

The TS EAMCET exam stands for Telangana State Engineering, Agriculture & Medical Common Entrance Test. It is a state-level annual entrance test which is conducted by the Telangana government to shortlist candidates for admission into undergraduate programs in Engineering, Agriculture, and Pharmacy. The exam is conducted by JNTU Hyderabad on behalf of Telangana State Council of Higher Education.

$N=M^2+M^4+.....+M^{98}$

$=\left(-{\alpha }^{2}I\right)+{\left(-{\alpha }^{2}I\right)}^2+....+{\left(-{\alpha }^{2}I\right)}^{49}$

$=I\left(-{\alpha }^2+{\alpha }^4-{\alpha }^6+....-{\alpha }^{98}\right)$

N =  $-I\left({\alpha }^2-{\alpha }^4+{\alpha }^6.......+{\alpha }^{98}\right)$

$=-I\frac{{\alpha }^2\left(1-{\left(-{\alpha }^2\right)}^{49}\right)}{1-\left(-{\alpha }^2\right)}$

N =  $\frac{-I{\alpha }^2\left(1+{\alpha }^{98}\right)}{1+{\alpha }^2}$  

Now  $\left(I-m^2\right)N=-2I$

$\left(I+{\alpha }^{2}I\right)\frac{\left(-I{\alpha }^2\cdot (1+{\alpha }^{98}\right)}{1+{\alpha }^2}=-2I$  

-> a¹⁰⁰ + a² = 2

->a = $\pm 1$