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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.

Candidates must have qualified in 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 will be out on Feb 19, 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 TS EAMCET exam are below - 

Application form filling

Downloading Admit Card

Appearing in the entrance exam

Results declaration and Rank card on display

Counselling Process

Seat Allotment and reporting to college.

TS EAMCET admission process has following stages 

Notification & application form release
Admit card release
Appearing in entrance exam
Result declaration
Counselling process which includes document verification, choice filling, seat allotment
Reporting to the college

The TS EAMCET exam dates for the 2026 session are out. The exam will be held from May 9 to 11, 2026.

The TS EAMCET official website is - eapcet.tgche.ac.in. 

The Telangana State Engineering, Agriculture & Medical Common Entrance Test is a state-level annual entrance test is conducted by Telangana government to shortlist candidates for admission into undergraduate programs in Engineering, Agriculture, and Pharmacy. 

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

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

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

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 = $\pm 1$