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The TS EAMCET cut off is the minimum marks which candidates need to score in order to qualify in the exam.

The Telangana State Engineering, Agriculture and Medical Common Entrance Test (TS EAMCET) has a qualifying cutoff of forty marks out of one hundred and sixty for candidates belonging to the general and backward class categories. Candidates belonging to scheduled caste and scheduled tribe categories have no minimum qualifying mark requirement. The admission cutoff for colleges varies every year based on factors such as the number of applicants, seat availability, and the difficulty level of the examination.

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$y^2\le 8xy>\sqrt[]{2}x$

$2x^2=8x\Rightarrow x\left(x-4\right)=0\Rightarrow x=0,x=4x=0,x=y$

Required area = ${\displaystyle \underset{1}{\overset{4}{\int }}\left(2\sqrt[]{2}\sqrt[]{x}-\sqrt[]{2}x\right)}$ dx

$=\frac{28\sqrt[]{2}}{3}-\frac{15\sqrt[]{2}}{2}=\frac{11\sqrt[]{2}}{6}$

The documents required for Telangana EAPCET counselling is mentoned below:

• TS EAMCET rank card 2025

• TS EAMCET 2025 hall ticket

• Aadhaar card

• SSC or its equivalent mark sheet

• 6th standard to inter-study certificates

• Transfer Certificate (T.C) and Intermediate certificate

• Income certificate (if applicable) and Caste certificate issued by competent authority

• Domicile certificate of either self or parents in Telangana for a period of 10 years in case of Non-Local candidates

• Eligibility certificate for Physically challenged (PH)/Children of Armed Personnel (CAP)/NCC/Sports/Minority certificate if applicable

For x < 0 0 < e^x < 1 Þ [e^x] = 0

$0\le x<1a{e}^x+\left[x-1\right]$  

$=a{e}^x+\left[x\right]-1$  

= a e^x – 1             b + [sin px]

$\therefore f\left(x\right)=\left[\begin{array}{l}0x<0\\ a{e}^x-10\le x<1\\ b-11\le x<2\\ -cx\ge 2\end{array}\right]$  

For f to be continuous at x = 0

   a – 1 = 0 Þ a = 1

$\begin{array}{l}\therefore a+b+c=1+e+1-e=2\\ a+b+c\ne 1\end{array}$  

Students have to pass the entrance test to participate in the TS EAPCET counselling. Qualifying students must register at the counselling portal. Then complete the document verification. After this, you have to exercise the web option entry. Seat will be allocated to students based om their prerformace.

The TS EAMCET result is declared on its official website a frw days after the answer key is out. Candidates who appeared in the exam, can download the result to know that they qualify in the exam or not.

${\displaystyle \underset{0}{\overset{1}{\int }}\left[-8x^2+6x-1\right]dx}$  

Let f (x) = -8x² + 6x – 1

f (0) = -1, f (1) = -3

max f (x) =  $-\frac{D}{4a}=-\frac{36-32}{-32}=\frac{1}{8}$  

= $-\frac{1}{4}-1\left(\frac{3}{4}-\frac{\lambda }{2}\right)-2\left(\frac{\sqrt[]{17}-3}{8}-\frac{3}{4}\right)-3\left(1-\frac{\sqrt[]{17}-3}{8}\right)$  

= $\frac{\sqrt[]{17}-13}{8}$  

Let $\frac{A_2}{A_1}=\frac{A_3}{A_2}=...=r$  

$\therefore A_1{A}_3{A}_5{A}_7=\frac{1}{1296}$

$\Rightarrow A_1{r}^3=\frac{1}{6}.......\left(i\right)$  

Again, A₂ + A₄ = $\frac{7}{36}$  

$A_{1}r=\frac{7}{36}-\frac{1}{6}=\frac{1}{36}........\left(ii\right)$

$\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$  

  $\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$  

The TS EAMCET is released after the registration window ends. Candidates whi have successfully submited their application form will be issued the admit card which they will be able to download from the official website and carry to the exam centre.

The TS EAMCET answer key is released on its official website after the exam is conducted. The TS EAMCET answer key for the 2025 is already released on its official website.

Sum of digits

1 + 2 + 3 + 5 + 6 + 7 = 24

So, either 3 or 6 rejected at a time

Case 1 Last digit is 2

……….2

no. of cases = 

  Case 2 Last digit is 6

……….6

= 4! = 24

Total cases = 72

|A| = 2

$\left|\left|A\right|adj\left(5adj{A}^3\right)\right|$

= $\left|A{P}^3\right|\left|adj\left(5adj\left(A^3\right)\right)\right|$

$\begin{array}{l}={\left|A\right|}^{15}.5^6\\ =2^{15}*5^6=2^9*10^6\end{array}$

System of equation is

$\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill \left|\lambda \right|\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left[\begin{array}{l}-2\\ 4\\ 4\lambda -4\end{array}\right]$

R₁ – 2 R₂, R₃ – R₂

$\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -2\hfill & \hfill \left|\lambda \right|-1\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}-10\\ 4\\ 4\lambda -8\end{array}\right)$

System of equation will have no solution for = -7.

$f\left(x\right)=[\begin{array}{ccc}\hfill 2n\hfill & \hfill \hfill & \hfill n=2,4,6....\hfill \\ \hfill n-1\hfill & \hfill \hfill & \hfill n=3,7,11,15....\hfill \\ \hfill \frac{n+1}{2}\hfill & \hfill \hfill & \hfill n=1,5,9,13....\hfill \end{array}$

for n = 2, 4, 6 ……

f (n) = 4, 8, 12, ….4 (n) form        

for n = 3, 7, 11, 15, ….

f (n) = 1, 3, 5, 7, …. (4n + 1) or, (4n + 3) from

f is one and onto.

Here are some tips for candidates preparing for the CUET PG exam in CBT mode:

Understand the exam format, make sure you have a good understanding of computer, choose a test center that is convenient for you, arrive early on the center.

Yes, for SC, ST, OBC, and local candidates as per state norms.